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V. Seleacu 


C. Dumitrescu 


The Smarandache Function 





Erhus University Press 


1996 


C. DUMITRESCU V. SELEACU 


THE SMARANDACHE FUNCTION 


ERHUS UNIVERSITY PRESS 


1996 


"The Smarandache Function", 
by Constantin Dumitrescu & Vasile Seleacu 


Copyright 1996 by C. Dumitrescu & V. Seleacu (Romania) 
and Erhus University Press 
13333 Colossal Cave Road, 
vail, Box 722, AZ 85641, USA. 


(Comments on our books are welcome! 


ISBN 1-879585-47-2 
Standard Address Number 297-5092 


Printed in the United States: of America 


Contents 


1 The Smarandache Function 5 
1.1 Generalised Numerical Scale .........2... 5 
1.2 A New Function in Number Theory........ 1] 
1.3 Some Formulae for the Calculus of S(n) ..... 18 
1.4 Connections with Some Classical Numencal Func- 
GODS So a es esl ot “Os en ok eer ine a e T A A 27 
1.5 The Smarandache Function as Generating Function 36 
1.6 Numenical Series Containing the FunctionS ... 48 
1.7 Diophantine Equations Involving the Smarandache 
Function ose an a oS a we es a A 62 
1.8 Solved and Unsolved Problems. .......... 73 
2 Generalisations of Smarandache Function TT 
2.1 Extension to the Set Q of Rational Nnumbers .. 77 
2.2 Numerical Functions Inspired from the Definition 
of Smarandache Function... ........... 86 

2.3 Smarandache Functions of First, Second and Third 
Be be des oS Wee AAS os a BG ee, 1 a 91 
2.4 Connections with Fibonacci Sequence. ...... 109 
2.5 Solved and Unsolved Problems........... 124 


Introduction 3 


Introduction 


The function named in the title of this book is onginated from 
the exiled Romanian mathematician Florentin Smarandache, who 
has significant contributions not only in mathematics, but also in 
literature. He is the father of The Paradozrist Literary Movement 
and is the author of many stones, novels, dramas, poems. 

The Smarandache function,say S, is a numencal function de- 
fined such that for every positive integer n, its image S(n) is the 
smallest positive integer whose factorial is divimble by n. 

The results already obtained on this function contain some 
surprises. Such a surprise is the fact that to expresse Sip) tne 
exponent «œ is written in a (generalised) numerical scale, say [p], 
and is read” in another (usual) scale, say (p) (eq. 1.21). More 
details on this subject may be found im section 1.2. 

Another surprise is that "the complement until the identity” 
(equation 1.34) of S{p*) may be expressed in a dual manner 


4 The Smarandache Function 


with the exponent of the prime p in the expression of n!, given 
by Legendre’s formula {eq. 1.15 and eq. 1.36). 

Finally, we mention that the Smarandache function may be 
generalised in various ways, one of these generalisations, the 
Smarandache function attached to a strong divmibility sequence 
{eq. 2.59}, and particulary to Fibonacci sequence, has a dual 
property with the strong divisibility {theorem 2.4.7). 

Of course, all these pleasant surprises ate “attractors” for us, 
the mathematicians, that we are in a permanent search for new 
wonderful results. 

But “the attraction” itself on the mitial concept, started by 
Florentin Smarandache, permitted to obtain the interesting re- 
sults mentioned above. Indeed, many mathematicians are al- 
ready inquired about this subject and have obtained these and 
other results, permitting the publication of the present book. 
Among these we mention here Ch. Asbacher, I. Balacenoiu, P. 
Erdos, H. Ibstedt, P. Gronas, T. Tomita. 

We mentione also two of the most interesting problems, still 
unsolved: 

i) Find a formula expressing S{n) by means of n itself and 
not using the decomposition of the number into primes. 

2) Solve the equation S(n) = S(n +1). 

The (positive) answer to first of these problems will permet 
to have more important information on the distnbution of the 
prime numbers. 

Let the future permit to reach the knowledge until these, and 
other, exciting results. 


THE AUTHORS 


Chapter 1 


The Smarandache 
Function 


1.1 Generalised Numerical Scale 


It is said that every positive integer r, strictly greater than 1, 
determine a numenical scale. That is, given r, every positive 
integer n may be written under the form: 


n = Cmr” + may tir + co (1.1) 
where m and c; are non-negative integers and 0 < œ < r—1, cm Æ 
0. 
We can attach a symbol to each number from the sequence 
0,1,2,...,7—1. These are the digits of the scale, and the equality 
(1.1) may be written as: 


Nr) =. fmm- NH (1.2) 


where y 1s the digit symbolsing the number c;. 
In this manner every integer may be uniquely written in a 
numerical scale (r) and if we note a; = r', one observe that the 


6 The Smarandache Function 
sequence (a;),en satisfies the recurrence relation 


Git = 7A (1.3) 


and (1.1) becomes 


Nn = CmOm + Cm—14m-1 +... +6141 + code (1.4) 


The equality (1.4) may be generalised in the following way. 
Let (6;)ien be an arbitrary increasing sequence. Then the non- 
negative integer n may be uniquely wntten under the form: 


n = chba + Ch—1bh—1 +... + cibi + Cobo (1.5) 


But the conditions satisfied by the digits in this case are not 
so simple as those from (1.3), satisfied for the scale determined 
by the sequence (a;)en . 

For instance Fibonacci sequence, determined by the condi- 
tions: 


R= =l, Fa = Fait & (1.6) 


may be considered as a generalised numerical scale, in the sense 


described above. 
From the inequality 


2 F: > Fiya 


it results the advantage that the corresponding digits are only 
O and 1, as for the standard scale determined by r = 2. 

So, using the generalised scale determined by Fibonacci se- 
quence for representing the numbers in the memory of computers 
we may utilise only two states of the circuits (as when the scale 
(2) is used) but we need a few memory working with Fibonacci 
scale, because the digits are less in this case. 


Generalised Numerical Scale 7 


Another generalised scale, which we shall use in the following, 
is the scale determined by the sequence 


:(p) = ——— 1.7 
a;(p) = (1.7) 
where p > 1 is a prime number. 
Let us denote this scale by [p]. So we have: 


[p]: 1, aa{p), as(p), ...-, ai(p), --- (1.8) 


and the corresponding recurrence relation is: 


Gi+1(p) = pa;(p) +1 (1.9) 
This is a relatively simple recurrence, but it is different from 


the classical recurrence relation (1.3). 
Of course, every positive integer may be written as: 


Nip] = CmOm(P) + Cm—18m—1(p) + ... + ciai(p) (1.10) 


so it may be written in the scale [p]. 

To determine the conditions satisfied by the digits c; in this 
case we prove the following lemme: 

1.1.1 Lemme. Let n be an arbitrary positive integer. Then 
for every integer p > 1 the number n may be written uniquely 
as: 


Nn = tian (P) + taan, (p) +... + tran, (p) (1.11) 
with nı > m >... > rų > 0 and 


1<tj;<p—1 forj=1,2,..,J-1, l<t;<p (1.12) 


8 The Smarandache Function 


Proof. From the recurrence relation satisfied by the sequence 
(a:(p))nene it results: 


a;(p) = 1, alp) =1+p, a3(p) =1+ptp”... 


So, because 


[a;(p), ai42(p)) N [ai (p), ai+2(p)) = 0 
it results 
{[a:(p), aimi) N N*} 


Then for every n € N® it exists uniquely n, > 1 such that 
n E [an (p), @n,4i(p)) and we have 


Nee 
t€ 


Ne 


n= ag eO 
where [z] denote the integer part of z. 
If we note 
kz a 
an, (p) 
it results 


n = tdm (p) +r with ry < an (p) 


H rı = 0, from the inequalities 


On; (p) sn Qn, +1(P) =o (1.13) 


it results 1 < ż; <p. 
If rı Æ 0, it exists uniquely ng E€ N* such that 


rE [Gna (p), An, +1(P)) 


Generalised Numerical Scale 9 


and because an (p) > riit results ni > nz. 
Also, because 


On, +1(P) — i hay | <p 
Gn, (p) 


from (1.13) it results 1 < tı < p—1. Now, it exists uniquely 
nz such that 


tis 


Ty = toda, (p) + r2 


and so one. After a finite number of steps we obtain: 


ri- = klan (p) +r; with r,=0 


and nı < 4-1, 1 < t < p, so the lemme is proved. 

Let us observe that in (1.11) unlike from (1.10) all the digits t; 
are greater than zero. Consequently all the digits c; from (1.10) 
are between zero and p — 1, except the last non-nul digit, which 
can take also the value p. 

If we note by (p) the standard scale determined by the prime 
number p: 


IPP eap (1.14) 


it results that the difference between the recurrence relations 
(1.3) and (1.9) induces essential differences for the calculus in 
the two scales (p) and [p]. 

Indeed, as it is proved in [1] if 


mis] = 442, nis] = 412 and ris] = 44 


then writing 


10 The Smarandache Function 


mtn+r= 442+ 
412 
—44_ 
dcba 
to determine the digits a,b,c,dwe start the addition from the 
second column (the column corresponding to a2(5)). We have 


Now, using a unit from the first column it results: 


5a,(5) + 403(5) = 03(5) + 4a (5) 


so (for the moment) b = 4. 
Continuing, we get: 


4a3(5) + 4a3(5) + a3(5) = 5a3(5) + 4a3(5) 


and using a new unit from the first column it results: 


4a3(5) + 4a3 (5) + a3(5) = a4 (5) + 4a3(5) 


soc=4andd=1. 
Finally, adding the remainder digits: 


4a1,(5) + 2a,(5) = 501 (5) + a, (5) = 5a, (5) + 1 = a2 (5) 
it results that the value of 5 must be modified, and a = 0. So 


mtn+r = 145055) 


A New Numerical Function 11 


1.2 A New Function in 
Number Theory 


This function is the Smarandache function S : N* — N* de- 
fined by the conditions: 


(s1) S(n)! is divisible by n, 
(32) S{n) is the smallest positive integerwith the property (s,) 


Let p > 0 be a prime number. We start by the construction 
of the function 


Sp: N — N" 
such that 


(ss) Sp(ai(p)) = p' 
(s4) Hn E N” is written under the form given by (1.11) then 
Sp(n) = t15,(an: (P)) + t2Sp(ana(p)) +... + tSp(an,(P)) 


1.2.1 Lemme. For every n € N” the exponent of the prime 
p in the decomposition into primes of n! is greater or equal to n. 
Proof. From the properties of the integer part we deduce: 


ato tte iS] [2 =| 
Se — — si — 
7 Sn) be 
for every a;,5 € N*. 


A result does to Legendre assert that the exponent of the 
prime p in the decomposition into primes of n! is: 


n 


e(n) = H + B TE (1.15) 


Then if n has the decomposition (1.11) it results: 


12 The Smarandache Function 


[ae tapitti] > [e=] es Bs ges cit [a2] 2 
= typ") 4 top} + + tpu} 


eee ee eee eee ere ere ere eee eee eee rer ee ee error re er eee eee ery 


fanaa > [oga + [a] + + [a = 
a typu—™ + top™—™ +... + trp? 


POCO OREO EOE ROE E RHE ERES ESE SEDO UEEE EEE SEH CEH ETO TH OREEO OTOH ESE DH OL EREHOE 


| eee eee | ae 
= tip? + [ae] +... + [see 


and so 


[e] + [3] +...+ [2] > tate + ph +... +p) +... 
+ti{p™ —1 +p™ -2 4.. +p? = 
= ilan (p) + t2am (p) ++ tian (p) = 


1.2.2 Theorem. The function S, defined by the conditions 
(s3) and (s4) from above satisfies: 
(1) S,{n)! is divisible by p” 
(2) S (n) is the smallest positive integer with the property (1). 


Proof. The property (1) results from the preceding lemme. 
To prove (2) let n € N* and p > 2 an arbitrary prime.Considering 
n written as in (1.11) we note 


z= typ™ + top™ + ..tp™ 


A New Numerical Function 13 


and we shall prove that the number z is the smallest positive © 
integer with the property (1). 

Indeed, if there exists u € N”, u < z such that u! is divisible 
by p”, then 


u <z => u<z-1= (z-1)! is divisible by p” 


But 


z—1=tp™ + fap" +... +tp™"— l 


and n >ng >... >n > 1. 
Because [k + a] = k + [a] for every integer k, it results: 





—] 
É p | =typ™—* + top) ++ typ th — 1 


Analogously we have for instance 





—1 
É | = typ ™ + top + tty pt! WM + tip? -1 


[a] = hpm Tl ipm tet pt -i Tmt 
+ bpa] apt ap nn 


because 0 < tp™ -l<p-p™—1< p™ +}, 
Also, 


[pete] = fp) + tap? + [aerial = 
=tp™T™ -1 + typ 


The last equality of this kind is: 


14 The Smarandache Function 


a] 





zi top™ +... + typ™ — 1 
f | =p? [ete = =tp 


p™ p™ 
because 


0 < tap™ +-+ tip™ < (p— 1)p™ +... + (p— 1)p™ -1+ 
p-p™—1<(p—-1) D pit pett—1<(p-1)emt= 


t=} Pat 
= pstl_j <p" —1< p™ 


Indeed, for the next power of p we have 


0 


— 


f = 7 = ae + top™ ak, + tip™ a 


p™ +1 p™ +1 


because 
O<tip™ +top* +... Htp —1<p“ t 1< pt 


From these equahties it results that the exponent of p in the 
dcomposition into primes of (z — 1)! is 


[252] + [st] +... + [252] = alph 2 + pu 2 +... +p") +... 
+tj1(p™ -171 +... + p?) + t¢(p™ t + +p)—np=n-—n<n 


and the theorem is proved. 
Now we may construct the function S : N* — N* having 
the properties (sı) and (s2) as follows: 


(t) S(1)=1 
(it) For every n= př - p5?...p%, with a; > 1, 
and p; primes, p; # p; we define: 


S(n) = max Sp, (æi) (1.16) 


A New Numerical Function 15 


1.2.3 Theorem. The function S defined by the conditions 
{t} and (ii) from above satisfies the properties (31) and (32). 

Proof. Let us suppose n # 1. We shall note by M(z) an 
arbitrary multiple of z and 


Spy (aig) = max Sp, (a) (1.17) 
Of course, 
Spa (2a)! = M(t 
and because S,;(a;)! = M(p%‘) for i = I, s, it results: 


Spa (cig)! = M (pf?) for i = 1,8 


Moreover, because p; A p = 1 it results: 


Spy (ig)! = M (pi p3p") 
and so (31) is proved. 
To prove (s2) let us observe that for every u < Sp, (ai,) we 
have u! # M(p;,*), because Sp, (ia) is the smallest positive 
integer with the property k! = M(p;,*). So, 


ul # M( pi -pi --p3') = M(n) 
and the property (s2) is proved. 
1.2.4 Proposition. For every prime p the function S, is 
increasing and surjective, but not injective. The function 5 is 
generaly increasing, in the sense thai: 


(V) ne N (A)REN* S(k) >n 


and it is surjective but not injective. 
1.2.5 Consequences. 1) For every a € N” holds: 


S,(a) = S°) (1.18) 


16 The Smarandache Function 


2) For every n > 4 we have: 


nis apime 4> S(n)=n 


Indeed, if n > 5 is a prime then S(n) = S,(1) = n. 

Conversely, if n > 4 is not a prime but S(n) =n, let n = 
pi + p?...po with s > 2, a; € N*, for i =1,s. Then if Sp, (a) 
is given by (1.17), from Legendre’s formula (1.15) it results the 
contradiction: 


Spi (Qin) < WigPig CN 
Also, if n = p° , with a > 2, it results: 
S(n) = S,(a)< p-a< p =n 


and the theorem is proved. 
1.2.6 Examples. 1) If n = 2% . 377 . 733 we have: 


S(n) = max{S3(31), S3(27), S7(13)} (1.19) 


and to calculate 53(31) we consider the generalised numerical 
scale 


[2]: 1, 3, 7, 31, 63,... 


Then 31 = 1- as(2), so S2(31) = 1- 25 = 32. 
For the calculus of $;(27) we consider the scale 


[3]: 1, 4, 13, 40,.... 

and we have 27 = 2- 13+ 1 = 2a3(3) + 2:(3) so 

53(27) = S3(2a3(3) + ai(3)) = 253(a3(3)) + 53(a1 (3)) = 
=2-334+1-3'=57 


Finally, to calculate S7(13) we consider the generalised scale 


A New Numerical Function 17 


[7]: 1, 8, 57, .. 
and it results 


13 = a,(7) + 5a,(7), 80 S7(13) = 1 - S$,(8) +5-S7(1) = 
—1.774+5-7= 84 


From (1.19) one deduces S({n) = 84. So 84 is the smallest 
positive integer whose factorial is divisible by 2°! - 377 . 738. 

2) Which are the numbers with the factorial ending in 1000 
zeros? 

To answer this question we observe that for n = 101°% it re- 
sults S(n)! = M (101%) and S(n) is the smallest positive integer 
whose factorial endsin 1000 zeros. 

We have S(n) = $(210°.51000) = max{53(1000), S;(1000)} = 
Ss{1000). 

Considering the generahsed numerical scale 


[5]: 1, 6, 31, 156, 781,... 
it results: 


Ss(1000) = Ss(as(5) + a4(5) + 2a3 (5) + ay (5)) = 
= 55 + 5t +2. 53+5 = 4005 


The numbers 4006. 4007, 4008, 4009 have also the required 
property, but the factonal of 4010 ends in 1001 zeros. 

To calculate S({p*) we need to writte the exponent œ in the 
generalised scale [p]. For this we observe that: 


Om{p) < a => (p™—-1)/(p— 1) < a> 
p™ < (p— ije +1 4 m < log, ((p — Iho + 3) 


18 The Smarandache Function 


and if 


Ap} = kyde(p) + ky—1đy—1 (p) +... + kıa (p) = hy Ky—1..-Ky (1.20) 


is the expression of the exponent a in the scale [p], then v is the 
integer part of log,((p — 1)a + 1) and the digit k, is obtained by 
the equahty 

E 


Using the same procedure for r,_; it results the next non-zero 
digit from (1.20) 


1.3 Some Formulae for 
the Calculus of S(n) 


From the property (34) satisfied by the function 5S,, one deduce: 


S(p*) = plop) (1.21) 


that is the value of S(p*) is obtained multiplying the prime p by 
the number obtained writing the exponent a in the generalised 
scale [p] and ”reading” it in the usual scale (p). 

