Skip to main content
Internet Archive's 25th Anniversary Logo

Full text of "Some problems on Smarandache Function"

See other formats


SOME PROBLEMS ON SMARANDACHE FUNCTION 


by 


Charles Ashbacher 


In this paper we shall investigate some aspects involving 


Smarandache function, S:N'-->N, s(n) = min {m | n divide m!}. 


1. THE MINIMUM OF S(n)/n 
Which is minimumum of S(n)/n if n > 1? 
1.1. THEOREM: 
a) S(n)/n has no minimum for n > 1. 
_b) lim S(n)/n as n goes to infinity does not exist. 
Proof: 

a) Since S(n) > 1 for nol it follows that S(n)/n > 0. Assume 
that S(n)/n has a minimum and let the rational fraction be 
represented by r/s. By the infinitude of the natural numbers, we 
can find a number m such 2/m < r/s. Using the infinitude of the 
primes, we can find a prime number p > m. Therefore, we have the 


sequence 


27D < 27m L T/S 


21 


We have S(p-p) = S(p) = 2p. It is known that S(p:p)s2p. The 


ratio of S(p°)/(p-p) is then 


2p/ (p?) = 2/p 


And this ratio is less than r/s, contradicting the assumption 
of the minimum. 

b) Suppose lim S(n)/n exists and has value r. Now choose, e > 0 
and e < 1/p where p is a twenty digit prime. Since Sip) = p, 
S(p)/p = 1. 

However, S(p-p) = 2p, so the ratio S(n)/n = 2p/(p-p) = 2/p. Since 


p is a twenty digit prime, 


| S(p)/p - S(p'p)/ (pp) | > e by choice of e. 


so the limit does not exist. 


2. THE DECIMAL NUMBER WHOSE DIGITS ARE THE VALUES OF SMARANDACHE 
FUNCTION IS IRRATIONAL. 
Unsolved problem number (8) in [1] is as follows: 
Is r = 0,0234537465114..., where the sequence of digits is 
S(n), n 21, an irrational number? 
The number r is indeed irrational and this claim will be 
proven below. 


The following well-known results will be used. 
DIRICHLET’S THEOREM: 


If d > 1 and a # 0 are integers that are relativey prime, then 


the arithmetic progression 


22 


A, ‘ay. +d; a+ 2d, a t 3075-2 


contains infinitely many primes. 


Proof of claim: 
Assume that r as defined above is rational. Then after some m 
digits, there must exist a series of digits tju, “Cs, Cin eae Enr 


such that 


ye 0023453746114 aabt Ott tea En 


where s is the m-th digit in the decimal expansion. 
Now, construct the repunit number consisting of 10n 1's. 
a = 11111 ... 111 
10n times 
and let d = 1000 ...`00 
10n + 1 0’s 
Since the only prime factors of d are 2 and 5, it is clear 
that a and d are relatively prime and by Dirichiet’s Theorem, the 
sequence 
a, a + d, a+ 2d, ... 
must contain primes. Given the number of 1's in a and the fact that 
S(p) =p, it follows that the sequence of repeated digits in r must 
consist entirely of i’s. 


Now, construct the repdigit number constructed from 10n 3's 


a = 2333-232 


10n times 


23 


and using 
d = 10000...00 
10n + 1 0's 
we again have a and d relatyvely prime. Arguments Similar to those 
used before forces the conclusion that the sequence of repeted 
digits must consist entirely of 3’s. 
This is of course impossibile and therefore the assumption of 


rationality must be false. 


3. ON THE DISTRIBUTION OF THE POINTS OF S(n)/n IN THE 


INTERVAL (0,1). 


The following problem is listed as unsolved problem number (7) 


in [1] 


Are the points p(n) =S(n)/n uniformly distributed in the 


interval (0,1)? 
The answer is no, the interval (0.5,1.0) contains only a 
finite number of points p(n). 


3.1. LEMMA: 


Sip) , sen 
p* p**t 


For p prime and k>0. 


24 


Proof: 
It ig well-known that S(p*)=j:p where jsk 


Therefore, forming the expressions 


— Á ee 


where m must have one of the two values {j, j+t} 
With the restrictions on the values of m and p, it is clear 
that 
ad 
Oo p 
which implies that 
Sip*) ‘ Sips?) 
p* pr 
which is the desired result. Equality occurs only when p=2, j=1 and 


m=2. 


