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Smarandache Notions Journal, Vol. 13, No. 1-2-3, Spring 2002, pp. 140-149. 
THE PSEUDO-SMARANDACHE FUNCTION 


David Gorski 
137 William Street 
East Williston, NY 11596 
(516)742-9388 
Gorfam@ Worldnet.att.net 


Abstract: 

- The Pseudo-Smarandache Function is part of number theory. The function comes from the 
Smarandache Function. The Pseudo-Smarandache Function is represented by Z({n) where n 
represents any natural number. The value for a given Z(n) is the smallest integer such that 
1+2+3+ . . . + Z(n) is divisible by n. Within the Pseudo-Smarandache Function, there are several 
formulas which make it easier to find the Z(n) values. 


Formulas have been developed for most numbers including: 
a) p, where p equals a prime number greater than two; 
b) b, where p equals a prime number, x equals a natural number, and b=p"; 
c) x, where x equals a natural number, if x/2 equals an odd number greater than two: 
d) x, where x equals a natural number, if x/3 equals a prime number greater than three. 


Therefore, formulas exist in the Pseudo-Smarandache Function for all values of b except for the 


following: 
a) x, where x = a natural number, if x/3 = a nonprime number whose factorization is not 
a | 
b) multiples of four that are not powers of two. 


All of these formulas are proven, and their use greatly reduces the effort needed to find Z(n) 
values. 


Keywords: 
Smarandache Function, Pseudo-Smarandache Function, Number Theory, Z(n), g(d), g[Z(n)]. 


Introduction. 

The Smarandache (sma-ran-da-ke) Functions, Sequences, Numbers, Series, Constants, 
Factors, Continued Fractions, Infinite Products are a branch of number theory. There are very 
interesting patterns within these functions, many worth studying sequences. The name “Pseudo- 
Smarandache Function” comes from the Smarandache function. [2] The Smarandache Function 
was named after a Romanian mathematician and poet, Florentin Smarandache. [1] The 
Smarandache Function is represented as S(n) where n is any natural number. S(n) is defined as the 
smallest m, where m represents any natural number, such that m! is divisible by n. 

To be put simply, the Smarandache Function differs from the Pseudo-Smarandache Function in 
that in the Smarandache Function, multiplication is used in the form of factorials; in the Pseudo 
-Smarandache Function, addition is used in the place of the multiplication. The 
Pseudo-Smarandache Function is represented by Z(n) where n represents all natural numbers. The 
value for a given Z(n) is the smallest integer such that 1+2+3+ . . . + Z(n) is divisible by n. 


140 


Background 


As previously stated, the value for a given Z(n) is the smallest 
integer such that 1+2+3+ . . . + Z(n) is divisible by n. Because 
consecutive numbers are being added, the sum of 1+2+3+...+Z(n) is 
a triangle number. Triangle numbers are numbers that can be written in 
the form [d(d+1)]/2 where d equals any natural number. When written 
in this form, two consecutive numbers must be present in the 
numerator. In order to better explain the Z(n) function, the g(d) 
function has been mtroduced where g(d)=[d(d+1)]/2. 


Figure 1: The first ten g(d) values. 





141 


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or, 
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Π| CO |] Uo p 






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N | bo 
SS | = 






— ees janah 


i 
HN | in| 60] 00 | ON 






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wo 
ON 






i 
~d | Go 
rom | ON 


20 Í 





Figure 2: The first 20 Z(n) 
and g[{Z(n)] 
values. 


a 


—" 
ho 
_) 


g[Z(n)] values are defined as 9(d) values where d 
equals Z(n). Because of this, it is important to note that all 
g[Z(n)] values are g(d) values but special ones because they 
correspond to a particular n value. Since g(d)=[d(d+1)}/2, 
g{Z(n) |=[Z(n)[Z(n)+1]/2]. Because g(d) is evenly divisible by 
n, and all g[Z(n)] are also g(d) values, g[Z(n)] is evenly 
divisible by n. Therefore, the expression [Z(n)[Z(n)+1]/2] can 
be shortened to n*k (where k is any natural number). If k=x/2 
(where x is any natural number) then 
[Z(n)[Z(n)+1]/2]-(n*x)/2, and the “general form” for a 
g[Z(n)] value is (n*x)/2. Again, since (n*x)/2 represents a 
g(d) value, it must contain all of the characteristics of g(d) 
values. As said before, all g(d) values, when written in the 
form [d(d+1)]/2, must be able to have two consecutive 
numbers in their numerator. Therefore, in the expression 
(n*x)/2, n and x must be consecutive, or they must be able to 
be factored and rearranged to yield two consecutive numbers. 
For some values of n, g[Z(n)}=(n*x)/2 where x is much less 
than n (and they aren’t consecutive). This is possible because 
for certain number combinations n and x can be factored and 
rearranged in a way that makes them consecutive. For 
example, Z(n=12) is 8, and g[Z(12)] is 36. This works 
because the original equation was (12*6)/2=36, but after 
factoring and rearranging 12 and 6, the equation can be 
rewritten as (8*9)/2=36. 