1.3.1 Example. To calculate $(11!™°) we consider first the 
generalised scale 


{11]: 1, 12, 133, 1464,... 


Using the considerations from the end of the preceding sec- 
tion we get: 


1000 = 7a3(11) + 5a2(11) + 9a,(11) = 759ny 


Some Formulae for Calculus 19 


so $(112) = 11(759) 1) = 11(7-11?+5- 11 +9) = 10021. Con- 
sequently 10021 is the smallest positive integer whose factorial 
is divisible by 111°. 

The equality (1.21) prove the importance of the scales (p) 
and [p] for the calculus of S(n). 


Let now 


u . v v p-l 
%) =D GPs =D, hair) =P ks U2) 
= gal j=l 


be the expression of the the exponent a in the two scales. It 
results: 


(p — la =) k;p — > k; 
j=l y=] 


Then noting 
Op) (a) =); Ci; Spile) =)> k; (1.23) 
sx=0 j=l 


v ; o—1 : 

and taking into acount that > kj = p }> kjp ìs exactly 
j=1 j=0 

Plae) one obtain 


S(p*) = (9 — iera (1.24) 
* Using the first equality from (1.23) we get: 


u 


pag) =P) alp? — 1)+ Dic 
s=0 


3=0 
or 





p = J 
aa =} cia; (p) + ogha) 


1=0 p= 


20 The Smarandache Function 


consequently 


—ł ] 
o = —— (ee) i 4 zaga) (1.25) 


where (œp) ip] denote the number obtained writing the exponent 


& in the scale (p) and reading it in the scale [p]. . 
Replacing this expression of œ in (1.24) we get: 








sie) = FF (ag) + ala) tola) (126) 


One may obtain also a connection between S(p°) and the 
exponent e,(a) defined by Legendre’s formula (1.15). It is said 
that e (œ) may be expressed also as: 


epla) = a (1.27) 
so using (1.25) one get: 


epla) = (apy) fp] e (1.28) 
An other formula for ep(œ) may be obtained as follows: if œ 
given by the first equality from (1.22) is: 


Op) = Cap” + Cup" +... + Cp + co (1.29) 
then because 


epla) = [a] + [a] +... + [E] = (cup? + capt? +... + ci) 
+(cup + Cu1) + Cu 


we get: 


epla) = ((a — co)cpy py = (( H Jo) (1.30) 


Some Formulae for Calculus 21 


where a») = GiCy—1-.-co 18 the expression of œ in the scale (p). 
From (1.26) and (1.28) it results: 





(pt) = P= (ea) +a) + Foyle) topla) (131) 


Using the equalities (1.21) and (1.26) one deduce a connection 
between the following two numbers: 


(œp) jip] = the number œ written in the scale (p) and readed 
in the scale [p] 

(2 1)(p) = the number œ written in the scale [p] and readed 
in the scale(p) 


namely: 


P (ano) — (2 — 1) (app = Palae) + (P — I a¢p)() (1.32) 


To obtain other expressions for S(p*) let us observe that from 
Legendre’s formula (1.15) it results: 


: ; ; œ— il 
S{p*) = pla — ip la)) with 0 < la) < TE (1.33) 
Then using for S(p®%) the notation S,(a) one obtain: 
1 . 
ea +i,(a)=a (1.34) 
and so, for each function S, there exists a function t,such that 
the linear combination (1.34), to obtain the identity, holds. 


To obtain expressions of i, let us observe that from (1.27) it 
results: 


22 The Smarandache Function 


a = (p— tep(a) + a¢(0) 
and from (1.24) it results a = (5,(a~) — of) (a))/(p — 1), 80 


(p — 1)ep(a) + ogla) = ‘lala alah 


S(p*) = (p — 1)°e,(a) + (p — I)o la) + ogla) (1.35) 


Let us return now to the function 1, and observe that from 
(1.24) and (1.34) it results: 


pla) = & = Iple) (1.36) 
P 
consequently we can say that there exists a duality between the 
expression of e (æ) in (1.27) and the above expression of t,(a). 
One may obtain other connections between 2, and e,.For in- 
stance from (1.27) and (1.36) it results: 


ipla) p (p = 1)e (a) Laai x Tla) (1.37) 
Also, from 


Op = Kykyt..ky = kolp! +p? +... +1) + koa (p? 7+ 
p+.. tpi)... + kapt ith 
one obtain 


o = (kop’) + koip”™? + ... + kop + ki) + ko (p? 7+ 
+p°" +... +1) + koa(p? > +p? * +... 1) +... 


+k3(p + 1) + ka = (apg) + [2] - [22] 


Some Formulae for Calculus 23 


that because 


[a] = kolp? + peo +... + p + 1) + E + kolp 
HP H. EPH A + AS to t kalp + 1)+ 
tatkipi yi 
and [n + 2] = n + [z]. 
One obtain 


a = (odo) + z| - za] 108) 


and we can wnitte: 


stor) =p- (|| - (h a9 


p 
and from 1.36) and (1.39) we dededuce 


tp(a) = H = on) e (1.40) 
P P 
This equality results also directly, from (1.36), taking into 
acount that 
stents. [fy 
P p P P 








consequently 


P p P 


An other expression of i (œ) is obtained from (1.21) and 
(1.36) or from (1.38) and (1.40). Namely 


sf 





la) =a— (ati) (1.41) 


24 The Smarandache Function 


From the definition of the function S it results: 


Slol) =p [2] = 0-0, 


where œp is the remainder of œ modulo p, and also: 


eolSp(a)) >a, ep(Sp(a)—1)<a (143) 


Sp(a) — a%)(Sp()) TA S la) — 1 — Tpl Spla) — 1) an 
p-l1 p-1 
Using (1.24) it results that S (œ) is the unique solution of 
the system: 


olz) < op (a) < alz- 1) +1 (1.43) 
At the end of this section we return to the function i,, to 
- find an asimpthotic behaviour for this “complement until the 
identity” of the function S,. 
From the conditions satisfied by this function in (1.33) it 
results for 


Q 


Alap) = [2] - ilo) 


that A(a,p) > 0. p 
To find an expression for this function we observe that: 








R 


and supposing that œ € [hp +1, hp+p-— 1] it results (= = [e] 
so: 


Some Formulae for Calculus 25 


Aam E — 4(a) = |) (145 
Also, if œ = Ap, it results 


PS ]= [BR ]nact ae [=a 


so (1.44) becomes: 








A(a,p) = Eg -1 (1.46) 


Analogously, if œ = hp + p, one obtains 


nn 


and [2] = h + 1, so (1.44) has the form (1.46). 

It results that for every œ for which A(a,p) has the form 
(1.45) or (1.46), the value of A(a, p) is maximum if of,);(q) is 
maximum, so for œ = ay, where 


au = (p - 1)(p—1)...e—- 1p 
Peet ae a 


v terms 





We have then 
ia GEO E eee E T 


(p — 1)(2S + P+ + ES) t+ p= 
(p° +p?) +... +p? + p) — (v — 1) = palp) — (v — 1) 
It results that ay is not divisible by p if and only if v — 1 is 
not divisible by p. In this case 


Talau) = (v-1)(p—1)+p=pv—vt+1 


26 The Smarandache Function 











and 
eT A 
° ilan) 2 kS i -v 
that is 








3 &u — 1 a 
E 7 V, 
son f= =n es 
If v—1€ (hp, hp + p) it results [=] = h, and 
h(p-1)+1< Afaw,p) < hlp- 1)+p+1 


hm A(ay,p) = œ 


aun oo 


We also observe that 


oe aad eo 


P P 











ofl _ _ ptl tpl 4° 

Boned ena a asian 
p-1 p p—i pal 

So, if a4 — œ as p” then A(ay, p) — œas z. 

Also, from 


ip(œ) is aop) — v 


a] afp) - [4] 


— } 





it results 


Connections with Classical Functions 27 


p(x) 
=i 


P 


=] 


1.4 Connections with Some Classical. 
Numerical Functions 


In this section we shall present some connections of Smarandache 
function with Euller’s totient function, von Mangolt’s function, 
Riemann’s function and the function H(z) denoting the number 
of primes not greater than z. 

1.4.1 Definition. The function of von Mangolt is: 


_j lan ifn=p™ 
A(n) = l gen (1.47) 

This function is not a multiplicative function, that is from 
n Ü m = 1 does not result A(n- m) = A(n)-A(m). For imstance, 
if n = 3 and m = 5 we have A(n) = In3,A(m) = In5 and 
A(m-n) = A(15) = 0. 

We remember the following results: 

1.4.2 Theorem. The following equalities hold: 


(2) p A(d)=Inn 
OA =P wld) in 3 


where u is Mobius function, defined by: 


1 uni 
p(n) = 0 if n is divisible by a square (1.48) 
(-1}* if n — py ` o....-Dk 


28 


The Smarandache Function 


1.4.3 Definition. The function Y : R —> R is defined by: 


W(z)= 9 bap (1.49) 


From the properties of this function we mention only the 
following two: 


1.4.4 The function WV satisfies: 


() (2) =F Aln) 


(ii) U(r) = In{1, 2, 3,..., [z] 


where [1, 2, 3, ..., [z]] denotes the lowest common multiple of 1, 2, 3, ..., [z] 


It is said that on the set N* of the positive integers one may 
consider two latticeal structures: 


No=(N",A,V) and Ng=(N"A,V) (1.50) 


where 


A= min, V = max 

A= the greatest common divisor 
d 
V= the lowest common multiple 


We shall note also n A m = (n,m) and n ¥ m = [n,m]. 


The order in the lattice N, is noted by < and the order from 
Na is noted by $ It is said that: 


ni L n > n divides n —> nifn (1.51) 
G 


and we also observe that the Smarandache function is not a 
monotonous function: 


nı < n does not implique S(n,) < S(nz) 
But, taking into account that 


Connections with Classical Functions 29 


d 
S(ny V ng) = S(mi) V S(n2) (1.52) 
we can consider the function S as a function defined on the lattice 
Nj with values in the lattice N,: 
S: Ni — N (1.53) 
In this way the Smarandache function becomes an order pre- 
serving function, in the sense that: 


nı 5 na =>> Shni) < S(n2) (1.54) 


It is said [31] that if (V, A, V)is a finite lattice, V = {21, 22,..., zn}, 
with the induced order <, then for every function f : V — R, 
the corresponding generating function is defined by: 


F(n)=)0 Fly) (1.55) 
yn 
Now we may return to von Mangolt’s function. Let us observe 
that to every function: 


f:N*—+N* (1.56) 


one may attach two generating functions, namely the generating 
functions F* and F° determined by the lattices N4 and Mh. 
Then, by the theorem (1.4.2), for f(z) = A(z) it results: 


F4(n) =Y A(k) =Inn (1.57) 


kin 
4 
and also 


F°(n) =$ A(k) = ¥(n) = of], 2, .. n] 


k<n 


30 The Smarandache Function 


Then it results the following diagram: 





It results a strong connection between the definition of the 
Smarandache function S and the equalities (1.1) and (2.2) from 
this diagram. 

Let f from (1.56) be the function of von Mangolt’s. Then 


Se eee j| = ef (n) = ef) . ef(2). ofl) = etn) 
n = saa = eF%X)). eF 42) _eF%n) 


and so, using the definition of S, we are conducted to consider 
functions of the form: 


Connections with Classical Functions 31 


y(n) = min {m / n <[1,2,..m]} (1.58) 


We shall study this kind of functions in the section 2.2 of the 
following chapter. 

Returning now to the idea of finding connections between the 
Smarandache function and some classical numerical functions, 
we present such a connection, with Euller’s function y. Let us 
remeber that if p is a prime number then: 


y(p*) = p” — p% (1.59) 


and for œ > 2 we have 


p?' =(p—I)oaa(p)+1 so op (p*") =p 
Using the equality (1.24) it results: 


Sp) = (p — Ip") + olp) = plp) +p (2.60) 


1.4.5 Definition. Let C be the set of all complex num- 
bers. Then the Dirichlet series attached to a function 


f:N* — cC 
D;(2z) p? fn) (1.61) 


For some z = z + ty this series may be convergent or not. 
The simplest Dirichlet series is: 


(=o = 


asl 


32 The Smarandache Function 


named the function of Riemann or zeta function. This function 
converges for Re({z) > 1. 
It is said that the Diriclet series attached to Möbius function 
u 18: 
D(z) = E for Re(z)> 1 
¢(z) 


and the Diriclet series attached to Euller’s function ¢ 1s: 


D(z) = aaa for Re(z)>2 





We also have: 


D(z) = (z) for Re(z)>1 


where T(n) is the number of divisors of n, including land n. 
More general, 


Dy, (z) = C{z)-((z—k) for Rez) >k+1 


where o;(n) is the sum of k“— powers of the divisors of n. 

In the sequel we shall writte a(n) instead of o,(n) and r{n) 
instead of co(n). We also suppose that z = z, so z is a real 
number. 


1.4.6 Theorem. H 


is the decomposition of n into primes then the Smarandache 
function and Riemann’s function are linked by the following 
equahty: 


C= 1) yn ty he -ri 
¢(z) ada p: 





(1.62) 


Connections with Classical Functions 33 


Proof. We have seen that between the functions y and ¢ 
there exists a connection given by: 





((z-1)_ > ofr) 
C{z) Bess nz (1.63) 


Moreover, 


oln) =Ï olp?) =i (Spl?) - pi) 


and replacing this expression of p(n) in (1.63) it results the 
equality (1.62). 
The Dirichlet series corresponding to the function S is: 





and noting by D Fa the Dirichlet series attached to the generating 
function Fé it results: 
1.4.7 Theorem. For every z > 2 we have: 


(1) ¢(z) < Ds(z) < ¢(z — 1) 
(tt) C*(z) < Drg(z) < C(z) e(z- 1) 


Proof. The inequalities (7) result from the fact that 


1 < S(n) <n for every n E€ N” (1.64) 
(tz) We have: 


C(e): Ds(z) =(X bÈ $) = 991) + TO 
+ Sts) + SOSO) +..... = Dpa(z) 


and the inequalities results using (2). 
One observe that (ti) is equivalent with 


34 The Smarandache Function 


D-(2) < Deg (2) < De(2) 


This equality may be also deduced observing that from (1.64) 
it results: 


YW 1<) Sk) <} k 
k<n k<n k<n 


and consequently: 


r(n) < F$(n) < a(n) (1.65) 
In [19] has been proved for Fé even that: 


r(n) < F(a) Sn+4 


To prove other inequalities satisfied by the Dirichlet series Ds 
we remember first that if f and g are two unbounded functions 
defined on the set R of real numbers satisfying g({z) > 0, and if 
there exist the constants C,, C2 such that 


/f(z)/ < Cyg(z) for every z > C3 


then the functions f and g are said to be of the same order of 
magnitude and one note 


f(z) = O(g(z)) 
Particularly, is noted by O(1) any function which is bounded 


for z > Ca. 
The fact that it exists 


f(z) 
Ae g(z) 





=0 


is noted by 


Generating Functions 35 


f(z) = o(9(z)) 


Particularly is noted by o{1) any function tending to zero 
when z tends to infinity and evidently we have: 


f(z) = ofg(z)) => f(z) = A(g(z)) 
It is said that Rieman’s function satisfies the properties given 


bellow: 
1.4.8 Theorem. For all complex number z we have: 


(2) ¢(z) + Ay + O(1) 
(ii) In¢(z) = ln +, + O(z - 1) 


(ii) C(e) = -rrr +001) 
Using the theorems (1.4.7) and (1.4.8) now we obtain: 


1.4.9 Theorem. The Dirichlet series Ds attached to the 
Smarandache function S and his derivative D% satisfy: 





(i) + Ons < Ds(z) < 45+ Ol) 
(i) — gy + O(1) < D; TA < -gy +0(1) 


The number of primes not exceeding a given number z is 
usually denoted by H(z}. In [39] is given a connection between 
the Smarandache function S and the function I. 

Starting from the fact that S(n) < n for every n and that, 
for n > 4 we have S(n) = n if and only if n is a prime, it is 
obtained the equality: 


H(z) = >> ; 2] - 1 


36 The Smarandache Function 


1.5 The Smarandache Function as 
Generating Function 


It is said that Mobius inversion formula permet to obtain any 
numerical function f from his generating function F?. Namely, 


f(n) =D wa) F4(5) (1.66) 
djn 
if 
F4(n) =F f(d) 
djn 


So, we can consider every numerical function f in two distinct 
positions: one is that in which we are interested to consider its 
generating function, and in the second we consider the function 
f itself as a generating function, for some numerical function g. 


a(n) = DEG) — D — Fin) =F fd) (1.67) 


djn 


For instance if f(n) = n is the identity map of N” we get: 


g(n) 2 MdG =el); Fn) =} d=old) (1.68) 


djn 


In the case when f is the Smarandache function S, it is dif- 
ficult to calculate for any positive integer n the value of F4(n). 
That because : 


F3(n) =$ S(d) =} max{s(5") (1.69) 


djn djn 


where 6; are the prime factors of d. 


Generating Functions 37 


However, there are two situations in which the explicite forme 
of F¢(n)} may be obtained easily. These are for n = p® and for 
n a square free number. 