3.2. LEMMA: 


The interval (0.5,1.0) contains only a finite number of points 


p(n), where 


p(n) = 312) and n is a power of a prime. 
Proof: 
If n=p S(p) 24 , outside the interval. 





pP 
Start with the smallest prime p=2 and move up the powers of 2 


S(2:2) L4 23 
(2'2) 4 


5 (2°22) 


ae 
(2'22) 8 


25 


S$(2°2:2°2) 6 = 
Si2222) 2 6 <0.5 | 
(2222) 16 R 


And applying the previous lemma, all additional powers of 2 


will yield a value less than 0.5. 


Taking the next smallest prime p=3 and moving up the powers 


of 3 
S(33) _ 6 
(3:3) 9 
S(333) _ 9 
22i ee 
Gaa a7 


and by the previous lemma, all additional powers of 3 also yield a 
value less than 0.5. 
Now, if p>3 and p is prime 


SPP) -20.5 
(pp) P l 
so all other powers of primes yield values less than 0.5 and we are 


done. 
3.3. THEOREM: 
The interval (0.5,1.0) contare only a finite number of points 
p(n) where 
p(n) = Sa 
Proof: 
It is well-known that if 


an 


N=D. p: p”. nEeD a then 
S(n) =max{S(P} ))} 
Applying the well-known result with the formula for p(n) 


S(p“ 
m ee Ł 


Di [| 2: 
ze: 


which 1s clearly less than 


26 


Theorefore, applying Lemma 2, we get the desired results. 


3.4. COROLLARY: 
The points p(n)=S(n)/n are not evenly distributed in the 


interval (0,1). 


4. THE SMARANDACHE FUNCTION DOES NOT SATYSFY A LIPSCHITZ 


CONDITION 


Unsolved problem number 31 in [1] is as follows. 
Does the Smarandache function veryfy a Lipschitz condition? In 


other words, is there a real number L such that 


| S(m) - S(n) |s L|[m-n| for all m,n i 107152730245) 


4.1. THEOREM 
The Smarandache function does not verify a Lipschitz 
cöndıition: 


Proof: 


Suppose that Smarandache function does indeed satisfy a 


Lipschitz condition and let L be the Lipschitz constant. 


Since the numbers of primes is infinite, is possible to fiind 


a prime p such that 


27 


p-(p+i)/2>L 


Now, examine the numbers (p-1) and (p+1). Clearly, at least 
one must not be a power of two, so we choose that one call it m. 
Factoring m into the product of all primes equal to 2 and 
everything else, we have 
m= 25n 


Then S(m) = max {$(2*),S(n)} and because $(2*) < 2* . 
we have 


m 
Sim) < 5 
And so, 
Isip) - S(m)|> |p - =| ee 
Since |p-m|=21 by choice of m, we have a violation of the 
Lipschitz condition, rendering our original assumption false. 
Therefore, the Smarandache function does not Satisfy a 


Lipschitz condition. 


5. ON THE SOLVABILITY OF THE EXPRESSION S(m) =n! 


One of the unsolved problems in [1] involves a relationship 
between the Smarandache and factorial functions. 


Solve the Diophantine Equation 
S(m) = n! 


where m and n are positive integers. 
This equation is always solvable and the number of solutions 
is a function of the number of primes less than or equal to n. 


5.1. LEMMA: Let be a prime. Then the range of the sequence 


28 


S(P SDH); S (DD D)pri 


will contain all positive integral multiples of p. 
Proof: It has already been proven [2] that for all integers 


k > 0, there exists another integer m > 0, such that 


S(p*)k = mp where msk 


and in particular 


S(p) =p 


So the only remaining element of the proof is to show that m 
takes on all possible integral values greater than 0. 

Let p be an arbitrary prime number and define the set 
M = { all positive integers n such that there is no positive 
integer k such that S(p*) = mp } | 
and assume that M is not empty. 

Since M is non-empty subset of the natural numbers, it must 
have a least element. Call that least element m. It is clear that 
aa 3° ck. 


Now, let j be the largest integer such that 
S(p) = (n= 1) -p 


and consider the exponent j + 1. 
By the choise of j, it follows that either 
1) S(p7*t) = mp 


or 


2) §{pi**) np where n> m 


in the first case, we have a contradiction of our choise of m, 


29 


so we proceed to case (2). 

