Download CONSTRUCTION OF A PRIME NUMBER FUNCTION It is well

CONSTRUCTION OF A PRIME NUMBER FUNCTION
It is well known that prime numbers N=p have the property that they can be divided only
by one and themselves. This suggests to us that if we take all the divisors of a number N
and subtract (1+N) from it we get a function which will vanish whenever N is a prime. To
prevent such a function from becoming too large, one can divide the result by N to
generate the normalized function-
f (N ) 
 ( N )  (1  N )
N
Here σ(N) represents the familiar divisor function of number theory. We first came up
with this quotient a little over a year ago(see http://www2.mae.ufl.edu/~uhk/NUMBERFRACTION.pdf ) and have explored its properties at length since that time. We term f(N)
for any given N the Number Fraction. If one takes a number such as N=60, one has
σ(60)=1+2+3+4+5+6+10+12+15+20+30+60=168, so that the number fraction becomes
f(60)=[168-61]/60=107/60. For a prime number such as p=127, we find σ(127)=128 but
f(127)=0. It is our purpose here to construct an additional new function which defines the
prime numbers as a function of f(Nn).
We start with the observations that for any prime p the values for f(p)=0, f(p2)=1/p,
f(p3)=(1+p)/p2, and f(p4)=1+p+p2)/p3. Recognizing that the numerator in these results is
simply a finite geometric series, we can generalize things to 1  p1  n
f (p ) 
p 1
n
which should hold for any positive integer n>1. Taking the ratio between n+1 and n, we
getf ( p n 1 ) ( p n  1)
 n
f ( pn )
( p  p)
On setting n=2 we findf ( p 3 ) ( p  1)

f ( p2 )
p
This can be re-written asf ( p2 )  1
1
pf ( p 3 )
So that every prime will have this quotient equal to one. If we now replace p by any
number N, we get a new functionP( N ) 
f (N 2 )  1
 (N 2 )  1

N f (N 3)  (N 3)  N 3  1
applicable for any positive integer N≥2. The second form involving σ makes the
evaluation of the function faster since σ is a built in function in most advanced math
computer programs. What is very interesting about this new function is that it will have a
value of one whenever N is a prime but will be less than unity when N is a composite. Let
us demonstrate for two casesP (64) 
[ f (642 )  1]
 0.3124260897.. so 64 is a composite
64 f (643 )
P (61) 
[ f (612 )  1]
 1.000000000... so 61 is a prime
[61 f (613 )]
and-
For obvious reasons, we call the new function P(N) the Prime Number Function. It is a
point function valid only for integers although there may possibly be a definite integral
representation for non-integer N similar to the relation between n! and the Γ(n+1)
function.(See if any of you can find it). To plot P(N) we can use the list-plot operation in
our MAPLE computer program. This draws straight lines between neighboring points.
We have done this for 3≤N≤50 and present the results here-
The graph clearly shows that the function has a value of unity for those numbers where
N=p. It can also be thought of as a criterion for primes. The criterion readsA number N is prime if P(N)=1 and composite if P(N)<1
Note that this statement is equivalent to just saying that f(N)=0 whenever N=p. The form
of P(N) however is somewhat more informative then a graph for f(N). It more clearly
indicates double-primes and semi-primes.
The values of N used in the P(N) function are only limited by one’s computer ability to
handle the multiplication and division of large numbers. Here is an example of finding
primes near one trillion using our home PC. It involves an evaluation of P(N) over the
range 1000000000<N<1000000015 to produce a double prime at one trillion and seven
and one trillion and nine-
The calculation took only a split second to carry out. Note how in the immediate
neighborhood of a prime the value of P(N) dips to very low values only slightly above
zero. Such values of N represent super-composites having numerous factors. The number
N=1000000008 is such a composite with P(N)=0.1374565066 x 10-8. It has a total of 144
divisors.
We next look at the semi-prime N=pq, where p and q are primes. We find thatf ( pq) 
and-
( p  q)
pq
p≠q
f [( pq) 2 ] 
[( p  p 2 )  ( q  q 2 )]  pq[1  ( p  q )]
( pq ) 2
p≠q
We can use the result for f(pq) and the fact that N=pq to obtain the quadraticp 2  pNf ( N )  N  0
This equation yieldsp
Nf ( N )
Nf ( N ) 2
 (
) N
2
2
So if we take N=232+1, which happens to be one of the Fermat numbers, we get for the
lower root p=641. That is the semi-prime –
232+1=4294967297=641 x 6700417
It should be mentioned that finding the p component by this approach is really equivalent
to finding the divisors for the numerator of f(N) and thus in effect one already has found
the factors of the semi-prime N=pq before actually evaluating the above equation for p.
The following calculation will work just as welldivisors{(N)}-(N+1)={641,6700417}
One could also combine f(pq) with f[(pq)2] , to find a quadratic in p2. This would
however not decrease the calculation efforts over the quadratic for p1 given above. The
expression is however handy when calculatingf (8281)  f (72  132 ) 
November 2013
[(7  49)  (13  169)  7  13(1  7  13)] 307

8281
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