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