In the first case we have 


Fi) =È Spi) =È (@ - i + ol) = 


a l (1.70) 
= (p— 1) D4 È opi) 
I= 
Let consider n = py - p2.....py a square free number, where 
pı < p2 <... < pk are the prime factors of n. It resulta: 


S(n) = pk and 
F#(1) = SQ) =1 
F§(p1) = S(1) + S(p) = 14+ pr 
F§$(p1 - p2) = S(1) + S(pi) + S(p2) + S(pr - p2) = 14+ pı +2p2 
F$(py + pa: ps) = 1+ pr + 2p: + 2 ps + F(p: p2) + 2 ps 
and also: 
Fé(n) =l + F4(p, * Pa.-..-- Phi) + 257o 


Then 


k 
Fé(n) = 1+ ye (1.71) 
One observe that because S(n) = ph, replacing the values of 
F(t) given by (1.71) in 


S(n) = 5y u(r) Fé(t) (1.72) 


rtan 
apparently we get an expression of the prime number p; by means 
of the preceding primes pı, 92,...pp-1. In reality (1.72) is an 


38 The Smarandache Function 


identity in which ,after the reduction of all similar terms, the 
prime numbers p; has the coefficient equal to zero. 
In [19] it is solved the equation 
Fé(n) =n (1.73) 
under the hypothesis 


SQ) =0 (1.74) 


and it is found the following result: 

1.5.1 Proposition. The equation (1.73) has as solutions 
only: all the prime numbers n and the composit numbers n = 
9, 16, 24. 

Proof. Because 


Fi(n) =F S(d) (1.75) 


djn 


under the hypothesis (1.74) one observe that every prime is a 
solution of our equation. Let now suppose n > 4 be a composit 
number: 


ko 8: 
n=I p;' 
sacl 
where the primes p; and the exponents r; are ranged such that 
(c:) piri > pir; for every t€ {1,2,...,4} 
(co) pi < pi+ı for 2 € {2,3,...,4— 1} whenever k > 2 
Let us suppose first k = 1 and r, > 2. From the inequality 


S(pi') < pis: 
it results 


Generating Functions . 39 


; d dt ri(ri +1 
pt =n = Fin) = FG?) =D Set) $b ps = BED 
a3 =0 s, =0 
so 
2p <rifri +1) if r, >2 (1.76) 


This inequality is not verified for pı > 5 and rı > 2, so we 
must have p, < 5. That is p, € {2, 3}. 

By means of (1.76) we can find a supremum for rı. This 
supremum depends on the value of pı. 

If pı = 2 it results for r, only the values 2, 3,4, and for pı = 3 
it results 7; = 2. 

So, for n = p}' there are at most four solutions of the equation 
(1.73), namely n € {4, 8,9, 16}. In each of these cases calculating 
the value of F2(n) we obtain: 


Fg(4) = 6, F4(8) = 10, F4(9) = 9, F4(16) = 16 


Consequently the solutions are n = 9 and n = 16. 
Let now suppose k > 2. Writing in the equation (1.73) the 
decomposition into primes of n we get: 


k , k tet rt Ti ep: oes 
Mo pi = FAI pi) =r S(d) = yp ..... 2 SIL pit) = 
t=1 s=1 djn 3; =0 =0 i=l 


33> 


ries 2 max {S(p}, S(pi?), ..., S(p) < 


33 Pa , 
ES losi > max {p131, P232, ---PkSk } < 
s, =0 3p =0 
fi Tè 
DES L max {piri, P22,- Phrk} = 
s1 =0 $,=0 
ri r k 
a ate D pin < pin D (rit) 
si= $,=0 t=] 


40 The Smarandache Function 
Consequently, the inequality: 


pi: piril +1) _ rfr +1) 
A ~ rI 


(1.77) 
= r; +1 Pi Py 


holds, and we are then conducted to study the functions: 





and = g{z)= 


z({z +1) 
— r fozr>0 


b=- 


f= 


where a, 5 > 2. 
The denvatives of these functions are: 


fi(z)= = syd [(z + 1)na — 1] and 
g{z) = (Finds a tabiet 
Because (z + 1)na—1 > (1+ 1)m2—1=2in2-1>0 


it results f(z) > 0 for x> 1. In addition the maximum of this 
function is obtained for z = max {1,z}, where 


` 2-Inb+ f(t)? +4 


2Inb 
and we deduce ,/(In 5)? + 4 < lnb +2, for b> 2, so 


< (2=Ind)+(mb+2)_ 2 2 2 
ee ee eee => Si5 
2in6 mp2“? 


We also have hm f(z) = g({z) = oo, and then pj! /(rı+ 


1) increase from p;/2 to aa “wheal rı € N*. Moreover, be- 
cause 


& 


IV 


MLS 


ifp >2 


Slo 


it results 


Generating Functions 41 


MED < mata È, B) o mala) 63 


Using (1.77) we obtain: 





k p k pi ri(ri + 1) e tum +4) i 
ata ad < — < <3 1.78 
al 2 sH rit] poo S 27:1-1 , ( ) 
for r, € N*, and so 
k P; 
— <3 
pis 2 = 
But we have also 
4p.23 5 «15 
Sa od LE E Mec | 
and then it results k < 3. 
For k = 2, using (1.77) and (1.78) it yendi! 
ae ela) and <3 
r+ 1 pi 2 


80 p < 6. 
If we suppose rz > 3, it results 


6 
pi-p2>2-3=6 or p > — 
Pi 


and then 


Pa P7 alr + }) 


6 
< 2, —} < max{2, = 
A < FZ DOED < marf2, Å) < maxt2, pa) = p2 


so it results the contradiction p3 < 4, and we have pz € {2, 3, 5}, 
r2 € {1,2}. Moreover, from 


42 The Smarandache Function 


z Pee mi(ri +3) © mili +1) 


i< 
rotl ` ppl 7 2n- 





JK 


it results rı < 6. 
Then, for fixed values of p2 and r3, the inequalities 


rri +1) a p? 


rı > par 
peo TET Piri > P2re2 


give us iformations for finding an upper bound of rı, for every 
value of pı. It results r, < 7 and the conclusions are given in the 


table bellow. 


P2 è m Pi ry n= pi P? F $(n) 
a) 2 1 3 Asn <3 2-32 24 3ri(ri +1) 
b 2 1 5 l<ry,<2 2-5% 245 (r: +1) 
c) 2 1 pT 1 2: pı 2+ 2p 
d 2 2 3 2 36 34 
e) 2 2 p>25 1 4p; 3p, +6 
f) 3 1 2 25755 32” 2r? — 2r, +12 
g) 3 1 p?>25 1 3pı 2p: +3 
h) 3 1 2 3 40 30 
If Fi(n) =n 
then 
a) 3 divides 2 
b) 5 divides 2 
: c) =2 
Conclustons: d) 34 = 36 
e) Pi = 6 
A on=3 
g) p =3 


h) 30 = 40 


Generating Functions 43 


It results that we must have 
m=3-2" or 7r4=3 


son = 3-23 = 24. That is for k = 2 the equation (1.73) has as 
solution only n = 24. 
Finally, supposing k = 3 , from 


aeg 


2 2 
it results p2- p3 < 12, so pg = 2 and p; E {3, 5}. 
Using (1.78) from 


ri(ri + 1) < ri(ry + 1) 


rol _ =ł 
Pi 37 


<2 (1.79) 


it results po = 3. 

Also, from (1.78) and (1.79) we obtain 

272 3n 
rot] rm +1 

and because the left hand side of this inequality is the product 
of two increasing functions on [0, 00), it results for rz and r3 only 
the values r3 = r, = 1. 

With these values in (1.77) one obtain: 


<2 








3 < ri(ry + 1) < ri(ri + 1) 
2 pmt E 5ri—1 


and so rı = 1. Consequently, the equation (1.73) is satisfied only 
for n = 2. 3- pı = 6p). 
But 
1 1 ee 1 1 a 
+ YY SY py) = 8+ E L max{5(2'- 9), pı} = 8 + 4p 
$= j= 


t=0 y=0 


44 The Smarandache Function 


because $(2* - 37) < 3 < p; for i,j E€ {0, 1}, and so it results the 
contradiction p; = 4. 

Then for £ = 3 the equation has no solution and the theorem 
is proved. 

1.5.2 Consequence. The solutions of the inequation 


Fin) >n | (1.80) 
result from the fact that this inequation implies (1.77). So, 


Fg(n) > n <=> n E {8, 12, 18,20} or n= 2p, with p a prime 
We deduce also that 
Fé(n) <n+4, for every n € N” 
Moreover, because we have the solutions of the inequation 


F4(n) >n 


we may deduce the solutions of the inequation Fé(n) < n. 
In [40] is studied the limit of the sequence 


T(n) = 1 — lIn F3(n)+ D? FAJ 


which contains the generating function. It is proved that 


im T(n) = —co 


In the sequel we focus the attention on the left side of (1.67), 
namely we shall regard the Smarandache function as a generating 
function of a certain numerical function s. 

By definition we have 


s(n) =} H(A) S(5 


djn 


Generating Functions 45 
If the decomposition into pnmes of the number n is 


n = pi Ds ape 
it results 


s(n)= E ays ——) 


Pii Pig---Pip Pi Pig-+-Ps, 


Let us consider that _ 


S(n) = max {S(p"*) = S(p;,") (1.81) 
We have the following cases: 
(a,) There exists io E {1, 2, ...,¢} such that: 
Spa?) > SM) for iF io 
The divisors d of n for which u{d) Æ Oare of the form: d= 1 
or d = Pi ` Dig----Di,- 


A divisor of the aecond kind may contains pi, or not. Using 
(ref 1510), with the notation Cf = gryp it results: 


s(n) = Slp (1 — Cii +C +.. + (ijaya 
Sipke -1+ 0L, -C +... + (10t) 


and so, we have: 


s(n) = { 0 if t > 2 or S(p;," ) = S(pz*~*) 


Dis otherwise 


(a2) There exists Jo € {1,2,...,¢} such that we have: 
Slp) < S(pj2) and Sees) > 2 S(p) fort ¢ {io, Jo} 


In this case, supposing in addition that 


S(p?) = max {S(p} / S32 T?) < S(p%*) } 


46 The Smarandache Function 


one obtain: 


a(n) = S(p;.°)(1 — Ch + CP, — e + (-1) NCEE )+ 
+S(p -1 +C} — C?_, +... + (—1)'7 10} 23)+ 
+S T A- Ola + COR- ... + (-1)*- 7723) 


jo 
and it results: 


s(n) = 0 if t>3 or S(p? = = S(p) 


— Pi otherwise 


Consequently, to obtain s(n} we construct, as above, a max- 
imal sequence 11, t2,...,2%, such that 


a: as a: ai 21l Qi; 
S(n) = S(p;,"), S(p” *) < 3 0 ess Er ) < S(p;* 
and it results: 


s(n) = | if t>k+1 or S(p) = S(p) 


—1}* +p; otherwise 


Now, because 


S(p*) = S(p*—*) <=> (p — 1) + opla) = (p — 1)(a — 1)+ 
tori (a — 1) > apla — 1)- agla) =p-1 


and 
SO JEST e Tpha —1)- Oty Co) =<] 
it results 


f t¢>k+1or 


0 
s(n) = Ciplak — 1) — oip (ak) = Pk - 1 
(—1)***p, otherwise 


Generating Functions 47 


1.5.3 Consequence. It is said [31] that if (V, A, V) is a finit 
lattice with the induced order <, then considering a function 
f : V — R as well as its generating function F', defined by the 
equality 1.55), and noting 


gis = F(z: A z;) 


it results 
det(gi;) = f(z1)- f(22).--f (zn) 


In [31] it is proved a generalisation of this result to an arbi 
trary partial ordered set, namely, defining the function g;; by: 


9;= DY f(z) 
I~ Ti 
ze Tj 
Using these results and noting A(r) = det(S(t A J)), for 


1,7 = ],r, we get: 


A(r) = s(1) - 9(2)...9(r) 
so, for sufficiently large r (in fact for r > 8) we have A(r) = 
0. Moreover, for every n E€ N” there exists a sufficiently large 
r € N” such that noting A(n, k) = det S({n + i) A (n + 7)), for 
i,j = 1,k, we have A(n, k) = 0 for k > r. Indeed, this assertion 
is valid because 


A(n, k) =i a(n +3) 


Ending this section we consider the Dirichlet series D, at- 
tached to the function s to prove the following result: 

1.5.4 Theorem. The Dirichlet series D, of the function s, 
given by 


48 The Smarandache Function 


D (z) a an ste) 


n=l 
satisfies: 
(i) 1 < D,(z) < D(z) forz >2- 
(ii) 1< D.) S ty 


for some positive constant A. 
Proof. (t) Using the multiplication of Dirichlet series we 
obtain: 


GPe = (5 EYE $B) = „aso + ASD 4 
posure) + -ORO OEO OEE ) 4 


=r ke Dale) 


seen 


and the afirmation results using the inequalities (i) from the 
theorem (1.4.7). The inequalities (11) also results using the same 
theorem. 


1.6 Numerical Series Containing 
the Function S 


It is difficult to study the variation of the function 5 on the set 
N* of all positive integers, because this function is not monotonous 
in the usual sense. Then the study of some numerical series in- 
volving this function may be an useful instrument to obtain new 
informations about the function. 

In this section we add to the study begun by the Dirichlet 
series, the study of some new series, which shall give us informa- 
tion about the order of average of the Smarandache function. 


Numerical Series 


49 
1.6.1 Theorem. The series 


È s (1.82) 


converges. If 2, is its sum, then £ € (e — 3, 5). 
Proof. Let us note 





1 z 
E,=1+— Ta a tenth 
Then we shall prove the inequality 


> (Sk) 1 
ee eat 3 





(1.83) 
Indeed, we have 


Ma 
tj- 
[l 


& Tey =% (a= EF) >L an 


P 
tt 
ps 
Pa 
tl 
j 


and from S{k) < k it results: 
*. Sk) > 1 1 1 
2 ‘ (k +1)! $b ares =3 (1) ~2 


On “a other hend, pa k > 2 we have S(k) > 1 and conse- 
quently 








= S(k) “ 1 oe 1 3 
= = 4 = +... + = Pa - - 
2 (k +1)! >, (k + 1)! a 3 vier (n+ 1)! Ta 


1.6.2 Proposition. The series 


rS r * and (1 ica) with r 
een oe EN a) 2 ee he thre N 


50 The Smarandache Function 


converges. 


Proof. We have 


x es R» myrar +. reen 


s a toim + (a t+4+- -+ = 


n—r)t 
TEn-r + En- r—i 
and it results: 
> Stk 
> a < <rkn-- + En- r—1 
so the series from (i) is convergent. Analogously one may prove 


the convergence of second series. 
1.6.3 Remark. Because if n > 3 and m= z! we have: 





Le a 
Sim} n! 2 
it results the divergence of the series: 
o k 
= (1.84) 
2 S(k)! 
We may consider the series: 
~ _ S(k) 
1.85 
fs{z ) =) ‘(k+ DI? ( ) 
For 
Sk) 
SO EHN 


it results ak +1/ak — 0. Indeed, 


Oke: S(k+1) k+l i 


a,  (k+2)S(k) ~ (k +2)S(k) ` S(k) 





Numerical Series 51 


and so the series 1.85 converges, for all z € C. 

1.6.4 Propoisition. The function fs from (1.85) satisfies: 
/ fs{z)/ < Bz on the unit disc u(0, 1) = {z / /z/ < 1}, where 
f is the sum of the series (1.82). 

Proof. A lemme does to Schwarz assert that if a function 
f is holomorphic on the unit disc u(0,1)and satisfies f(0) = 
0, /f’(z)/ < 1 on this disc, then /f(z)/ < z on u(0,1) and. 
/f(0)/ <2. 

For fs it results 


ffs{z)<B tf /z/<1 


so the function (1/8) fs satisfies the conditions of Schwarz’s lemme. 
The connection between the function S and the factorial jus- 
tifies to consider the complement of a number until the most 
appropriate factorial. 
So, let us consider the function: 


b: N* —+ N* 
defined by the condition that 


Sín)! 
bays Sey )! 
1.6.5 Proposition. The sequences (6(n))nen- and (b(n) /n*)nens, 
for k > 0, are divergent. 
Proof. Of course, b(n!) = 1, and if (pa)nen- is the sequence 
of all the primes, we have 


(1.86) 


lpn) = “Wal — Pel = (pn ~ 1) 


Noting 


b(n) 
S 


52 The Smarandache Function 


for fixed k > 0 it results: 





_ S{n)! 
eR eat 
and so 
a ni)! ii eh 
1 = 
n! ni oF mS nije tt 
= = ait > E > Pa 


because it is said [33] that for fixed k and sufficiently large n we 


have 


Pi- P2---Pn-1 > pet? 


1.6.6 Proposition. The sequence 


T(n)=1+ > TO — In b(n) (1.87) 


has no limit. 
Proof. Let us suppose that fim. T(n) =1 < oo. From (1.84) 
it results 


Sa | 
2 


and then by the hypothesis, using (1.87) it results 


fim, In b(n) = co 


If we suppose lim T(n) = —oo, using the expression of b(n) 
from (1.87) it also resulta im In b(n) = 

We can’t have lim T(n) : = oo, because e T(n) < 0 for in- 
finitely many n. Indeed, komi < S(:)!, it results 


Numerical Sertes 53 


i i 
— < > 
Oy <1 fori >2 


Tipa) = 1+ san + say t+ + sey — allpa — 1)!) < 
1+ (Pa — 1) —In((pp — 1)!) = pn — In((pn — 1)!) 


But for sufficiently large k we have e* < (k — 1)!, and con- 
sequently there exists m € N such that p, < In((p, — 1)!) for 
n > m, and the proposition is proved. 

Let us consider now the function 


Ai(z)= $ b(n) (1.88) 


2<n<s 


1.6.7 Proposition. The series 


>> Ay*(n) (1.89) 
n=2 i 
converges. 