However, it is a direct consequence of the definition of prime 
numbers that if ((m - 1)-p)! contains j instances of the prime p, 
then m-p is the smallest number such that (m-p)! contains more than 
j instances of p. Then, using the definition of Smarandache 
function where we choose the smallest number having the required 
number of instances we have a contradiction of case (2). 

Therefore, it follows that there can be no least element of 
the set M, so M must be empty. 

5.2.THEOREM: Let n be any integer and p a prime less than or 


equal to n. Then, there is some integer k such that 


S(p*) =n! 


Therefore, each equation of the form S(m) = n! has at least “’p 
solutions, where “p is the number of primes less than or equal 
ton. 

Proof: 

Since n! is an integral multiple of p for p any prime less 
than or equal to n, this is a direct consequence of the lemma. 

Now that the question is known to have multiple solutions, the 
next logical question is to determine how many solutions there are. 

5.3. DEFINITION: Let NSF(n) be the number of integers m, such 


that S(m) = n!. 


From the fact hat s(n) = max {S(p,‘)} , we have the following 
obvious result. 


30 


Corollary: 


Let n be a positive integer, q a prime less than or equal to 


n and k another positive integer such that s(q*) =n! . Then, all 


numbers having the prime factorization form m = g*p.*p,"p,?..-D.* 


where §(q*) > $(p;‘) will also be solutions the equation S(m) = n! 


To proceed further, we need the following two obvious lemmas. 
5.4. LEMMA: If p is a prime and m and n nonnegative integers m > 
n, then S(p") s S(p”). 
5.5. LEMA: If p and q are primes such that p < q and k > 0, then 
S(p“) < S(q*). 
The following theorem gives an initiai indication regarding 
how fast NSF(n) grows as n does. 
5.6. THEOREM: Let q be a prime number and k an exponent such that 
S(q) = n! Let p,,P),.--,P, be the list of primes less than q. Then 
the number of solutions to the equation S(m) = n! where m contains 


exactly k instances of the prime q is at least (k +1)". 
Proof: Applyng the two lemmas, the numbers m = p,'p,’p,°...p,’g* 


where all of exponents on the primes p; are at most solutions to the 
equation. Since each prime pi can have (k + 1), {0,1,2,...,k} 
different values for the exponent, simple counting gives the 
result. 

Since this procedure can be repeated for each prime less than 
or equal to n, we have an initial number of solutions given by the 


formula 


31 


> 
DE a en 
ž-2 
where s is the number of primes less then or equal to n, k is the 


integer such that 


S(p;*) =n! 
And even this is a very poor lower bound on the number of 


solutions for n having any size. 

5.7. COROLARY: Let q be a prime such that for some k S(q‘) = n!. 
Then if p is any prime such that there is some integer j such that 
S(p) < S(qt), then the product of any solution and p any power less 
than or equal to j will also be a solution. 

Proof: Clear. 

If q is the largest prime less than or equal ton, it is easy 
to show for "large" n that there are primes p > n > q that satisfy 
the above conditions. If p.is any prime, then by Bertrand’s 
Postulate, another prime r can be found in the interval p > r > 2p. 
Since q < n < 2n < n! for n > 2 and S(p) = p, we have one such 
prime. Expanding this reasoning, it folows that the number of such 
primes is at least j, where j is the largest exponent of 2 such 
that q2 s n!, or put another way, the largest power of 2 that 1s 
less than or equal to n!/q. 

Since there are so many solutions to the equation S(m) = n!, 
it is logical to consider the order of growth of the number of 
solutions rather than the actual number. 

It is well known that the number of primes less than or equal 
to n is asymptotic to the ratio n/in(n). Now, let p be the largest 


prime less than n. As n gets larger, it is clear that the factor m 


32 


such that mp = n! grows on the order of a factorial. Since ms k, 
where k is the exponent on the power p, it follows that the number 
grows on the order of the product of factorials. Since the number 
of items in the product depends on the number of primes q such that 
q < mp =n!, it follows that this number also grows on the order of 
a factorial. 

Putting it all together, we have the following behavior of 
NSF (n). 

NSF (n) grows on the order of product of items all on the order 
of the factorial of n, where the number of elements in the product 
also grows on the order of a factorial of n. 