The Pseudo-Smarandache Function specifies that only 
positive numbers are used. However, what if both d and n 
were less than zero? g(d) would then represent the sum of the 
numbers from d to —1. Under these circumstances, Z(n) 
values are the same as the Z(n) values in the “regular” system 
(where all numbers are greater than one) except they are 
negated. This means that Z(-n)=-[Z(n)]. This occurs because 
between the positive system and the negative system, the g(d) 
values are also the same, just negated. For example, 
2(4)=4+3+2+1=10 and g(-4)= -4+ -3+ -2+ -1=-10. Therefore, 
the first g(d) value which is evenly divisible by a given value 
ofn won’t change between the positive system and the 
negative system. 


142 


Theorem 1 


If ‘p’ is a prime number greater than two, then Z(p)=p-1 


Proof: 


Since we are dealing with specific p values, rather than - 
saying g[Z(n)]=(n*x)/2, we can now say g[j(p)|=(p*x)/2. 
Therefore, all that must be found is the lowest value of x that is 
consecutive to p, or the lowest value of x that can be factored 
and rearranged to be consecutive to p. Since p is prime, it has 
no natural factors other than one and itself. Therefore, the 
lowest value of x that is consecutive to p is p-1. Therefore 


Z(p)}=p-1. 





Figure 3: The first 10 Z(p) 
values. 


Theorem 2 


If x equals any natural number, p equals a prime number greater than two, and b equals p*, then 
| Z(b)=b-1 | 


Example: 





bo 
as 
a 


343 342 
2401 
16807 
117649 


w 
-15624 
















16806 
117648 





Figure 4: the first Z(b) values for different primes. 





NO) 
D 
O 
© 





143 


Proof: 


The proof for this theorem is similar to the proof of theorem 2. Again, the g(d) function is 
made up of the product of two consecutive numbers divided by two. Since b’s roots are the same, 
it is impossible for something other than one less than b itself to produce to consecutive natural 
numbers (even when factored and rearranged). For example, g[Z(25)]=(25*x)/2. When trying to 
find numbers less than 24 which can be rearranged to make two consecutive natural numbers this 
becomes g[Z(25)]=(5*5*x)/2. There is no possible value of x (that is less than 24) that can be 
factored and multiplied into 5*5 to make two consecutive natural numbers. This is because 5 and 
5 are prime and equal. They can’t be factored as is because the have no divisors. Also, there is no 
value of x that can be multiplied and rearranged into 5*5, again, because they are prime and equal. 


Example: 


a 














| E 
a: EL 
| 4127 
m8 
| 2048 | 4095} 


sioa] 16383 


~} 


~d U| a| Cr 





Figure 5: The first six Z(x) 


values. 


3 
l 


Theorem 3 


If x equals two to any natural power, then Z(x)=2x-1. 


Proof: 


According to past logic, it may seem like Z(x) would equal 
x-1. However, the logic changes when dealing with even numbers. 
The reason Z(x)#x-1 is because (x-1)/2 can not be an integral value 
because x-1 is odd (any odd number divided by two yields a number 
with a decimal). Therefore, [x(x-1)]/2 is not an even multiple of x. In 
order to solve this problem, the numerator has to be multiplied by two. 
In a sense, an extra two is multiplied into the equation so that when the 
whole equation is divided by two, the two that was multiplied in is the 
two that is divided out. That way, it won’t effect the “important” part 
of the equation, the numerator, containing the factor of x. Therefore, 
the new equation becomes 2[x(x-1)]/2, or [2x(x-1)]/2. The only 
numbers consecutive to 2x are 2x-1 and 2x+1. Therefore, the smallest 
two consecutive numbers are 2x-1 and 2x. 
Therefore, Z(x)=2x-1. 


144 


Theorem 4 


If ‘P is any natural number where j/2 equals an odd number greater than two then 













Example: 

i oje [Gay 
D E E E 
E a A 
E E a 
E L E 
EE E L G 
a E E 
ECNE E 

Figure 6: The first 


twenty j(z) values. 