Proof. The sequence (6(2) + 6(3) = ...b(n))n>2 strictly in- 
crease to infinity and 


S(2) $(3)! S(2)! 
Sy a a 


S(2)!  S(3}}  S(4)! _ S(3)! 
a ase ean e A 


S(2)! | S(3)! S(4)! (5)! _ S(5)! 
Sag ge cage a 


S2} S3) s(a)! S(5)1 — S(6)! _ S(5)! 
a e a a a ee 


54 The Smarandache Function 


S(2)! 3)! 4)! 5)! 6)! 7)! SCT)! 
SOL 4 SDH A y Spy SOE 4 E > se 


CHAM e eee He Se He BERET HEH RES OEE ES EDOEO HERO ODESD 


so it results: 


2H) = sa + sara + + ee _ ee << 
<rt art grt - .+ £ METEOR < 


<14 © Baim) 142 ++ È alan) 


k=2 


But (p, — 1)! > pi: po...pp for n > 4 and so 


$ Arn) < 3 Bey a 


n=2 kzt 
where 
_ Pkípk+ı — pk) _ (Pe+i — Pr) < Prti Pk Pr+i 
i= a ee ee 


Pr! (pe — 1)! Fi Pa Pe Pi ` P2---Dk 


Because for sufficiently large k we have p; - p2...pk > Py) it 
results: 


l Pr+1 1 
ak < = 
Phat Phot 


and then the convergence of the series (1.89) results from the 
convergence of the series 








1 





k>ko Phos 


We shall give now an elementary proof of the series 


Numerical Series 55 


= 1 

Y —— m, vith a> 1 (1.90) 
k=2 (S(k)*)\/ S(k)! 

and using this convergence we shall prove the convergence of the 
series 


ee 1.91 
2 SUF oH 
1.6.8 Proposition. The series (1.90) converges, for all œ > 


Proof. We have succesively: 


a 1 ee 1 1 1 
Py (SSR) 2 V2! taat naty Jat 
1 1 1 _= 
taza T 7a JT T 40 V4! P =L ga Ja 
where mu ìs the cardinal of the set 


M: ={k/ S(k)=t}= 


Sih [divides iland doa not dvit- 1°?) 


It results that M: C {k / k divides t! }, so mu is lowest than the 
number of divisors of t!. So we have 


m, < r(t!) 


But it is said that r(n) < 2,/n, for every positive integer n, 
consequently 


= Mua 2V4! 3 < 
L TE eva ee 


and the proposition is proved. 


56 The Smarandache Function 


1.6.9 Consequence.From the convergence of the series (1.90) 
it results the convergence of the series (1.91). To prove this we 
shall use the following result: 

1.6.10 Proposition. For a > Olet us note 


= Coa | 


Then the inequality t= Vt! < t! holds for every t >t”. 
Proof. We have 


(t2) Vil < t! <=> (Eej! < (tY? <> t < t! 
On the other hand 
pa < (E)E ep (08) < (EE c lE < (E S 
<=> e7 < (+) 


But 
t> etl o (t j= Ja > (St yt- -la — 
sE e e 


Now, for z > 0 we have e? > 1 + z, and so, taking z = 2a + 1 
and t > 2a + 1, it results 


t 
(1) T7 > e*t > e? 
e 
Then for t > t” we get 
2a o fÊ\t—2a la o (fyt 
< (-) => iů <(5) <t! 
e 


It results £2 < t! if t > t. 
Using this result we may wnitte: 


>= for t > t 





CEG o S? 


Numerical Series 57 


and from the proposition (1.89) it results the convergence of the 
series 


and of course we have 
3 S(k)! 2 t! 


1.6.9 Theorem. Let f : N* — R be a function which 
satisfies the condition 


c 
t) < — 
ft) s t«(d{t!) — d((¢ — 1)!) 
for ¢ E€ N* and the constants a > 1,c > 0. Then the series 


2 f(S(k)) 


is convergent. 
Proof. For M, given by (1.92) we have M; = d(¢!)—d((t—1)!) 


and 


FSC) => M.f(t) 


k=1 = 
Then because M: f(t) < Mi- a, = & it resultsthe convergence 
of the series. 
1.6.10 Proposition. If (za),en- is any strict increasing 
sequence of positive integers, then the series 
5 Intl — In 
n=1 S(zn) 


is divergent. 


58 The Smarandache Function 
Proof. Let consider the function 


eam reel oe R, f(z) =Innz 


From the theorem of Lagrange it results that there exists c, € 
(Zn, Zn41) such that 





In ln zn4: — In ln z, = (Za4i — Zn) 


1 
Cn IN Can 


and because ry, < Ca < Zn+1, we have 


Tn — 2 Zn+1 — 2 
— < lnInz,4, —hlnz, < A 
+ 


1.93 
Taji MN n41 InN Zp l ) 


for every n E€ N*.Then for n > 1 


an S(n) < 1 


<1 => 92 ae 
ars alan = tan 


That is 
Sr) _ 
n—onlnn 
and hence for every n € N* there exists k > 0 such that Sin) < k, 
or ninn > Ia. Then 
REE O ca 
Inin zn  S(zn) 
Introducing (1.94) in (1.93) we obtain 


(1.94) 


ln in za — ln ln za <b 


for every n > 1. Summing it results 


Numerical Series 59 


= In+1—- In ji 
— n > -(lnla zm} — inln 
3 S(zn) ; (in Tm+1 zı) 
and the divergence of the series results from the fact that ln ln zm 
tends to infinity. 

Consequences. 1) For z, = n it results the divergence of 
the seres 


= ] 


2) If z, = pa (the n—th prime), it results the divergence of 
the series 


$ Pn+1 — Pn 
n=1 Pn 

3) If (t,)nen+ is an arithmetical progression of positive inte- 
gers then the series 


Eag 
n=l S(Zn) 
is divergent. 


1.6.11 Proposition. The series 


1 


2 S(4)S(2)...5(n) 


n=1 
is convergent to a number s € (1.71, 2.01). 
Proof. From the definition of the Smarandache function it 
results the inequality 
1 


KOM 


3 | 


and summing we get 


The Smarandache Function 


60 
= 1 <] 
2 S(1)S(2)...S(n) an 


n=l 


On the other hand the product §(1)S(2)...S(n) is greater 
than the product of primes from the set {1,2,..n}, because 


S(t) = i if i is a prime. Therefore 
1 


1 
n . k 
H St) Üp: 
t=1 t=1 


where p; is the greatest prime number not exceeding n. Then 


S i = i 
5 =2 ma = sin + aiken t+ 
+ SIystay SR) Oa ane ATE 
St... f Pee +, 


2-3-5-7 P1 P2---Ph 
and using the inequality pip2...p% > p4, for every k > 5 (see 


[33]) it results: 
PEEN += = + tas at oe (1.95) 

23 T 3 aa 
let us note P TT Ft and observe that P < r+ gr + 


grt... 
It results 
x? ] 1. 1 
Pep cUt at ate tag 
because = = 1+4 +++- 
Introducing in a 95) we obtain: 
1 1 1 2 x 1 1 1 


(te ae E AS 
Sia 2 gh ge" Io 


Numerical Series 61 


Estimating with an approximation of order not more than 
107? it results s € (1.71, 2.01). 
1.6.12 Proposition. For every œ > 1, the series 


2 S(1)S(2)...5(n) 


converges. 
Proof. If (pr)een- is the sequence of primes, we can wnitte: 


A es -l 
ay = a = idl 
dat 
Na as Taps 
or 
4 < B 
TEST < TA pipz 
< 8l < P 
COORONG nen Pi paps 
P 
GONOCONONO) < Aan. a * Pipaps 


POSSE CE DO e eee oD OeeEteoeereree 


nm nt Pest 
KOKOREC < PIPr.--Pe < PIP? --.pr 


where p; < n for i =1,k , and pg} > n. 
Therefore 


= (Prl =P) Pest < 


2 —~1 
E aaa <1 + 2° HÈ aak; 
Se Piz 


Pip2.--Pa 
Because it es ko € N* such that for any k > ko we have 


PiP2...Pk > pet}, one have 


3 n° < 1 +204 D Pee ae = 1 
n=l S(4)S(2)...S(n) =, PiP2---Pk k=ho Phi 


and so our series is convergents. 


62 The Smarandache Function 


Consequences. 1) There exists no E N* such that $(1)S(2)S(3)...S(n) > 
n“ for every n > ng. Indeed, 


om. 5) 5(DS3).5(n) O BUSS) Si) <P TNS 


2) It exists no E N* such that 


S(1) + S(2) + S3) +- + S(n) > n™ for n> no 
Indeed, we have: 


S(1) + S(2) +... + S(n) > ni/'S(1)S(2)...5(n) > nent =n 


for n > no. 


1.7 Diophantine Equations Involving 
the Smarandache Function 


The formula (1.21) may be used to solve certain diophantine 
equations involving the Smarandache function. 
1) The equation 


S(z-y) = S(2) + Sty) (1.96) 


has an infinity of solutions. 
Indeed, from (1.16) it results that if zo and yp are solutions 
of the above equation then zp A yo # 1. That because 


d 
S(zo + yo) — S(zo V yo) = max {S(z0), S(yo) } 
Let now z = p°A, y = pê B be such that 


Diophantine sequations 63 


S(z) = S°), Sly) = S°) 
Then S(z - y) = S(p*+°) and the equation becomes p((a + 
bilo) = plapo) + PCE py» or 


((a + bio) = (epee) + pe) (1.97) 

There are infinitely many values of a and 6 satisfying this 

equality. For instance, a = a3(p) = 100p} 6 = (p) = 10, , for 
which (1.97) becomes: 


(1105) )¢p) = (1005o) + Ope) 
2) The equation 


S(z-y) = S(z) - Sty) (1.98) 
has no solutions z, y > 2. 
Indeed, let us note m = S(r) and n = S{y). It is sufficient 
to prove that S(z-y) Æ m-n. But it is said that m! n! divides 
(m+n)!, so 


en)! < +n)i<mi-nl<c- 
(m n)! $ (m n= m ngay 


and consequently S(z -y) < m-n. This is a strict inequality if 
m:n > m+n, so it is for m,n > 2. 

Consequently the equation (1.98) has as solutions only the 
numbers z,y < 2. 

3 The equation: 


zAy= S(z) A S(y) (1.99) 


also has infinitely many solutions. 
Indeed, because z > S(z), and the equality holds if and only 
if z is a prime or z = 4, it results that the equation (1.99) has 


64 The Smarandache Function 


as solution every paire of prime numbers, as well as every paire 
of square free numbers. 
Let now z and y be such that z Ay = d > 1 and 


S(z) = plagio Sly) = abiada) 
Because p Aq=1, noting a, = (ap})(p) and bı = (bike) if 
we have p A bı = a; A q = 1, the equation becomes: a; A 6; =d. 


This equality is satisfied for many values of a and b. For instance, 
if z = 2- 3% and y = 2- 5° it results d = 2 and we have 


(atada) A (ès) = 2 
for many vralues of a and b. 
4) Let now consider the equation: 


2Vp=S(z) ¥ SG) 


Every pair of primes is a solution of this equation, and if z, y 
are composite numbers, we observe that if we note 


S{z) = S(pi*) 5 Sly) = S(p;’ ), with pi # pj 


it results that the pair (z, y) is not a solution of the equation, 
because: 


2 Vy > pë -pY 2 S(z)- S{y) > S92) Ý Sty) 
Finally, if z = p*A,y = p’B, with S(z) = S(p*) and S{y) = 


S(p*). it results 
d d : d 
S(z) Y S(y) = plap g) Y plopp) = PCat ey V (pie) 


d d 4 
and z V y = p™*+4}(4 V B), consequently the equation als 
has many other solutions, which are not relatively prime. 


Diophantine sequations 65 
5) The equation 


S(z) + y=24+ S(y) (1.100) 
has as solution every pair of prime numbers, but also every com- 
posit numbers z = y are solution. It may be found other kind of 
composit numbers as solution for this equation. For instance, if 
p and g are consecutive primes and we note 

q—-p=h (1.101) 
taking z = pA, y = qB, the equation becomes: 


y — z = Sy) — S(z) (1.102) 


Considering the diophantine equation qB — pA = h, it results 
from (1.100) that Ap = Bo = 1 is a particular solution for this 
equation, and then the general solution is 


A=1l+rq, B=1+rp, for arbitrary re N 


Talking r = 1 it results z = p(1+q),y = q(1 + p), and 
y— zr =h. In addition, because p and g are consecutive primes, 
of course p+ 1 and q + 1 are composite numbers and then 

S({z) =P, S(y) =q, S(y) TA S(z) =h 
so the equation (1.102) is verified. 
6) To solve the equation 
S(m-z)=m- S(z) (1.103) 
let us observe that S(m-z) < S(z) +m. This fact results from 
the equahty 


(S(z) +m)! = S(z)!(S(z) + 1).....6S5(2) + m) 


‘TT 
at 


ct 
CT: 
oo 
O 
ta 
z 
R 
53 
E 
3 
Gy 
Q 
Q 
c 
e 
3 
Ca 
ce 
© 
3 


talang mto account that S{z)! is divisible by z and the product 
of m consecutive integers is divisible by m. 

If z is a solution of the equation it results m-S(z) < S(z)im, 
30 

(m— 1){S{z)-1})}< 1 (1.104) 

Then we have to analyse the following cases: 

(a) If m = i, the equation becomes S{(z} = z and has as 
solution every A integer. 

(5) H m = 2, it results we can have S(z) € {1,2}, and then 
z E€ {1,2}. 

(c) H m > 3, we must have S(z) = 1 so z = 1. 

7) For the equation 


Stat aa (1.105) 


lei us observe that S{z?) < y-z, because (yz)! = 1-2....2....(22).... (yz). 
Then, if the pair (z, y) is a solution for the equation, we must 
have y7 < yz. That is 


yo USS (1.106) 


H z = 1, the above condition is satisfied, and the equation 
becomes S(1) = y. Consequently, the pair (1,1) is a solution of 
the equation. 

For z > 2, only the pair (2, 2) verifies the inequality (1.106), 
so it is a solution of the equation. 

Indeed, for z > 3 we have z < 2771 4> lnz < (z — 1)ln2, 
and considering the function 


f(z}=(z—1)}b2-—lnrzr 


it results f(z) = {zla 2 — 1V/z, so f’{fz) = 0 4> z = 1/2. 


Dtophantine sequations 67 


For z > [1/ln 2] +1, hence for z > 2, this function is increas- 
ing, and in addition f(2) = 0. Then for z > 3 the inequality is 
strict. 

Let us now consider the equation 


A (1.107) 


n 


where k € (0, 1ļis a rational number. In [48] there are answered 
the following questions: 

(q1) For every k € (0, 1] there exists solutions of the equation 
(1.107) ? 

(q2) Find the values of k for which the equation has infinitely 
many solutions in N*. 

The answer to (q1) is negative, and the values of k for which 
the equation has an infinity of solutions are the following: 


k= with r € N* and 
kEQN(0 1], k=}, with p,geN*,0<q<p,pAq=1 


Indeed, if n is a solution of our equation, let 


and let d = n A S(n). Then, from the definition of d and 


from the fact that p and qare relatively prime, it results that 
S(n) = qd, n = pd and we have 


S(pd) = qd (1.108) 
Using the definition of S it results (gd}! = M(pd) and 


, — (92)! _ M(pd) _ M(p) 
ae aaa 7 


68 The Smarandache Function 


Because p and q are relatively prime, it results that (gd — 1)! is 
divisible by p and consequently 


S(p) <qd-1 
Let us prove also that S(p) > (q — 1)d. 
But, if the inequality S(p) < (q — 1)d holds, it results ({¢ — 
i)d — 1)! divisible by p. Then from d < (q — 1)d, it results 
d 


pd S ((¢ ~ 1)d)!, and so S(pd) < (q —1)d. This inequality is a 


contradiction of the fact that S(pd) = gd > (gq — 1)d. 
So, we have 


(g-1)d < S(p) <qd-1 (1.109) 


Taking q > 2, from the first of the above inequalities, it 
results d < S(p)/{q — 1), and from the second it results that 
(S(p + 1)/¢) < d, hence 


Sip +1) <d< Sle) (1.110) 
q q-1 
For q > 2 and k = p/q it results a necessary condition for the 
existence of at least a solution of the equation (1.107), namely 
the existence of an integer between S(p + 1)/q and S(p)/(q—1). 
But this condition is not a sufficient condition, as we can see 
from the examples listed bellow. 
Examples. 1) For k = 4/5 we have S(p + 1)/q = 3/2 and 
S(p)/(q — 1) = 5/3, so the equation has no solution. 
2) For k = 3/10 we have S(p + 1)/g = 11/3 and S(p)/(¢ — 
1) = 5/2, with the same conclusion as in the preceding example. 
3) For k = 3/29 it results S(p + 1)/q = 5/3 and S(p)/(g — 
1) = 14.5, so between S(p + 1)/q and S(p)/(q — 1) there exist 


more than one integer. However, the equation 


S(n) 3 
n 29 


Diophantine sequations 69 


has no solutions. Indeed, the number of the solutions equals the 
number of values of d for which (1.110) and then (1.108) holds. 
But it does not exist any integer between 2 and14 satisfying these 
conditions. 

Let us study now the equation (1.107) for k = 1/p, with p 
€ N*. We shall prove im this case that the equation has infinitely 
many solutions. 

Indeed, let po be a prime number greater than p and let 
n = ppo. It results S(n) = S(ppo) = po, and S(n)/n = 1/p=k. 

In [48] it is also answered the following question, posed by F. 
Smarandache: 

(q3) There exists infinitely many positive integers z such that 


| o? iiy Z (22) (1.111) 





where {z} = z — {z}? 

The system (1.111) of inequations has only one solution, 
namely z = 9. To prove this we shall prove first that the in- 
equation 


I 
t36) 

has infinitely many solutions. 
The inequality holds for z = 9, because 





}< (72) (1.112) 


9, 9,1 s(9), 2 
{sayt = tet = 2 and Fora haar 


At the same time one observe that any prime p is not a 
solution of the inequation. 
Let now z be of the form: 


z=py'-p5?...pet, with t > 2 
We have 


70 The Smarandache Function 


S(z) = max S(p,") 


1< k <t 


and let us put S(z) = S(p*}, where p° is one of p+, for i = 1,f. 
Then if z is a solution for (1.112) the number {say} may take 
one of the following values: 


2 s9- 
S(z)’ Sz) ° S(z) 


For such an z we have 
S(z) 1 
> 
z ~ S(z)’ 


It is said that from Legendre’s formula (1.15) it results S(p*) < 
œp. Then using (1.112) we deduce a?p? > p%, so 





so (S(p*))? > z > p° (1.113) 


ee a (1.114) 


If p > 2 then the last inequality holds only for integers œ < 

Qo. 