Cleary, this function grows at an astronomical rate. 


6.THE NUMBER OF PRIMES BETWEEN S(n) and S(n+1) 


I read the letter by I.M.Radu that appeared in [3] stating 
that there is always a prime between S(n) and S(n+1) for all 
numbers O<n<4801, where S(n) is the Smarandache function. 

Since 1 have a computer program that computes the values of 
S(n), I decided to investigate the problem further. The serch was 


conducted up through n<1,033,197 and for instances where there is 


no prime p, where S(n)<ps<S(n+1) . They are as follows: 
n=224=2:2:2:2'27 S(n) =8 n=2252=3°3°5'5 S(n) =10 


nm=2057 =1111°17 S(n) =22 n=2058=2:37'°77 S$(n) =21 


33 


27265225 75°5°203'103 S(n) =206 m=265226=2'13°1017101 S(n) =202 


N= 843637 =37°151°251 S(n) =302 


2=843638 =2°19°149°149 S(n) =298 


AS can be seen, the first two values contradict the assertion 
made by I.M.Radu in his letter. Notice that the last two cases 


involve pairs of twin primes. This may provide a clue in the search 


for additional solutions. 


7. ADDITIONAL VALUES WHERE THE SMARANDACHE FUNCTION SATISFIES THE 
FIBONACCI RELATIONSHIP S(n)+S(n+¢1) «S (n+2) 


In [4] T.Yau poses the following problem: 


For what triplets n, n+l and n+2 does the Smarandache function 


satisfy the Fibonacci relationship 
S(n)+*S(n+1) = S(n+2) ? 
Two sciutions 
S$(9)+S5({(10) = S5(11) 6+5 = 11 


S(119) +9(120) = 5(121) 17+5 = 22 
were found, but no general solution was given. 


To further investigate this problem, a computer program was 
written that tested all values for n up to 1,000,000. Additional 
Solutions were found and all known solutions with their prime 


factorizations appear in the table below. 


S(9)+S(10) =S(11) 9 =3°3 10 


2°5 11 = 11 


S(119) + §(120) = S(121) 119 = 7-17 120 


27235. T21 = 11-11 


34 


$(4900) + $(4901) = $(4902); $(26243) + S (26244) = S$(26245) 


$(32110) + S$(32111) = S$ (32112) ; S{64008) + $(64009) = S$ (64010) ; 
$ (368138) + $(368139) = $(368139); S$(415662) + $(415663) = 
S (415664) ; 


I am unable to discern a pattern in these numbers that would 
lead to a proof that there is an infinite family of solutions. 


Perhaps another reader will be able to do so. 


8. WILL SOME PROBLEMS INVOLING THE SMARANDACHE FUNCTION ALWAYS 


REMAIN UNSOL ED? 


The most unsolved problems of the same subject are related to 


the Smarandache function in the Analytic Number Theory: 


g:Z---N , S(n) is defined as the smallest integer such 


that S(n)! is divisible by n. 

The number of these unsolved problems concerning the function 
is equal to... an infinity!! Therefore, they will never be all 
solved! 

One must be very careful in using such arguments when dealing 
with infinity. As is the case with number theoretic functions, a 
result in one area can have many aplications to other problems. The 
most celebrated recent instance is the "prof" of "Fermat's Last 
Theorem". In this case a result in elliptical functions has the 
proof as a consequence. 


Since S(n) is still largely unexplored, it is quite possible 


35 


chat the resolution of one problem leads tc the resolution of many, 
perhaps infinitely many, others. If that is indeed the case, then 


ail problems may eventually be resolved. 


REFERENCES 

1. R. Muller : Unsolved Problems Related to Smarandache Function 
(Number Theory Publishing Co., Glendale, Az, USA, 
1993). 

2. P Gronas : A note on S(p‘) (Smarandache Function J., V. 2-3, 
Nr.1, (1993), 33). 

3. M. Radu : Letter to the Editor (Math Spectrum, V.27, Nr.2, 
(1994/95), 44). 

4. T. Yau : A probiem Concerning the Fibonacci Series 


(Smarandache Functicn J., V. 4-5, Nr.1, (1994), 42). 


Current Address: Decisionmark, 200 2nd Ave. 
SE Suite 300, Cedar Rapids, 


IA 52401, USA.