—], 7 —1 is evenly divisble by 4 
Zi) = | 
—1 is not evenly divisble by 4 


? 


J 
2 
J 
2 


NIJS. 


Proof: 


When finding the smallest two consecutive numbers that 
can be made from a j value, start by writing the general form but 
instead of writing n substitute j in its place. That means 
g[Z@)]=G*x)/2. The next step is to factor j as far as possible 
making it easier to see what x must be. This means that 
g{ZQ)|=(2*j/2*x)/2. Since the equation is divided by two, if left 
alone as g[Z(j)]=(2*j/2*x)/2, the boldface 2 would get divided out. 
This falsely indicates that j/2*x (what is remaining after the 
boldface 2 is divided out) is evenly divisible by j for every natural 
number value of x. However, j/2*x isn’t always evenly divisible 
by j for every natural number value of x. The two that was just 
divided out must be kept in the equation so that one of the factors 
of the g(d) value being made isj. In order to fix this the whole 
equation must be multiplied by two so that every value of x is 
evenly divisible by j. In a sense, an extra two is multiplied into the 
equation so that so that when the whole equation is divided by two, 
the two that was multiplied in is the two that gets divided out. | 
That way, it won’t effect the “important” part of the equation 
containing the factor of two. Therefore it becomes 
g[Z(@)]=(2*2*)/2*f)/2 where f represents any natural number. This 
is done so that even when divided by two there is still one factor of 
j. At this pomt, it looks as though the lowest consecutive integers 
that can be made from g[Z(j)]=(2*2*j/2*f) are (j/2) and(j/2)-1. 
However, this is only sometimes the case. This is where the 
formula changes for every other value ofj. If (j/2)-1 is evenly 
divisible by the ‘2*2? (4), then Z(j)=(j/2)-1. However, if (j/2)-1 is 
not evenly divisible by 4, then the next lowest integer consecutive 
to ¥2 is (/2}+1. (Note: If(j/2)-1 is not evenly divisible by 4, 


then the next lowest integer consecutive to j/2 is (j/2)+1. (Note: If G/2)-1 is not evenly divisible by 
four, then (j/2)+1 must be evenly divisible by 4 because 4 is evenly divisible by every other multiple of 
two.) Therefore, if (j/2)-1 is not evenly divisible by 4 then 21ZQ)]=[G/2)[G/2}+-1]]/2 or Z()=/2. 


Theorem 5 


If ‘p’ is any natural number where p/3 equals a prime number greater than 3 then 


3 


f aps Lis evenly divisible by 3 
Zínì — 3 


145 


Example: 


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~I Ww =j af AA 


Gd 


T 
$ 
jand =à 
2l] QAI NI © 


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promt Í pomah Ne 
Nj mi N 





pd 






WOI oo AI Ch 
CONi m=! CO! W! oO] ~) 





paved 
pom 
bmn 





Figure 7: The first ten 
Z(p) values. 


Proof: 


The proof for this theorem is very similar to the proof for 
theorem 4. Since p values are being dealt with, p must be 
substituted into the general form. Therefore, 2(Z(p)]=(p*x)/2. 
Since what made p is already known, p can be factored further so 
that g{[Z(p)]=(3*p/3*x)/2. At this point it looks like the 
consecutive numbers that will be made out of (the numerator) 
3*p/3*x are p/3 and (p/3)-1 (this is because the greatest value 
already in the numerator is p/3). However, this is only sometimes 
the case. When p/3-1 is divisible by 3, the consecutive integers in 
the numerator are p/3 and (p/3)-1. This means that Z(p)=p/3-1 if 
p/3-1 is evenly divisible by 3. However, if p/3-1 is not divisible by 
three, the next smallest number that is consecutive to p/3 is 
(p/3)+1. If (p/3)-1 is not divisible by 3 then (p/3)+1 must be 
divisible by 3 (see *1 for proof of this statement). Therefore, the 
consecutive numbers in the numerator are p/3 and (p/3)+1. This 
means that Z(p) = p/3 if (p/3)-1 is not evenly divisible by three. 


Note: Although there is a similar formula for some multiples of the first two primes, this formula 
does not exist for the next prime number, 5. | 





Figure 10 


_*1—“If (p/3)-1 is not divisible by 3, then (p/3)+1 must be divisible by 3.” 