Indeed, we have p*~? > 2%-? and 2%—? > o? holds for œ > 8 
(the function f(a) = 2*—? — a? is increasing and f(8) = 0) 

We have to prove only that for a € {1,2,...,7} the system 
(1.111) has no solutions. 

(a) fœ = 1 it results S(z) = S(p) = p, and because p 
divides z we have z/p € Z, first of the considered inequalities is 
not satisfied. 

Let us observe that there exist solutions for the second in- 
equahty.Indeed, noting p = pı, the number z is of the form 
Z = pi ' p? -Pi , 80 


(5) = {2} = {P7 prp} = 0 and 


Loney ae Oe O 
{ z } {z iP } Pa si, >00 


Diophantine sequations 71 


Example. For z = 23 . 2" . 39, we have S(z) = 23 and 


Ja we 
919. 39 





{apy} = {2-3} =9; { 


(b) For œ = 2 let us note z = p*-z,. Then S(z) = S(p?) = 2p 


and 








HX) = {F} € {0,5} 
so we must have 
pz 4 S(z), 2 
gi ge ae 


and it results pz; < 4, that is p € {2, 3}. 

If p = 2, it results z} = 1 and so z = 4, which is not a 
solution for the inequation (1) from (1.111) because S(4) = 

H p = 3, it results also z, = 1, so z = p? = 9. 

Lfet us observe that the second inequation from (1.111) has 
also solutions. Indeed, with the notation p = p; we have: 


Pz? - p3 epe" S(r), _ 2 
a d e aE 


consequently the inequation is verified for z > 2 even number. 
Example. For z = 2°. 37 . 11? we have S(z) = 19 and 


z 25. _ ats) 1 


‘3a! = e lsn he EE TET 








(c) Let now be a = 3. We have seen that in this case if 
S(z) = S(p%), it results p < 7. 
If p = 2 it results S(r) = S(2°) = 4 and then 


72 The Smarandache Function 


p =e 


consequently the mequation nee from (1.111) has no solutions. 
However, there exist solutions of the second inequation. Indeed, 
considering for instance z of the form 


me 268? bs 72 (1.115) 


with a,b,c,d E€ N” such that d = a,(7) = (7* — 1)/(7 — 1) and 
S(z) = S(74) it results S(z) = 7” and so z/S(z) is an integer. 
If p = 3, we have S(z) = S(3°) = 9 and also 


(sey 


The inequation (2) has solutions in this case too. For in- 
stance z = 3° - z, are solutions, because 


yez (1.116) 


Ehe je] 


332, 32, 


If p = 5, we have S(z) = S(5°) = 15 and (1.111) becomes: 





0< (221) < foe =h with 2 A5=1 
From the first of these es it results: 


= 





(She {35 


so we must have 1/3 < one That is 57z, < 9, which is an 
imposibility. 

If p = 7, it results S(z) = S(7°) = 21 and 
Tz, 3 


0<{ 3 C Tr, 








Solved and Unsolved Problems 73 


(ye (43) 


Analogously it results the contradiction 3/({7*z;) > 1/3. 

If œ = 4one obtain p € {2,3}. For p = 2 it results S(z) = 
S(2*) = 6 and because z = 2*z,, with 2 Az = 1, the system 
(1.111) becomes: 


From the condition 3/(82,) > 1/3 it results z, = 1, so z = 16. 
But for this value of z we have 


(gta 95> gat 


For p = 3, we have S(z) = S(3*) = 9 and one arrive at the 
condition (1.115). 

For a € {5,6, 7} we get only p = 2 satisfying the condition 
(1.114), so z = 2%z, and because $(2°) = S(2°) = $(2”) it results 
for all the cases S({z) = 8. The condition (1.116) is verified again 
and the system has no solutions. 





1.8 Solved and Unsolved Problems 


In the sequel we indicate by a star (*) the unsolved problems. 
For the solutions of solved problems see the collection of Smaran- 
dache Function Journal and its extension The Smarandache No- 
tion Journal 

1*) Find a formula for the calculus of S(n), containing in- 
stead of prime divisors of n the number n himself. 

2) Prove that S(p?t!) = p’ 

3) Inddicate the number of solutions of the equation 


74 The Smarandache Function 


S(r} =n. 


4) Prove that the equation S(z) = p, where p is a given 
prime, has exactly d{{p — 1)!) solutions, all of them between p 
and p!, where d(z) is the number of divisors of z. (A. Stuparu) 

Generalisatton: The number of solutions of the equation 
S(z) = nis d(n!) — d((n — 1)!). 

5) Prove that max{ 2 / n > 4 is a composite number} = 2, 
(T. Yau) 

6) Let q be a prime number and k be an exponent such that 
S(q*) = n!. Let pi, pa,...p, be the list of primes less than q. 
Then the number of solutions of the equation S{z) = n!, where 
z contains exactly k instance of the prime g, is at least (k +1)’. 
(Ch. Asbacher) 

7) For every prime p and k > 1 prove that 





S(p*) > S(p**") (Ch. Asbacher) 
pF p*+ti 

8) Is the number r = 0.1234574651..., where the digits are 
the values of S(n) for n > 1, an irational number ? (F. Smaran- 
dache) 

9) Find the largest strictly increasing series of integers for 
which the Smarandache function is strictly decreasing. (J. Ro- 
driguez) 

10) Find a strictly increasing series of integer numbers such 
that for any consecutive three of them the Smarandache function 
is neither increasing nor decreasing. (J. Rodriguez) 

11) Are the points p(n) = a uniformly distributed in the 
mterval (0, 1]? 

12) Prove that 

fen SP) 


3 aan = nmn 
t———> 00 Pi 


Solved and Unsolved Problems 75 


where pı < po < ...pg... ìs the sequence of prime numbers. (P. 
Melendez) 

13) For every composite integer n > 48, between S(n) and n 
there exist at least five prime numbers. (L. Seagull) 


14*) Calculate > Ofp](t) using 2 O(p)(t). 
15) If we note 


re | 
T(n) = 1— hb S(n)+ Y -= 
(n) (n) L 
prove that 
im T(n) = 


16) If (pn )nen+ denote the sequence of all the prime numbers 
then the sequence {say} is unbounded. (M. Popescu. P. 
Popescu) 

17) For every k € N there exists a sequence ny < ng < ny... 
of positive integers such that 


üm ——~->k (Th. Martin 
lim, gay >k ( ) 


18*} Solve the following equations: 


() S(T) S(27).-S (r3) = Sleat 


(ii) SEP) S(2P)...S(e22,) = S) P7) 
19) Solve the equations: 
r?) = S(z)}? 
z5) = S(y)* (L. Tutescu, E. Burton) 


22) + S(z)S(z)? +2 


20) For all positive integers m,n, r,s holds: 


76 The Smarandache Function 


(i) S(mn) < mS(n) 
(tt) S(mn) > max{S(m), S(n)} 
(i) max{S(m), S(n)} < ANA) (S. Jozsef) 
(iv) m = n= Ain) > 
(v) S(mn) + S(rs) > max{S(m) + S(r), S(n) + S(s)} 
Consequence. For all composite numbers m,n > 4 holds 
S(mn) > S(m) + S(n) <2 
mn — m+n 73 


21*) Find n such that the sum 


(S. Jozsef) 


15D 4 D a bt (nn — 1) 4 


is divisible by n. (M. Bencze) 
22*) May be written every positive integer n as 


n = (S(z)y + 2(S(y))? +3(S(2)F? (M. Bencze) 


23*) Prove that 


= 1 
2 (S(k)} — S(k) +1 


is irrational. (M. Bencze) 
24*) Solve the equation S(z) = S(z + 1). 
25) Prove that 


n Sn) 


D 


is convergent, for every p > 1. 


Chapter 2 


Generalisations of 
Smarandache Function 


2.1 Extension to the Set Q of 
Rational Nnumbers 


To obtain such a generahsation we shall define first a dual func- 
tion for the Smarandache function. 

In [15] and [17] it is make evident a duality principle by means 
of which, starting from a given lattice on the unit interval [0, 1], 
there may be constructed some other lattices on the same inter- 
val. We mention that the results of these papers have been used 
to construct a kind of bitopological spaces and to introduce a 
new point of view in the study of fuzzy sets. 

In [16] the method to construct new lattices on the unit inter- 
val, proposed in [17] has been extended to a general lattice. But 
the main ideas from these papers may be used in various domains 
of mathematics. We shall use here to construct a generalisation 
of Smarandache function to the set Q of all rational numbers. 

In the sequel we adopt a method from [16] permitting to 


TT 


78 Generalisation of Smarandache Function 


construct all the functions linked, in a certain sens of duality, 
with the Smarandache function. 
One observe that if we note 


Raln) = {m/n S mi}, Ly=im | mi Sn} 
Rin) ={mf/n<mi}, L(n)={m/m! <n} 


we can say that the Smarandache function is defined by means 
of the triplet (A, E€, Ra), because one can write: 


S(n) = A{m / m € Ra(n)} 


We may also create all the functions defined using the triplets 
(a,b, c), where: 

-a is one of the symbols: V, A, A, and v 

- 6 is one of the symbols: € and ¢ 

-c is one of the sets: Rafn), La(n), R(n), L(n) defined 
above 

Not all of these functions are not-trivial. As we have already 
seen the triplet (A, E, Ru) defines the function S,(n) = S(n), but 
the triplet (A, €, Cz) defines the function 


Safn) = A{m / m! <n} 
d 


which is the identity. 

Many of the functions obtained using this method are step 
functions. For instance if we note by S, the function obtained 
from the tnplet (A, €, R), we have: 


S3(n) = {m/n < mi} 


so S3(n) = m if and only if n € [(m— 1)!+1, ml]. 
In the following we focus the attention on the function S4, 
defined by the triplet (Vv, €, La) : 


The Extension to the Rationals 79 


- S(n)=V{m f/m! = nr} (2.1) 
which is, in a certain sense, a dual of Smarandache function. 
2.1.1 Proposition. The function S, satisfies: 
Safna n na) = Safn) A Sa (na) (2.2) 
so is a morphisme from (N*, A) to (N™, A). 
Proof. If pi, po, ..., pi, -..18 the increasing sequence of all the 
primes and 
ni = Up? , no = Ip? with ai, pi € N 


only a finite number of œ; and f; being non-nulls, we get: 


n A no = Tpminte:, Bi) 


If we note Safni na = m, Sy(nj} = m, for i = 1,2 ,and 


supposing Mı < mz, it results that the right hand side of (2.2) 
18 Mı Amy. 
From the definition of S, we get for the exponent ep; (m) of 
the prime p; in the factorisation of m! the following inequality: 
ep: (m) < minfa; P:i) fori >1 


and at the same time it exists 7 > 1 such that 
ep; (m +1) > min(a;, ;) 
Then it results: 
Ai 2 ep (m) Pi 2 ep, (mm) fora 2-1 


We also have: 


80 Generalisation of Smarandache Function 


epi (M1) <a, ep, (m2) Sai 


and in addition it exists A and k such that: 


ep, (Mı +1) > oR, ep, (M2 +1) > a 


So, because m, < my, it results 


min(a;, 8;) > min(e,, (M1), ep, (m2)) = ep, (Mı) 


and then mı < m. If we suppose the inequality is stricte, it 
results m! < nı, so it exists A such that e,, (m) > a, and we get 
the contradiction: 


ep, (M) > min(aa, Br) 


Remark. For many positive integers n we have S,(n) = 1. 
For instance, 54(2n + 1) = 1 for all n € N and S,(n} > 1if and 
only if n is an even number. 

2.1.2 Proposition. Let pi, p2,...,p;,.-. the sequence of all 
consecutive primes and 


n= pi ppp gf! i gh 


the decomposition into primes of a given number n € N”, such 
that the first part of the decomposition is formed by the (even- 
tually) first consecutive primes. If we note: 


j a if ep, (SO) > a (2.3) 
| Sle) +p 1 if ep, (S(p )) = o; 


Safn) = min {titz -k Pea -1 } (2.4) 


The Extension to the Rationals 81 


Proof. H e,;,(S(e*)) > œ; from the definition of Smaran- 
dache function we deduce that S(p% )— 1 is the greatest positive 
integer m such that ep, (m) < a. Also, if Ce: (S(p™) = a; 
then S(p?') + pi — 1 is the greatest positive integer m such that 
ep, (m) = Q; 

It results the number min {t1, t2, ..., tx, Pe+1—1} is the great- 
est positive integer m for which e,.{m)} < œ; for all t = 1, 2,...,k. 

2.1.3 Proposition. The function S, satisfies: 


d 
Saf, + no) A Safna Vv na) = Safn) A Safna) 


for every nı, nz E N”. 
Proof. The equality results from (2.2) taking into account 
that: 


d 
(ri +a) A (mV na) = mAn 


Before to construct the extension of the Smarandache func- 
tion to the set Q, of all positive rationals we shall make evi 
dent some morphism properties of any functions defined by the 
triplets (a, b,c). 

2.1.4 Proposition. (:) The function S; : N* — N* , 
where 


Ss(n) =V{m/misn} 


satisfies: 


Ss(nz n na) = Ssni) f Ss(n2) = Ss(n) A Ss (n2) (2.5) 


(11) The function Sg: N* — N*, defined by: 


82 Generalisation of Smarandache Function 


Se{n) =v {m/n<ml} 


satisfies: 


d d 
Se(ny V nza) = Se(na) V Se(n2) l (2.6) 
(111) The function S; : N* — N”, defined by: 


Sp(n) =V {m/m <n} (2.7) 


satisfies: 


Sr{n A na) = S7(n1) A Sr(nz), Szem V na) = Sr(n2) V Srna) 


(2.8) 
Proof. (1) Let 


A= {a /a! Sm}, B= {b; / bj! S na}, C = {cr / cn S m Ara} 
Then we have A C B or BC A. Indeed, let 


A = {01,02,00}, B= {b1, b2, ...,5,} 


be the elements of A and B writen in increasing order. That 
is d; < aj4; and b; < bj+ for i = 1,h — 1 and j = 1,r — 1. Then 
if aa < b,, it results a; < b, for: = 1,A, so a;! > 6! S nq. 
Consequently A C B. 

Analogously, if b, < ap pb results B C A, and of course we 
have C = AN B. So, £ A C B it results 





d d 
Ss mr A na) =V eo V a= Ss(n1) = 
min{ Ss (n1), Ss(n2)} = Ss(n1) A Ss (na) 


The Extension to the Rationals 83 


Considering the function Sg defined on the lattice Nj, from 
(1.100) it results that it is order preserving. But if we consider 
this function defined on the lattice M, it is not order preserving, 
because 


mi<mi+1 but Ss(m!) =[1,2,...,.m] and Ss(m! +1) =1 


(iz) Let us observe that 


Se{n) =v {m / (B) €1,t such that ep; (m) < a;} 
If we note a = V{m / ns mi} then n < (a+ 1)! and 
atl=a{m /n< mi} = 5(n) 
so 


Se(n) = [1, 2,..., S(n) — 1] 


and then 


d d 
Se{ni V nz) = (1, 2, ne (Ta V n2)—1 = Srni Yna) = S7(n2)VS7(n2) 
Also, we have: 
d b a 
Sefni) V Se{ne) = if], 2,..., Sna) — 1], 11, 2,..., S(m_) — I] = 
(1, 2,..., S(nmi) V Sm) — 1] 


(itt) The equalities results from the fact that if mis given by 
(2.7) then 


S7(n) = [1, 2,....m] 4> n E [m!, (m+ 1}! — 1] 


84 Generalisation of Smarandache Function 


Let us now define the extension of the Smarandache function 
to the set Q, of positive rationals. 

It is said [25] that every positive rational a may be written 
under the form 


a =Q p°? (2.9) 


with p a prime, a, € Z and only a finite number of the exponents 
are non-nulls. Taking into account this equality one may define 
the divistbility of rational numbers as follows: 

2.1.5 Definition. The rational number a =H p% divides 


the rational number b =[ p? if a, < pp for all prime p. 


P 
The equality (2.9) implies that the multiplication of rational 

numbers is reduced to the addition of some exponents. Con- 

sequently the problems on the divisibility of these numbers are 

reduced to order problems between exponents. 

_ The greatest common divisor d and the smallest common 

multiple e for rational numbers are defined [25] by: 


a= (Gb, Ja primar A CR I e pmoxtos Bp} 
P P 


(2.10) 

Moreover, between the greatest common divisor d and the 

smallest common multiple of any rational numbers there exists 
the relation: 


1 
(5 p sea) 
Of course, every positive rational a may be written under the 
form: 


EATA (2.11) 


a= — with n E€ Nin, E€ N*, and (n,n) =1 
n 


L 


The Extension to the Raticnals 85 


2.1.6 Definition. The extension S : Q% —» Q* of the 


Smarandache function to the positive rationals is: 
s2 _ (r) 
Ni ~ Galm) 


A consequence of this definition is that if ny and ng are pos- 
itive integers then: 








(2.12) 


l d 1 1 1 - 
S(— N ma Sel Ye) (2.13) 
Indeed, 
S(= ż)= Sara) = = many = Sar ASI) E 


= Sim) V hy = S +)v S(t) 


For two arbitrary positive rationals we have: 


(= 9 2) = (Sin) Evs 2.14) 


This formula generalise the equality (1.16). 
2.1.7 Definition. The function 5 : Qj. — Q3, defined by: 


5(a) = (2.15) 


ae SÈ) 
is called the dual of Smarandache function. 
2.1.8 Proposition. The dual S of the function S satisfies: 
(i) Sima A na) = Sfm) A Slr) 
(it) SGA) ESAS 


for all positive integers n, and nz. Moreover, we also have 


86 Generalisation of Smarandache Function 


~n m ~ ~ x, l x, 1 
SE 9 T) = Fen) a Sm) GA) a 5h) 

The proof is evident. 

