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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- (N ) (1 N ) f (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/NUMBER- FRACTION.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 p1n f ( pn ) p 1 which should hold for any positive integer n>1. Taking the ratio between n+1 and n, we get- f ( pn1) ( pn 1) f ( pn ) ( pn p) On setting n=2 we find- f ( p3 ) ( p 1) f ( p2 ) p This can be re-written as- f ( p2 ) 1 1 pf ( p3 ) So that every prime will have this quotient equal to one. If we now replace p by any number N, we get a new function- f (N 2 ) 1 (N 2 ) 1 P(N ) 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 cases- [ f (642 ) 1] P(64) 0.3124260897.. so 64 is a composite 64 f (643 ) and- [ f (612 ) 1] P(61) 1.000000000... so 61 is a prime [61 f (613 )] 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 reads- A 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 that- ( p q) f ( pq) p≠q pq and- [( p p2 ) (q q2 )] pq[1 ( p q)] f [( pq)2 ] p≠q ( pq)2 We can use the result for f(pq) and the fact that N=pq to obtain the quadratic- p2 pNf (N ) N 0 This equation yields- Nf (N ) Nf (N ) p ( )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 well- divisors{(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 calculating- [(7 49) (13 169) 7 13(1 7 13)] 307 f (8281) f (72 132 ) 8281 1183 November 2013 .
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