In the table to the left, the underlined values are those that are divisible by 
three. The bold numbers are those that are divisible by two (even). Since p/3 is 
prime it cannot be divisible by three. Therefore, the p/3 values must fall 
somewhere between the underlined numbers. This leaves numbers like 4,5, 7, 8, 
10, 11, etc. Out of these numbers, the only numbers where the number before (or 
(p/3)-1) is not divisible by three are the numbers that precede the multiples of 
three. This means that the p/3 values must be the numbers like 5, 8, 11, etc. 
Since all of these p/3 values precede multiples of 3, (p/3)+1 must be divisible by 3 
if (p/3)-1 is not divisible by 3. 


Theorem 6 


If ‘n’ equals any natural number, Z(n)=*n. 


Proof: 


Theorem 6: Part A 
146 


Ifr is any natural odd number, Z(r)r<1 
Proof: 


When r is substituted into the general form, g[Z(r)=[r*(r-1)]/2. Since r is odd r-1 is even. 
Therefore, when r-1 is divided by two, an integral value is produced. Therefore, (r*r-1)/2 is an 


even multiple ofr and it is also a g(d) value. Because of this, Z(r)pe1. Since Z(r)r<1, Z(r)=r. 
Theorem 6: Part B 
Ifv is an natural even number, Z(v)*v. 
Proof: 


If Z(v) = v, the general form would appear as the following: g[Z(v)]=[v(v+1)]/2. This is 
not possible because if v is even then v+1 is odd. When v+1 is divided by two, a non-integral 
value is produced. Therefore, (v*v+1)/2 is not an integral multiple of v. Therefore, Z(v)=v. 


Theorem 7 


If w is any natural number except for numbers whose prime factorization equals 2 to any power, 
Z(w)<w. 


Proof 


As in several other proofs, this proof can be broken down into two separate parts, a part for 
r values (r is any natural odd number) and one for v values (v is any natural even number). As 
_ proven in Theorem 6: Part A, Z(r)r<1. This proves that Z(r) is less than r. 

For v values, v must be substituted into the general form in order to be able to see patterns. 
Therefore, g[Z(v)]=(v*x)/2. Since v is even it must be divisible by two. Therefore, v can be 
factored making g[Z(v)]=[2*(v/2)*x]/2. Since the numerator is being divided by two, when done 
with the division, one whole factor of v will not always be left. Therefore, an extra two must be 
multiplied into the equation so that even when divided by two, there is still one whole factor of v 


left. Therefore, g[Z(v)][4*(v/2)*x]/2. At this point, the equation can be simplified to 


g[Z(v)]*x. Therefore, x=v-1, and Z(v)<v-1. Z(v) is less than v-1 rather than less than or equal 
to v-1 because as proven in theorem 4, Z(v)*v-1. 


147 


Conclusion 





Through researching the relationships between different groups of natural numbers, patterns and 
formulas have been developed to find Z(n) values for most numbers. Formulas have been developed for 
most numbers including: 

a) p, where p equals a prime number greater than two 

b) b, where p equals a prime number, x equals a natural number, and b=p" 

c) x, where x equals a natural number, if x/2 equals an odd number greater than two 

d) x, where x equals a natural number, if x/3 equals a prime number greater than three 


In fact there are only two remaining groups of numbers for which there are no formulas or 
shortcuts. Formulas exist in the Pseudo-Smarandache Function for all values of b except for the 
following: 

a) multiples of four that that are not powers of two (figure 8) : 

b) x, where x = a natural number, if x/3 = a nonprime number whose factorization is not 3 


(figure9) 


If p equals a prime number greater than two then Z(p)=p-1. If p equals a prime number greater than two, 
x equals a natural number, and b=p* then Z(b)=b-1. However, if p=2 then Z(b)=2b-1. Ifx equals a 
natural number, and x/2 equals an odd number greater than two then if (x/2)-1 is evenly divisible by four 
then Z(x)=(x/2)-1. Otherwise, if x/2-1 is not evenly divisible by four then Z(x)=z/2. Ifx equals a 

natural number, and x/3 equals a prime number greater than three then if (x/3)-1 is evenly divisible by 
three then Z(x)=(x/3)-1. Otherwise, if x/3-1 is not evenly divisible by three then Z(x)=x/3. All of these 
formulas are proven, and their use greatly reduces the effort needed to find Z(n) values. | 


References: 
[1] Ashbacher, Charles, “Introduction to the Smarandache Function”, J. 
Recreational Mathematics (1996-1997): 249-251. 


148 


[2] Vasiliu, F., “Florentin Smarandache, a poet with the Dot under the 1” [Online] 
http://www.gallup.unm.edu/~smarandache/fstext.htm. 


149