Remarks. 1) The restriction of the function S$ to the set of 
the positive integers coincide with the function S4. 

2) The extension of the function S : Q% —> Q? to the set 
Q*of all non-nulls rationals may be made for instance by the 
equality: 


S({—a) = S(a) for al a € QO. 


2.2 Numerical Functions Inspired 
from the Definition of 
Smarandache Function 


In this section we shall utilise the equalities (2.1) and (1.58) to 
define, by analogy, other numerical functions. 

Let us observe that if n is any positive integer then n! is the 
product of all positive integers not greater than n in the lattice 
£. Analogously the product pm of all divisors of a given m, 
including 1 and m, is the product of all positive integers not 
greater than m in the lattice £4. So we can consider functions 
of the form: 


a(n) =A{m/ ns olm)} 
It is said that if 


— wi 
m = Py 


23 


*P2 


zt 


Di 


Functions Inspired from S 87 


is the decomposition into primes of a given number m, then the 
product of all the divisors of m is 


p(m) = Vmr (m) (2.16) 


where 7(m) = (z1 +1)(z2+))...(z_+1) is the number of divisors 
of m. 
if n has the decomposition 


n = py) - p>?...pe (2.17) 
then the inequality n . p(m) is equivalent with: 


gi(z) = zi(zı + 1)...(2, + 1) — 20, > 0 
ga(z) = za(zı +1)...(2, + 1) — 2a, > 0 


ge(z) = ze(z1 + 1)...(ze+ 1) — 2a, > 0 


(2.18) 


So, Of (n) may be deduced solving the tollovang non-linear 
programmimg problem: 


(min) f(z) = py! - p3?...py" (2.19) 
under the restrictions (2.18). 

The solution of this problem may be obtained applying for 
instance the algorithm SU MT (Sequential Unconstrained Min- 
imisation Techniques) does to Fiacco and Mc. Cormick [18]. 

Examples. 1) For n = 3*. 5%? the equalities (2.18) and 
(2.19) become: 


(min) f(2) = 3" 5% 


with the restrictions 


l gi(z) = zi(z1 +1)(22 +1) > 8 
g2(z) = zo(z; + 1)(z2 + 1) > 24 


88 Generalisation of Smarandache Function 


Using the algorithm SU MT we consider the function 


u(z,n) = f(z)—r 3 In gi{z) 


and the system 


ĝu — 0 
Seg (2.20) 


z2 


In [18] it is shown that if the solution z,(r),z2{r) of this 
system can’t be found explicitely from the system, we can take 
r — 0. Then the system becomes: 


zo(z, + 1)(z2 + 1) = 24 


and has the solution zı = 1, r2 = 3. So we have: 


l 2y(2y + 1)(z2 + 1) =8 


min{ m / 3*- 52? < p(m)} =m =3-5° 


Indeed, p{mo) = Vmi (mo) — Mas 5 =n, 


2) For n = 37-5’, from (2.20) it results for z3 the equation 


223 + 9254+ 7z2 — 98 =0 


with a real solution in the interval (2, 3). It results z, € (4/7, 5/7). 
Considering z} = 1 we observe that for z = 2 the pair 
(z1, 22) is not an admisible solution of the problem, but z2 = 3 
give 6(37 - 57) = 3*. 57, 
3) In general, for n = pf! - p>? it results from the system 
(2.20) the equation: 


3 2 2 — 
aza + (a + a@2)23 + agz2 ~ 2a, = 0 


with the solution given by the formula of Cartan. 


Functions Inspired from S 89 


Remark. Using "the method of triplets” we may attache to 
the function 0 defined above many other functions. 

Starting from the function v, given by (1.58), we may also 
obtain numerical functions by the same method. 

In the following we shall study the analogous of Smarandache 
function and its dual in this second case. 

2.2.1 Proposition. If n has the decomposition (2.17) then: 


(2) u(n) =max pri, (at) u(n v na) = y(n) V u(n) 


Proof. (+) Let be p?" = maxp?'. Then př < p%* for all 


¿= 1,t, so 


pes [2 -pa ] 
But (pf, p3 ) = 1 for i # j and then 


cata t R N ol 


H for some m < pg" we have n < [1,2,...m], it results the con- 
d 


tradiction 
pe* < [1, 2,...m] 
d 
(ii) If 
nı = Hp% j Ta = Tp? 


then 
nı ý n = Lpmertor Ae} 


d 
so v(m V nz) = maxp™ {9942} = max{ max p% , maxp*}and 
the property is proved. 


90 Generalisation of Smarandache Function 


Of course, we can say that the function v, = v is defined by 
the triplet (A, €, Rig}, where 


Rg={m]/n [1, 2,..., m]} 


Its dual, in the sense defined in the preceding section, is the 
function defined by the triplet (Vv, €, Lig), where 


Lig = {m / [1, 2, sesy m] < n}. 
a 

Let us note by v, this function: 

n(n) = V{m / [1,2,..., m] S n} 

Then 1(n) is the greatest positive integer having the prop- 
erty that all positive integers m < n(n) divide n. 

Let us observe now that a necessary and sufficient condition 
to have u.(n) > 1 is the existence of m > 1 such that every 
primes p < m divide n. 

From the definition of 1% it also results 
vn) = m <=> n is divisible by every i < m,but not by m+1 

2.2.2 Proposition. The function v, satisfies: 

n(n p na) = n(n) A v(m) 

Proof. Let us note 


n= n Am, n(n) =m, n(m) =m: fort = 1,2 


H mi = mi A ma, we prove that m =m. Indeed, from the 
definition of v, it results: 


S-functions of First, Second and Third Kind 91 


Baten, eS 
<=> {(V¥)i < m; => n is divisible by m; but not by m; +1 } 


If we have m < m, then m+1 < my < my, so m41 divides 
nı and ng, and so m+ 1 divides n. 

If m > m, then mı + 1< m, so m, +1 divides n. 

But n divides nı, so m, + 1 divides n,, and the proposition 
is proved. 

Let us observe that if we note 


to = max{t / j < i => n is divisible by 7} 


then 14(n) may be obtained solving the lmear programming 
problem 


tg 
(max) f(z) >L z; ln pj; 
_ b 
z; So; fori = 1,to; D ziln p; <ln pup 
s=1 


If fo is the maximum of f from this problem, then n(n) = 
efo, 

For instance 14(2° - 37.5.11) = 6. 

Of course, the function y may be extended to the set of all 
rational numbers by the same method as Smarandache function. 


2.3 Smarandache Functions of First, 
Second and Third Kind 


Let X be an arbitrary nonvoid set, r C X x X an equivalence 
relation, X the corresponding quotient set and (J, <) a total 
ordered set. 


92 Generalisation of Smarandache Function 


2.3.1 Definition. If g : X — J is an arbitrary injective 
function then the function 


f:X — I , defined by f(z) = 9(z) (2.21) 


is said to be a standardisation. About the set X we shall say in 
this case that it is (r, (J, <), f) standardised. l 

2.3.2 Definition. If r, and rz are two equivalence relation 
on X, the relation r = r, A rz is given by: 


rry <= rry and gry (2.22) 


Of course, r defined as above is an equivalence relation. 
2.3.3Definition. The functions f; : X — I, i = ], 3 are of 
the same monotonicity if for every z,y € X we have: 


flr) < fly) <=> fi(z) < fly) for k,j=1,s 
2.3.4 Theorem. If the standardisations f; : X — I, corre- 


sponding to the equivalence relations r; (for i: = 1, s) are of the 
same monotonicity then the function 


(2.23) 


f = max f; 

is a standardisation, corresponding to r =A ri, and it is of the 
same monotonicity as the functions f;. 

Proof. We give here the proof when s = 2. For an arbitrary 
value of s the assertion results then by induction. 

Let z,,, 2,, and 7, be the classes of equivalence c of z corre- 
sponding to the relations r;,r2 and r =r Arg. If X,,, 8n Xe 
denote the quotient sets induced by these relations then: 


filz) = 9:{2,,), for t= 1,2, where g;: Xe :—~ I are injective 


S-functions of First, Second and Third Kind 93 


The function g : X, — I defined by g(2,) = max(g,(Z,, ), ga(Z,, )) 
is injective. Indeed, if = Æ 22 and 


max{gi(Z}, »ga( Zt, )) = max( gi EA A g2(223,)) 


then from the injectivity of gı and gz it results for instance: 


max(91(Z7, ), 92(Z,)) = 91(Z}, ) = g2(Z?,) = max(gi(Z?, ), g92(22,)) 


and we have a contradiction, because 


filz?) = 91(22,) < o1(2},) = Az’) 
falz?) = ga(Z},) < g2(Z?,) = fa(z?) 


That is fı and f are not of the same monotonicity. 
From the injectivity of g it results that the function 


f:X — I f(r) =9{2,) 


is a standardisation. Moreover, we have: 


f(z?) < f(z?) <== g(t) < 9(2?) <> max(g:(Z!, ), 92(F4,)) < 
< max(9i(Z?, ), 92(Z7,)) => max{fi(z"), fo(z")) < 
< max(f,(z7), fa(z?)) 4> filz?) < f(z?) and fo(z') < f(z?) 


because f) and fz are of the same monotonicity. 

Let us now consider two algebraic lows T and L on X re- 
spectively on J. 

2.3.5 Definition. The standardisation f : X — I is said 
to be L-compatible with the lows T and L if for every z,y E€ X, 
the triplet (f(z), fly), f(zTy)) satisfies the condition E. In 
this case we shall also say that the function f L-standardise the 
structure (X, T )on the structure (I, <, L). 

Example. If the function f is the Smarandache function S : 
N” — N”, one can make evident the following standardisations: 


94 Generalisation of Smarandache Function 


(a) The function S, ©,-standardise (N*,-) on (N*, <, +) be 


cause we have: 


(E1) : $(a-5) < S{a) + 5(8) 
(b) The function S also satisfies: 


(£2) : max(S(a), S(b)) < S(a-b) < S(a), S(b) 
so this function }3-standardise the structure (N*, -} on the struc- 
ture (N", <,-). 

Now we may define the Smarandache function of first kind. 
We have already seen (section 1.2) that the Smarandache func- 
tion is defined by means of the functions Sp. We remember that 
for every prime number p the function S, : N* — N” is defined 
by the conditions: 

1) S,(m)! is divisible by p”, 

2) S,(n) is the smallest positive integer with the property 1). 

Using the definition of a standardisation in [2] there are given 
three generalisations of the functions Sp- 

To present these generalisations let us note by M{n) any 
multiple of the integer n. 

2.3.6 Definition. The relation r, C N* x N*is defined for 
every n € N” by the conditions: 

(i) H n =u‘, with u = 1 or u = p (a prime) and 1,a,b € N*, 
then: 


ar,b <=> (4) K € N*, such that k! = M(u'*), k! = M(u®) 
and k is the smallest positive integer with this property. 
(it) If 


n= pi! + ph ap: (2.24) 


is the decomposition of n onto primes, then: 


S-functions of First, Second and Third Kind 95 


Ta Era Nra AA. ANT 
n pi! p? py 


2.3.7 Definition. For every n € N” the Smarandache func- 
tion of first kind is the function S, : N* —+ N* satisfying the 
conditions: 

(i) H n = u, with u = 1 or u = p, then S, (a) is the smallest 
positive integer k having the property k! = M(u'*). 

(ii) If n = p} - pi?...pt then 

Sala) = max (Ss (2)) 

Remarks. 1. The functions 5, are standardisations corre- 
sponding to equivalence relations r, defined above. If n = 1, it 
results =,, = N”, for every z € N*, and S,(n) = 1 for every 
ne N”. 

2. If n = pis a prime number then S,, is just the function S, 
defined by F. Smarandache. 

3. All the functions S, are increasing and so are of the same 
monotonicity, in the sense of definition 2.3.3. 

2.3.8 Theorem. The functions S, have the properties that 
Dy-standardise (N*,+) on (N*, <, +) by the relation: 


(21): max(S, (a), S,(6)) < Sanla +) < Sala) + Salb) 


for every a,b € N*, and also £2-standardise the structure (N*, +) 
on the structure (N°, <,-) by: 


(£2): max(S,(c), 5.(6)) < Sala +b) < Sala) - Salb) 
for every a,b € N”. 


Proof. Let p be a prime and n = p', with 1 € N”. Let also 
be a°” = Spi (a) , 6* = Sp: (b) , k = Spi (a +b). Then from the 


96 Generalisation of Smarandache Function 


definition of S,, it results that a*,b* and k are the the smallest 
positive integers satisfying the properties: 
ath M(p a, bt! = M(p*), k= M (p+) 


From k! = M(p'*) = M(p*) it results a* < k and b* < k, so 
max({a*, 5") < k and the first inequality from (£i), as from (Fa), 
is proved. 

Because 


(a* +b°)! = at !(a* + 1)...(a* + 6") = M(a7!6"!) = M(piet5)) 
it results k < a* + 6*, so (£1) is satisfied. 


If n = pit, - p}...p* , taking into account the above consider- 
ations we get: 


(2) : max(S ij (a), S iy (6)) < S ij (a+ b} < S i; (a) + Ș§ ij (b) 
Pj Pj Pj Pj Pj 
for j = 1,3 and consequently: 
max(max; Si; (a), max; Si , (b)) < max; Si slato 
< max; Si ey + max; Sy (6). 
for 1, s ,80 


max( Sala), Sn(b)) < Safa + 6) < Sala) + Salb) 


To prove the second inequality from (£2) we remember that 
(a + 6)! < (ab)! if and only if a > 1 and b > 1. Our inequality is 
satisfied for n = 1, because 


Sla + 6) = Sla) = S(b) = 1 


S-functions of First, Second and Third Kind 97 


Let now be n > 1. It results for a" = S,(a) that a’ > 1. 
Indeed, if n has the decomposition (2.24) then: 


d = 1 <> S,(a) = max S i; {a)=1 
J 


and that imphes pı = pı = ... = p, =l, son= 1. 
Consequently for every n > 1 we have 


S,(a) =a" > 1 and S,(b) =8* > 1 
Then (a* + 6*)! < (a° - 5°)! and we get: 


Sala +5) < Sala) + Sal) < Sala) - Saf) 


In the sequel we present some results on the monotonicity of 
Smarandache functions of the first kind. 

2.3.9 Proposition. For every positive integer n the Smaran- 
dache function of first land is increasing. 

Proof. If nis a prime and ky < kz from (S,(k2))! = M(n™) = 
M(n*) it results S,.(k1) < Sn(k2). 

If n is an arbitrary positive integer let 


Som (im ky) = Maxi <j<k Sp; (ijk) = Sn (ky) 
Da (teka) = MaXı<j<r Sp; (2;k2) = Sn (k3) 


From 


Spm (tmé1) < Spm (tmk2) < Spe (t4k2) 


it results S,(k,) < S,{kz) and the proposition is proved. 
2.3.10 Proposition. The sequence of functions (5,:);en- is 
monotonously increasing, for every prime number p. 
Proof. For every t1,12 E€ N* , with i < i2 and for every 
n € N” we have: 


Spi (n) = Sp(t1 - n) < S,(t2- n) = Spo (n) 


98 Generalisation of Smarandache Function 


o S,: 


pit pa and the proposition is proved. 
2.3. 


<5 
11 Proposition. Let p and q be two given primes. Then: 


p < q => S,(k) < Sík) for every k € N* 


Proof. The arbitrary integer k € N* may be written in the 
scale [p] as 


k = tya,(p) + tea,_1(p) +... + ta; (p) (2.25) 


It is said that 0 < t; < p—1 for i = 1, s and the last non-null 
digit may also be p. 

Passing from k to k + 1 im (2.25) we can make evident the 
following algonthm: 

(i) t, increases with unit. 

(ii) if t, can’t increase with unit, then t,_, increase with an 
unit and t, take the value zero. 

(iti) if neither ¢, nor t,_, cam increase with an unit then 
t, 2 crease and t, as well as t,__; becomezero. 

The processus is continued until we get the expression of k+1. 

Noting 


Ax(Sp) = Sp(k + 1) — Sp(k) (2.26) 


the increment of function S, when we pass from k tok + 1, 
following the above algorithm one obtain: 

- if (t) holds then A;,(S,) = p, 

- if (12) holds then A,(5,) = 0, 

- if (tit) holds then A,(S,) = 0. 


and it results 
Sp{n) =) Ax(Sp) + Sp(1) 
k=l 


Analogously: 


S-functions of First, Second and Third Kind 99 


S,(n) =) Alsa) + S,(1) 


Taking into account that S,(1) = p < q = S,({1) and using 
the algorithm mentioned above it results that the number of 
increments of value zero of the function S, is greatest than the 
number of increments of value zero for the function S,, and the 
increments with value p of S, are smaller than the increments of 


value q of S}. So: 


E A3) + 30) <E AlS) + S_(1) (2.27) 
k=1 


k=1 


and then S,(n) < S,(n) for every n € N*. 
Example. The values of S, and S, are listed bellow. 


k 1234 5 6 7 8 9 10 
increment 2022 0 02 2 9 
So(k) 2446 8 8 8 10 12 12 
increment 330 3 3 3 0 3 3 
S3(k) 3 6 9 9 12 15 18 18 21 24 
k 11 12 13 14 15 16 17 18 19 20 


increment 2 2 0 0 0 2 0 2 2 2 
Salk) 14 16 16 16 16 18 18 20 22 24 
increment 3 0 0 3 3 3 0 3 3 3 
53(k) 27 27 27 30 33 36 36 39 42 45 


and one observe that S2(%) < 53(&), for k = T, 20. 
Remark. For every increasing sequence 


DiS Do Se as 


100 Generaltsation of Smarandache Function 
of prime numbers it results: 


IEE Op. Oy <i pe Sie 


and if n = (p; - pz...p:)' with p) < pz <... < pz, then 
Salk) = max Sp; (E) = Spi (E) = Sp (i) 
2.3.12 Proposītion. If p and q are prime numbers and 
p:i<q, then Spi < S4- 
Proof. From p- 2 < g it results: 
Spi (1) < pi < g = S4(1) and Sp: (k) = S,(tk) < iSp(k) (2.28) 
Passing from k to k + 1, from (2.28) one deduce: 


Arl Spi) < Ak(Sp) (2.29) 


The proposition (2.311) and the equality (2.29) imply that 
passing from k to k + 1 we get: 


Ak( Spi) < AlS) Si-p <q, 3 Ax(Sp) aS A;(Sq) (2.30) 


Because we have 


S,3(n) = Sy (V+ > AnSe) < Se (D423 Au(S) 
and 


San) = S((a)+ E AlS) 


S-functions of First, Second and Third Kind 101 


from (2.28) and (2.30) it results Sp: (2) < S,{n) for every n € N*, 
and the property is proved. 

2.3.13 Proposition. If p is a prime number, then S, < S, 
for every n < p. 

Proof. If nis a pnme, from n < p and the proposition (2.3.11) 
it results S,{k) < Sp(k) for every k € N*. If 


— atl i3 i 
n= pi: p--pPr 


is a composit number then: 
Salk) = max Sy (k) = Spi (k) 


and from n < p it results pt” < p. So, using the preceding 


proposition and the inequality p, < p!” < p, one obtain 


Spir (k) < S,(k) 
That is S,,(k) < S,(k) for every k € N”. 


We shall present now the Smarandache function of second 
kind, defined in [2]. 

2.3.14 Definition. The Smarandache functions of second 
land are the functions 


S*:N* — N" , defined by S*(n) = S,(k) 


for every fixed k € N*, where S, is a Smarandache function of 
first kind. 

From this definition it results that for k = 1, S* is just the 
function S. Indeed, for n > 1 we have 


Sn) = S,(1)= max S i (1) = max S,, (t;) = S(n) 


102 Generalisation of Smarandache Function 


2.3.15 Theorem. Every Smarandache functions of second 
kind ©5-standardise the structure (N*,-) on the structure (N°7, < 
, +) by: 


(Zs): max(S*(a), S*(6)) < Fla- 8) < (a) + S*(6) 


for every a,b € N*. At the same time these functions L4- 
standardise the structure (N*,-) on (N°, <,-) by: 


E): max(S* (a), S*(b)) < S#(a-b) < S (a) - S*(d) 


for every a,b € N*. 
Proof. The equivalence relation r* corresponding to S* is 
defined by: 


ark bh ==> (3ja* € N” a! = M(a*), d! = M(b*) (2.31) 


and a” is the smallest positive integer satisfying (2.31). Conse- 
quently we may say that S* is a standardisation attached to the 
equivalence relation r*. 

Let us observe that the Smarandache functions of second kind 
are not of the same monotonicity, because, for instance, S? (a) < 
S?(b) <=> S(a*) < S(b?) and from this it does not result S"(a) < 
S1(6). 

For every a,b € N” let us note a* = S*{a), 6* = S*(b), 
c* = S*(a-6). Then a*, b”, c” are the smallest positive integers 
with the properties: 


at! = M{a*), ot! = M(b*), ct! = M(a*- 6*) 


and so c*! = M(a*) = M(b*). It results a” < c*, b* < c*, and 
then max{a*, b*) < c*. That is: 


S-functions of First, Second and Third Kind 103 


max{S*(a), S*(b)) < S*(a- b) (2.32) 
But from (a* + 5°)! = M(a*!b*!) = M(a*5*), it results c* < 
a* + 6*, s0 
S*(a +b) < S*(a) + S*(b) (2.33) 
From (2.32) and (2.33) one obtain: 


max(5*(a), S*(6)) < S*(a) + S*(6) 
so (13) is verified. 
Finaly, because (a*b*)! = M(a*!b"!), we have also: 
S*(a-b) < S*(a) - S(b) 
and (X4) is proved. 
2.3.16 Proposition. For every k,n € N* we have 
S¥(n) <n-k (2.34) 


Proof. Let us consider n = pt - p?...pit and 


S(n) = max(5p; (i) = S(p) 


Then because 
S¥(n) = S(n¥) = maxı<j<t Sp, (1; k) = S(p F) < kS(pir) < 
< Spin) = £5(n) 


and S(n) < n, it results (2.34). 
2.3.17 Theorem. Every prime number p > 5 is a local 
maximum for the functions S*, and 


S*(p) = plk — i(k) 


104 Generalisation of Smarandache Function 


where t, are the functions defined by the equality (1.33). 
Proof. If p > 5is a prime, the first part of the theorem results 
from the inequalities 


Sp-1(k}) < Sp(k) and Sp41{%) < Sp(x) 
satisfied by the Smarandache function of first kind. 


The second part of the theorem results from the definition of 
functions S*: 


S*(p) = S,(k) = p(k — i(k)) 

and the theorem is proved. 

Remark. For p > k we have S*(p) = pk. 

2.3.18 Theorem. All the numbers kp, with p a prime and 
p > k are fixed points for the function S*. 

Proof. Let m = pf" -p3?...p¢t be the decomposition of a given 
m into primes and p > 4 be a prime number. Then p;-a; < p% < 
p for i = 1,t, so we have 


S*(mp) = S((mp)*) = max(Sp; (ai), Sp(k)) = Sp(k) = kp 

For m = k it results S*({kp) = kp, so kp is a fixed point for 
SE. 

2.3.19 Theorem. The Smarandache function of second kind 
has the properties: 


(i) S*(n} = of{nit+*) for every e > 0 
(12) „Dm sup ee) =k 
Proof. We have 


< lim ksin) _ on Sina) _ 5 








S-functions of First, Second and Third Kind 105 


and (2) is proved. 
Also, 
k k k 
wn akana ea a 
nh CO n n n 


——t OD Rm SO Pn 


where (pa)new* 18 the increasing sequence of all the primes. 
2.3.20 Theorem. The Smarandache functions of second 
kind are generaly increasing, in the sense that 


(Y) n € N* (3) mo € N* (Y) m >=> S*(m) > S*(n) 


Proof. It is said [44] that the Smarandache function is gen- 
erally increasing, in the following sense 


(Y) t € N* (3) ro E€ N” (Y) r > ro => S(r) > S(t) (2.35) 


Let t = n* and ro be such that S(r) > S{n*), for every r > ro. 
Let also mo = [2/ro] + 1. Of course, mo > Yro => mk > ro, 
and m > mo <=> mf > mk. 

From m* > mk > ro, it results S(m*) > S(n*), so S*(m) > 
SF(n). 


Then we have: 


(Y) n € N° (3) mo =[Yro] +1 (Y) m > mo => S*(m) > S*(n) 


where ro = ro(n*) is given by (2.35). 

2.3.21 Theorem. H p > max{3,k) is any prime number, 
then n = p! is a local minimum for S*. 

Proof. Let p! = pi! . pi?...pim - p the factonsation of p!, such 
that 2 = pı < po <..., Pm < p. Because p! is divisible by p7 , it 
results S(p ) < p = S(p) for every j = 1, m. 


106 Generalisation of Smarandache Function 


Of course, 


S*(p!) = S((p!)*) = max (S; *), S) 
and l , 

S(pj 4) < S(p¥) < kS(p) = kp = S(p*) 
for k < p. Consequently, 


S*(p!) = S(p*) = kp fork <p (2.36) 
If the decomposition of p! — 1 into primes is 


pi~1= gp - ag 
then we have q; > p for j = It. 
It results: 


Pa = ij — im 
S! ~ 1) = max (S47 )) = Slax 
with qm > p, and because S(gim ) > S(p) = S(p!) it also results 


S(p! — 1) > S(p!) 


Analogously it can be proved that S(p!) + 1 > S(p!). 
Of course, 


S*(p! — 1) = S((p! — 1) > S(t”) > S(qk) > S(p*) = kp 
(2.37) 
and 


S*(p! +1) = S((p! + 1)*) > kp (2.38) 


From (2.36), (2.37) and (2.38) it results the assertion. 
Now we present the Smarandache function of third kind [2]. 
Let us consider two sequences: 


S-functions of First, Second and Third Kind 107 


Cay Taa Gas lap 
(b): 1 = bi, bg,..., day. 


satisfying the properties: 


Bin = Gh Onr Oka Oke On (2.39) - 


Of course there exist infinitely many such sequences, because 
chosing an arbitrary value for az, the next terms of the sequence 
(a) are determined by the recurrence relation (2.39). 

Let now be the function 


fÈ: N* —+ N* defined by fèln) = Sa, (dn) 


where Sa, is the Smarandache function of first kind. 
One observe easily that . 


(i) -if an =1, and b, = n for every n E N", then f= Sı 
(ii): ifa, =n and b, = 1 for every n € N”, then fè =S 
(2.40) 
2.3.22 Definition. The Smarandache functions of third 
kind are the functions defined by any sequences (a) and (b), 
different from those of (2.40), such that: 


S= fa 


2.3.23 Theorem. All function f’, Es - standardise the 
structure (N*,-) on the structure (N*, <, +,-) by: 


(Zs): max(f,°(k), f(r) < Fèl n) < bafa (E) + bn fa (n) 


Proof. Let us note 


108 Generalisation of Smarandache Function 


falk) = Say (a) =F, FPO) = Salba) = n°, 
f,°(nk) = ay (Doce) =f" 


Then $°, n* and ¢* are the smallest positive integers for which 


k*! = Mat), n’! = M (abr) , t= M (a*r? ) = M ((apan)® ‘bn ) 
so 


max(k*, n*) < ¢* (2.41) 
Moreover, because (d,-n*)! = M((n*!)°+), (bn:k*)! = M((k*!)% ) 


and 


(bn n + bn RY) = M( (bn n)a -*)!) = 
= M{((n*!)'s ; (k*!)en ) = M{(abn)>e i (ae jén) = M ((a; ' an Pe n ) 
it results 
E Sbn- kh + bn” (2.42) 
From (2.41) and (2.42) one obtain: 


max{k*,n*) <t* < bn- kt +5,-n” (2.43) 


From the last inequality it results (Ds), so any Smarandache 
function of third kind satisfies: 


(Ze): max( Sg (k), 5. (n)) < S (kn) < ba S E(k) + de S Hn) 


for every k,n € N°. 


Connetions with Fibonacci Sequence 109 


Example. If the sequences (a) and (b) are determined by 
the condition a, = bn = n, for n € N”, then the Smarandache 
function of third kind is: 

S2: N* — N*, S2(n) = S,(n) 
and (6) becomes: 


max(5;(k), Sn(n)) < Sh.n(k- nr) < nS,(k) + kS,(n) 


for every n € N*. This relation is equivalent with the following 
relation, written using the Smarandache function: 


max(S(k*), 3(n*)) < S((kn)M) < nS(kF) + kS(n") 


2.4 Connections with 
Fibonacci Sequence 


In the Introduction of the Proceedings of the Conferences ”Ap- 
plications of Fibonacci numbers ” [3], [36], [38], it is mentioned 
that the sequence: 


1, 1, 2, 3, 5, 8, 13, 21, 55, 89,......... (2.44) 


known as the Fibonacci sequence, was named by the nineteenth- 
century French mathematician Edouard Lucas, after Leonard Fi- 
bonacci of Pisa, one of the best mathematicians of the Middle 
Ages, who referred to him im this book Liber Abaci (1202) in 


connection with his rabbit problem. 


110 Generalisation of Smarandache Function 


TheGerman astronomer Johann Kepler rediscovered Fibonacci 
numbers, independently, and since then several renowned math- 
ematicians, as J. Binet, B. Lamé and E. Cartan, have dealt with 
them. 

Edouard Lucas studied Fibonacci numbers extensively, and 
the simple generalisation: 


2:134 7-11 18,2947 762125, a, (2.45) 


bears his name. 
It said that there exists a strong connection between the Fi- 
bonacci sequence and the gold number: 


1+ 75 


2 


For instance noting by F(n) the n — th term of Fibonacci 
sequence (2.44) one has: 


$ = 





F(n+1) 7 
ji Fo o } (2.46) 


and so, 


„üm VF(n) =¢ 


Let us now remember some of the properties of Fibonacci 
sequence. 

It is said that Fibonacci sequence satisfies the recurrence re- 
lation 


Fin+2)= F(n+1)+F(n), with F(1) = F(2)=1 (2.47) 


and also the properties: 


Connetions with Fibonacci Sequence 111 


(a) F(n) = KE - A 

(p2) F) + F(2) +... + F(n) = F(n+2)-1 

(p3) FO) + F(3) +... + F(2n — 1) = F(2n) 

(p) F(2)+ F(4) +... + F(2n) = F(Qn4+1)-1 

(os) F(2) — F(3) + F(4) —... + (-1)"F(n) = (-1)"F(n — 1) 
(ps) F°(1) + F°(2) +... + P(n) = F(n)- Fin +1) 

(yr) F(n)- F(n +2) = Fn + 1) + (-1)"*? 

(ys) F(2n) = F? (n) + F?(n — 1) 

(p) F(2n+1) = F?(n) + F?{n+)) 

(gio(F(n — 1)- F(n +1) — F?(n) = (-1)" 

(gu) F(n—2)- F(n+2)— F (n) = (-1)"*? 

(p12) F(n—1)- F(n+1)— F?(n — 2)- F(n +2) = 2(-1)" 


T. Yan [50] has posed first a problem concerning a connec- 
tion between Fibonacci sequence and the Smarandache function. 
Namely, for whath triplets (n — 2,n — 1,n) of positive integers 
the Smarandache function venfies a Fibonacci-hke equalhty: 


S(n — 2) + S(n — 1) = Sn) (2.48) 


Calculating the values of S(n) for the first 1200 positive inte- 
gers he found twosuch triplets, namely (9, 10, 11) and (119, 120, 121). 
Indeed, we have: 


S(9) + S(10) = S(11), and S$(119) + $(120) = $(121) 
More recently H. Ibstedt [26] showed that the following num- 


bers generating such tnplets are: 


n = 4,902; n = 26, 245; n = 32, 112;n = 64, 010; 
n = 368, 140; n = 415, 664 


112 Gceneralisation of Smarandache Function 


and proved the existence of infinitely many positive imtegers sat- 
isfying the equality (2.48). 

Indeed, excepting the triplet generated by n = 26, 245 the 
other triplets (S(n— 2), Sín — 1), S(n)) satisfy the property that 
oue of terms is the duble of a prime number, and the other two 
are prime numbers. For instance taking n = 4902 = 2. 3. 19-43 
we haven—J] = 490] = 137-29. n—2 = 4900 = 22.52.72 and the 
equality (2.48) becomes 2-7 + 29 = 43. Also, for n = 32,112 = 
2*- 3-223 it results n— 1 = 32, 111 = 163-198, n— 2 = 32, 110= 
2.3-137-19, so (2.48) oe 2. 13 + 197 = 223. 

Using this remark, H. Ibstedt proposed [26] the folowing 
algorithm: 

Let us consider the triplets (n — 2,n — 1, n) satisfying the 
relations: 


n=zr: p", with a< pand S{z) <a-p (2.49) 
m-l=y-g’, with b< gand Sly) < bq (2.50) 
m-2oz-r°%, with c< rand S(z) <e-r (2.51) 


where p, g, 7 are prime numbers. In these conditions it results: 


S(n) =a- p, S{n—1)=b-q, S(n-2)=c-r 


Substracting (2.50) from (2.49), and (2.51) from (2.50) we 
get the system: 


zept—y-P=1 (2.52) 


y:g—z-r=l (2.53) 


Connetions with Fibonacci Sequence 113 


a-p=b-gte-r (2.54) 


Every solution of the equation (2.54) generate an infinity of 
solutions for (2.53) which may be written under the form 


r=r +É- t, y=yo-—p*-t (2.55) 


where ¢ is an integer parameter and (zo, yo) is a particular 
solution (such a solution may be found by means of the algonthm 
of Euclid). 

The solutions (2.55) are then introduced in the equality 


ga V2 
re 
for obtaining integer values of z. 
H. Ibstedt in [26] give a very large list of triplets (n — 2,n— 
1, n) for which (2.48) is verified. These solutions have been gen- 
erated for 


(a,b,c) = (2,1, 1), (a,b,c) = 4, 2,1) and (a,6,c) = (1,1, 2) 


with the parameter t restrained only to the interval —9 < ¢ < 10. 
To make now in evidence an other connection between the 
Smarandachr function and Fibonacci sequence we return to the 
twoo latticeal structures defined on the set N* of positive inte- 
gers. 
We have already seen that the Smarandache function etab- 
lishe a connection of these lattices by the equality: 


Sn v ng) = S(n,) V S(m) 


and so we are conducted to consider S : Nz —3 N,. 
2.4.1 Definition. The sequence o : M, — Nz is said to be 


multiphcatively convergent to zero (m.c.z) if: 


114 Generalisation of Smarandache Function 


(V) n € N? (3) mn E N? (V) m > ma => o(m) >n (2.56) 


In [10] a sequence o : N — N satisfying (2.56) is named 
multiplicatively convergent to infinity. We prefered the above 
definition which is in connection with the fact that zero is the 
last element in the lattice Ni. 

The {m.c.z) sequences having also the property of monotonic- 
ity are used in [10] to obtain a generalisation of p— adic numbers. 

The set Z, of p — adic numbers may be considered as an 
inverse limit (see [10] ) of the rings En = Z/p"Z of integers 
"modulo p"”, where p is a prime number. 

Considering, mstead of the sequence (p*)nen an arbitrary 
(m.c.z) and monotonous sequence (c(n))nen there are obtained 
the sets En = Z/a(n)Z whose inverse limit is a generalisation of 
p — adic numbers. 

Let us observe that the monotonicity for a sequence c : 
AN, — Nj is expressed by the condition 


(moa) n < m= > a(n) = alm) 


The sequence o{n) = n!is a {m.c.z)}sequence and for every 
fixed n € N* the smallest m, given by (2.56) is exactly the value 
S(n) of the Smarandache function. So, we can pose the problem 
of generalisation of Smarandache function in the following sense: 

To each (m.c.z) sequence o : No —> Ng one may attach a 
function 


fo: N* — N*, f.(n) = the smallest m, given by (2.56) 


and we observe that if n = p{'- p3?...p¢t is the decomposition 
of n € N* into primes then: 


Connetions with Fibonacci Sequence 115 


feln) = max f,(p;*) (2.57) 


This formula generalise the formula (1.16) of the calculus 
of S{n). But the efective calculus of f,(p%') depends on the 
particular expression of the sequence vg. 

We have also the properties: 


(fi) felm V m) = f(m) V folna) 
(fo) mi = na => feln) < folm) 


which entitle us to consider 


fo : Na — No 


Now, we may also consider the sequence 


Soc: N, _ N, 
or, more general, if c and @ are two (m.c.z) sequences, then there 
exist the sequences: 
fz08: No — Nn, hoo: M — N, (2.58) 
8o fp: Nia — Nz go fg: Nia — Ng ` 


2.4.2 Proposition. If the sequences o,9 : Mo — Ni 
are monotonous, then the sequences defined by (2.58) are also 
monotonous, in M, and N; respectively. 

Proof. For an arbitrary n € N* one has @(n) = O(n + 1) and 


f- satisfies (f2), so: 


(fo o8)(n) = f.(A(n)) < f-(A(n + 1)) = (fe 0 A)(n +1) 
For the second kind of sequences let n S na. Then folni) < 


f,(ng) and so 


116 Generalisation of Smarandache Function 


(8.0 e)ra) = (falra) $ folna) = (6 fo)(r2) 


The two latticeal structures considered on N* justify the con- 
sideration of the following kind of sequences: 


(ti)  (o,0)sequences: aa: No — N, 
(22) (o, d)sequences: Soa: No — Niu: 
(iit)  (d,o)sequences: C4: Na — N, 
(iv) (d, d)sequences: cag: Ni — Nu: 
For each of these sequences one may adapt the definition of 
monotonicity and of the limit. We have so the following situa- 
tions: 


1) For an (0,0) sequence coo the condition of monotonicity 
is: 


(moo) (Y) Ni, Na E€ N”, rey < nm => Toolni) < Tool na) 


an this sequence tends to infinity if: 


(coo) (Y) n € N* (3) m, € N* (Y) m > mn => Colm) > n 


2) The (0, d) sequence coa is monotonous if: 


(moa) (Y) mi, na E N*, ni < ny => acalm) $ Toa{N2) 


Connetions with Fibonacci Sequence 117 


and it is (multiplicatively) convergent to zero if 


(Cod) (V) n € N” (3) ma E N° (Y) m2 ma = calm) > n 


3) If og, is a (d, o) sequence, it is monotonous if 


(mao) (Y) nnz E N* ny = na ==> F4o(1) L Tao(n2) 


and tends to infinity if 


(cdo) (V) n E N* (4) ma E N? (V) m > mn = oam) > n 


From the properties of the Smarandache function it results 
that the sequence (S(n))ew- is a (d, o) sequence, satisfying the 
conditions (mdo) and (cao). 

4) The condition of monotonicity for a (d, d) sequence cq is 


(Maa) (Y) 1, M2 € N” m $ n2 => Calm) $ Taal na) 


N. Jensen in [5] named divisibility sequence a sequence satis- 
fying the condition (maa). This concept has been introduced by 
M. Ward [51], [52]. 

Moreover, the sequence cda is said to be strong divisibility 
sequence ( shortly (sds), see [5] pg. 181) if the equality 


Taalma A na) = oaa(r) A Taal na) (2.59) 


holds for every nı, na E N”. 

The term of (sds) has been used first in [28]. It is easely to see 
that if a sequence is (sds) then it is also a divisibility sequence 
(shortly, (ds)). 


118 Generalisation of Smarandache Function 


It is proved [12] that the Fibonacci sequence is (sds). 
On the sequence oz; we shall say that it is (multiplicatively) 
convergent to zero if: 


(ca) (V)n € N* (3) m, EN’ (V) m2 m, => aam) > 


To each sequence o;;, with i,j € {o, d}, satisfying the con- 
ditions (m,;) and (c;;) we may attach a sequence f;; defined 
by: 


fij(n) = min{m, / m, is defined by (c; ;)} (2.60) 


2.4.3 Proposition. Each function f,, defined by (2.60) bas 
the properties: 


(:) foo satisfies the condition (meo) of monotonicity 
(i2) foolna v no) = fool) Vv foo{n2) 
(222) footy A na) = foo{m1) A foo{n2) 


Proof. (1) We have: 


folna) = min{ Mny / (v) MN Z Mn => Too(™) > n} 
foo(na) = min{ m,, / (Y) M > Mm => Foo(m) > na} 


so, for every m> f..(nz) it results: colm) > mi > na. 
The assertions (72) and (iči) are consequences of (2). 
2.4.4 Proposition. Each function f,4 has the properties: 


(:v) foa satisfies the condition (mea) of monotonicity 


(u) falm V na) > falm) V foa(rea) 
(vi) foan A na) < foala) A foa(na) 


Connetions with Fibonacci Sequence 119 


Proof. (iv) Let be n, S na. Then from 


foa(ri) = min{ Mn; / (V) M Z Mn, => Toalm) > ni for: = 1,2 


it results colm) 2 na 2 ny, for m > falna). So, foa{ri) < 
foa{n2). 
The properties (v) and (vz) result from (iv). 
2.4.5 Proposition. Every function fz, has the properties: 
(vit) is (only) (0,0) monotonous 
oe a 
(viii) falm V ma) $ falm) Ý fiolm) 
(iz) fao(ra Ana) 2 fao(ra) A fao(na) 
Proof. (vii) If ny < na then for every m > m,, we have 
Taom) > m > m, and so fa,(m) < fao{n2). 


(vin) For : = 1,2 one has: 
far) = mia mn; | (Y) m > ma, => ouslrn) > r 


Let us suppose nj < nz, 80 m, V na = nz and fa.{ny V m) = 
fao(n2). Then if we note 


d 
mo = fao(m) V fao(ne2) 
for m > mpg it results og,(m) > n,, for i = 1,2, so ag,{m) > 
d 
ni V m and so 


fol V ma) = faala) $ faolm) Ý falna) 


Consequences. From (vii) it result the following properties: 


120 Generalisation of Smarandache Function 


faolny V n2) = faolri) V fao(na) 
fao(m1 A na) = fao(mi) A fao{ne) 


and so: 


faolm) A faoln2) < fao(r1) A fao(n2) = faon Ara) < 
< fao(m) V faoa) = fao(ri V n2} < faon) v fao(ra) 


2.4.6 Proposition. The functions fz, satisfie: 


(2) falma Vm) < fulm) Ý falma) 
(z) H ni s Nz OF na Sn, then fag{ny V n2) = 


= faln) V Bie) 
(zi) faa(nr A na) < faala) A fear) 


Proof. It is analogous with the proof of above propositions. 
2.4.7 Theorem. H the sequence caq is (sds) and satisfies 
the condition (c44), then: 


(a) fulna Ý ra) = fulm) Ý fulna) 
(b) m < no => falna) S faalna) 


Proof. (a) It is sufficient to prove the inequality 


faln) S$ faa{ry ý n2) for 1=1,2 (2.61) 


If, for instance, this inequality does not hold for nı, it results: 


fulma) A fulm V m) = do < fulm) 


and we have 


Connetions with Fibonacci Sequence 121 


aualdo) = osal faala) A faa(m V m)) = 


= ox faam)) A oad falm Ý m) > m Arg =m 
d d 


d d 
because oal faa{m)) 2 nı and ni 5 nı V na < caalfaa{ni V 


n2)). So, one obtain the contradiction fza(mi) < do < faa(n1). 
(6) This condition is the (d, d) monotonicity. If nı < n then 
d 


d ‘ ; 
na = nı V m, and using the property (a) it results: 


fara) = faam Ý m) = fulma) Ý faa(n2) 
so faslni) = falna). 


Remarks. 1) Even if c44 is (sds), does not result the sur- 
jectivity of fag, in general. Indeed, the function fy, attached 
to Fibonacci sequence is not surjective, because, for mstance, 
fu (2) = 9. We also remember that the Smarandache function is 
the function fod corresponding to the (0, d} sequence opa(n) = nl, 
and it is surjective. 

2) One of the most interesting diophantine equations associ- 
ated to a function f,;, for 1,7 € {1,2}, is that giving its fixed 
points: 


frer (2.62) 
The function f;; attached to Fibonacci sequence has n = 5 
and n = 12 as fixed points, but tbe problem of finding the gen- 
eral solution of the equation (2.62) corresponding to this famous 
sequence is an open problem,until now. 
In the section 1.6 there has been studied the convergence of 
some numerical series involving the Smarandache function. Such 
kind of series may be attached to all (generalised) sequences f; ;. 


122 Generalisation of Smarandache Function 


In the sequel we focus the attention on the analogous of the 
series 


1 æ ) 
Steyr ond è 5(k)* - /S(k)! 


im the case when the function S is replaced by an arbitrary func- 
tion fdo, corresponding to a (m.c.z) sequence. 

2.4.8 Theorem. If o is a (m.c.z) sequence satisfying the 
condition (moa), let us denote by f, the corresponding fod se- 
quence and by g, the sequence c o f,. Then for every a > 1 the 
series 





EMs 


Od (fo AT (a) D 


are convergent. 
Proof. To prove these assertions we use the same method as 
for the series (1.90) and (1.91). 
(:) We have: 
Yo ae = M 
1 (fol(k))*- Vgolk) kai trV/o(t) 
where m; = card{k / f,(k) =t}. But 


k . a(t) => m < d(o(t}) 


where d(n} is the number of divisors of n. 
From the inequality d(c({t)}) < 2\/o(t) it results 








oo mM © ? a(t) o 1 
=? a 
2 t*,/o(t) <È ta /o(t) 2 ta 


(i1) If we note a(n + 1)/a{n) = kn+1, it results successively: 


Connetions with Fibonacct Sequence 123 








tl gol(k) mE alt) t=1 \/o(t) 


and putting z; = 1/c(t), it results 2,434 /2, = 1/ Ake 41. 
As m = 0 f & = 1, it results that when m; Æ 0 we have 
k; > 1, so the senes D (1/,/o(t)) is convergent, as well as the 
t=1 
series (72). 
Example.Let the sequence a be defined in the following way: 
a(t) =k! if and only if k! < t < (k+4+1)!. 
It results that c is a (m.c.z) sequence satisfying the condition 
(Mod) and we have: 


o(1) = 1, c(2) = 2!, (3) = o(4) = 3!, 0 (5) =... = o (10) = 4! 
all) =o(2) =.= (26) = 5! 
Then 
f.(1) = 1, fe{2) = 2, fo(3) = 3, f(A) = 5, fo(5) = 11, 
fo(6) = 3, fo(7) = 71, f-98) == 5, ... 


and so 





es Oe: See Oe PN a a a 
Lame aM tag tay tat ant 


Ee + EEEN eee E2 =A 
From the fact that 


ma = 0, me = my =... = Mio = 0, miz = mis = ...m = 


it results: 


124 Generalisation of Smarandache Function 


t=1 a(t) 2) o({3) o{5) 11 T x21) OF rosea: = 
ie oe ae a ee ae a les S 
T OPETE- SE 
SLG LET 


which is a convergent series. 
Remark. As one can see from the above example, the func- 
tions f, are, in general, neither one-to-one, nor onto. 


d 


2.5 Solved and Unsolved Problems 


As in the section 1.8 we note by a star (*) the unsolved problems. 
By pı < pa < ... < px... is denoted the increasing sequence of all 
the prime numbers. For the solutions of solved problems see the 
collection of Smarandache Function Journal. 

1) Prove that the Smarandache function does not verify the 
Liepschitz condition 


(3) M>0(¥) mneN* = /S(m) — S(n)/ < M/m—n/ 
2) The functions SC!) and St?) defined by: 


SH (n) = a ; Sn) = HO) 


ce y” 
verify the Liepschitz condition, but the function SO(n) = 5a) 
does not verify this condition. (M. Popescu. P. Popescu) 

3) If 


Solved and Unsolved Problems 125 


o5{z) = S(d), and 


T(n) = 1—Ings(n)+ 3 D FEN 


t=1lk=1 
then im,—... T(n) = —oo. 
4) If i(z) = card{p / p is a prime, p < z}, prove that the 
following numerical functions: 


O(2) 
(i) Fs : N* —> N, Fs(z) =u S(p?), 
(it) 6: N* — N, Oz) =¥ S(p?), 
pst 
(iii) @:N* — N, 6(z) = 2 S(p?) 
Piz 
pi not divides z 


which involve the Smarandache function, do not verify the Liep- 
schitz condition. (M. Popescu,P. Popescu, V. Seleacu) 
5) Let a: N* — N" be the function defined by: 


a(n} = k <=> k is the smallest positive integer such that 
nk is a perfect square. 


Prove that: (1) If n has the factorsation n = gf" - q%?...q2", 
then a(n) = qf! - qË ...q8*, with 
pas 1 if a; is odd number 
‘1 0 if a; is even number 


(zt) The function a is multiplicative, that is a(zy) = a(z)a(y) 
for all z, y E€ N° such that z A y= 1. 


(iii) The series D “ diverges.(J. Balacenotu, M. Popescu, 
n>l 
V. Seleacu) 


126 Generalisation of Smarandache Function 


6) For the function a defined in the preceding problem prove 
that: (i) if z,y > 1 are not perfect squares and z A y = 1, then 


the diophantine equation a(z) = a(y) has no solution. 

(22) a(zy?) = afz), for z,y > 1. 

(iti) a(z”) = 1 if n is even and a(z”) = a(z) if n is odd. 

(tv) for every perfect square m € N* the equation za(z) = m 
has 2* different solutions, where k is the number of prime factors 
of m. 

(v) solve the equations: 


sae a ded 


za} + Falah = al) 
Aa(z) + Bafy) + Ca{z) = 0, Aa(z) + Bafy) =C 


(I. Balacenoiu, M. Popescu, V. Seleacu) 

7) For the same function a defined above prove that if F? 
denote the generating function associated to this function by 
means of the lattice Ma, then: 


~ pape, | ¥(¢+1)=1 if wis even 
(2) Fila =f aa if œ is odd 


(i) File) =i (H(@)(q+ 1) + BG 


where n = gj’ -g”...g¢" is the decomposition of n into primes and 
H(a) = card{z / z < a, z is odd}. (I. Balacenoiu, M. Popescu, 
V. Seleacu) 

8) The Smarandache no-square digits sequence is defined as 
follows: 2, 3, 5, 6, 7, 8, 2, 3, 5, 6, 7, 8, 2, 2, 22, 23, 2, 25, 26, 27, 
28, 2, 3, 3, 32, 33, 3, 35, 36, 37, 38, ... (take out all square digits 
of n). It is any number that occurs infinitely many time in this 
sequence ? 

9*) Let n be a positive integer with not all digits the same, 
and let n’ its digital reverse. Then let ny = /n — n'/, and ni its 


Solved and Unsolved Problems 127 


digital reverse. Again, let no = /n, — ni /, and n} be ita digital 
reverse. After a finite number of steps one finds an n; which 
is equal to a previous n;, therefore the sequence is periodical 
(because if n has, say, k digits, all other integers n; following it 
will have & digits or less, hence their number is limited and one 
apphes the Dinchlet’s box principle). 

Find the length of the period (with its corresponding num- 
bers) and the length of the sequence’till the first repetition oc- 
curs for the integers of three digits and the integers of four digits. 
Generahsation. (M. R. Popov) 

10) Let c : N — N be a second order recurrence sequence, 


defined by: 


a(n) = Aa(n — 1) + Bo(n — 2) 


where A and B are fixed non-zero coprime integers and o{1) = 
1,0(2) = A. We shall denote the roots of the characteristic 
polynomial 


P(z) = 27+ Az + B 
by œ and 8. Prove that: 
(7) if the sequence is non-degenerate (that is AB £ 0, A? + 
4B # 0 and is not a root of unity) then the terms a(n) can be 
expressed as: 


a” — p” 
a— Pp 
for all n € N*, and if pis a prime such thatpA B= 1 then there 


are terms in the sequence o divisible by p. (The least positive 
index of these terms ts called the rank of appantion of p in the 
sequence and it is denoted by r({p). Thus r(p) =n if p = a(n) 


holds, but p < (n + 1) does not hold). 
d 





oín) = 


128 The Smarandache Function 


(2z} there is no term of the sequence a, divisible by the prime 
pif p divide B and AA B= 1. 

(iii) if p does not divides B and we note: D = A?+4B and 
(D/p) = the Legendre symbol, with (D/p) = 0if p $ D, then 


1) (p) < (p — (D/p)) 
2) p $ a(n) > rp) Sn 


11*) Find a formula for the calculus of Smarandache gener- 
ahsed function f, corresponding to Fibonacci sequence. 


Bibliography 


[1] 


T 


M. Andrei, C. Dumitrescu, V. Seleacu,L. Tutescu, St. Zan- 
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dache Function Journal, V. 4-5, No. 1, (1994), 1-5). 


I. Balacenoiu: Smarandache numerical functions (Smaran- 
dache Function Journal, V. 4-5, No. 1, (1994), 6-18). 


E. Burton: On some series involving the Smarandache func- 
tion (Smarandache Function Journal, V. 6, No. 1, (1995), 
19-15). 


I. Balacenoiu, V. Seleacu: Some properties of Smarandache 
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{1 
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The function named in the title of this book is originated from 
the exiled Romanian mathematician Florentin Smarandache, who 
has significant contributions not only in mathematics, but also in 
hterature. He is the father of The Paradoztst Literary Movement 
and is the author of many stories, novels, dramas, poems. 

The Smarandache function,say S, is a numerical function de- 
fined such that for every positive integer n, its image S(n) is the 
smallest positive integer whose factorial is divisible by n. 

The results already obtained on this function contain some 


THE AUTHORS 


$ 17.95