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ON PRIME NUMBER FUNCTIONS a Necessary and Sufficient Condition

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ON PRIME NUMBER FUNCTIONS a Necessary and Sufficient Condition MORE ON PRIME NUMBER FUNCTIONS In several recent notes we have derived and discussed some properties of the point function- f (N 2 ) 1 F(N ) Nf (N 3 ) , where f(x)=[(x)-(x+1)]/x is a function we have designated as the number fraction, is the familiar sigma function of number theory, and x=N represents any positive integer starting with two . We can generate the following table for the functions f(N) and F(N) for integers 2 through 15- Integer, N Number Fraction, f(N Prime Number Function, F(N 2 0 1 3 0 1 4 1/2 15/31 5 0 1 6 5/6 90/383 7 0 1 8 3/4 21/85 9 1/3 40/121 10 7/10 216/1339 11 0 1 12 5/4 134/1117 13 0 1 14 9/14 398/3255 15 8/15 201/1432 What is immediately obvious from this table is that the function F(N) equals unity only if N is a prime number and is found to be less than one for composite numbers. Also the number fraction f(N) has zero value when N is a prime but a finite value when composite. For obvious reasons we refer to f(N) and F(N) as the Prime Number Functions. It is our purpose here to discuss some additional properties of these functions. Again, from earlier notes, we recall that any prime number above N=3 has the form 6n±1 without exception. We call such prime numbers Q Primes. They represent all possible primes N=5 or greater. However, it must be noted that not all N=6n±1 numbers need to be primes. Coming to mind are the composites 25, 35, 49, etc. How does one know when N is indeed a prime? The answer can be summarized by the following - A necessary and sufficient condition for N>3 to be a prime is that N=6n±1 and that F(N)=1 and f(N)=0 In actual calculations one can distinguishes between prime and composite numbers by evaluating just one of these functions since the second will follow. The required calculations efforts to show f(N)=0 is somewhat simpler than determining F(N)=1. However, the graph of F(N) versus N in a given range of N more clearly shows the locations of primes as indicated by the following graph over the range3<N<50- The primes clearly stand out in this figure. What is important to emphasize that all primes above N=3 have the form N=6n±1 without exception. For example the prime N=127 has the form N=6(21)+1 which also means that 127 mod(6)=1. Let us work out a few more examples and present the results in the following table- N N mod (6) f(N) F(N) Category 3247 1 0.064059 0.0048308.. Composite 1797629 5 0.0016388 0.000339558.. Composite 659083149 3 - - Composite 4294967297 5 0.0015602 0.14923 x 10-6 Composite 1763524987135789 1 0 1 Prime 5343721409546135798791 0 1 Prime The last 22 digit long number lies at the limit of my PCs capability. It does clearly show that this number is a prime. Note that the mod(6) operation comes in very handy for telling one if the number N has the possibility of being a prime. The requirement is N mod(6) =1 or N mod6)=5 followed by the F(N)=1 or f(N)=0 test. If an odd number has the form N mod(6)=3 it implies N=6n+3 and hence can the number can never be a prime. The 4th number given in the above table is the Fermat number 232+1. It was first shown to be composite by Leonard Euler. We next look at semi-primes. These have the important property that N=pq , where p and q are different prime numbers. Thus N=403=13x31 is a good example of a semi-prime. Such semi-primes, when involving numbers of one hundred digit length or so, play a central role in present day public key cryptography. The number fraction for any semi-prime equals- (1 p)(1 q) pq 1 ( p q) 1 1 f ( pq) pq pq p q This will generally be a very small number much less than one but greater than zero. We have f(403)=44/403=0.10918114… Since N=pq, we have the quadratic equation- p 2 pf (N ) N 0 with the solution- Nf (N ) Nf (N ) p [ ]2 N 2 2 So, once f(N) for a semi-prime is known, the two prime number components p and q can be determined. Lets try this for the semi-prime N=455839=6(75973)+1. Here we have f(N)=1360/455839=0.0029835.. and the quadratic solves as p=599 and q=761. This particular semi-prime is often used to demonstrate the elliptic curve factorization method of Lenstra. We can also use a graphical method to factor semi-primes. Take for example the semi-prime N=pq=120643 for which f(N)=0.006332733.. Using the above identity for f(pq) we have- x 1 f (N ) where x p or q N x Plotting this equation and the line f=0.00633.. over a large enough range produces two curves which intersect at x=p and x=q as shown- One can also work out the value for the function F(N) when N=pq. This however leads to a rather complicated expression since- ( p p 2 ) (q q2 ) pq(1 p q) f [( pq)2 ] ( pq)2 and- ( p p 2 p3 ) (q q2 q3 ) pq(1 p p 2 q q2 ) ( pq)2 (1 p q) f [( pq)3 ] ( pq)3 It produces- [( p p 2 q q2 ) pq(1 p q) ( pq)2 ] F( pq) [( p p 2 p3 q q2 q3 ) pq(1 p p 2 q q2 ) ( pq)2 (1 p q) That this equality is correct may easily be checked by setting p=2 and q=3 so that N=6. We find- f (36) 1 (6 12) 6(6) 36 90 F(6) 6 f (216) (14 39) 6(7 12) 36(6) 383 The values of f(N) and F(N) can often be written down in analytic form. Take the case of N=2n for n two or greater. We have f(4)=1/2, f(8)=3/4, f(16)=7/8, and f(32)=15/16. From this we see that- f (2n ) 1 21n Thus f(1024)=f(210)=1-(1/512)=511/512. Substituting into the definition equation for F(N) also yields- (22n 1) F(2n ) (23n1 1) As n gets large f(2n) approaches the value of one and F(2n) approaches a value of zero. Thus N=2n will never satisfy the criteria for a prime number. This is of course obvious since 2n is an even number . However, the slight modification f(2n-1) changes things completely. This new number represents a Mersenne Number. If n takes on certain prime number values, the Mersenne number can become a prime. For example, N=213-1=8191 is a Mersenne Prime. Continuing on to f(3n), we find f(9)=1/3, f(27)= 4/9, and f(81)=13/27. So the general term will be- 1 f (32 ) (1 31n ) 2 Also one finds that- (32n 1) F(3n ) (33n1 1) Thus- 38 1 6560 F(81) F(34 ) 0.0370316... 311 1 177146 By looking at the forms F(2n)and F(3n), one can postulate that for any number rn we have- (r 2n 1) F(r n ) (r 3n1 1) In testing out this equality, we find the expression in indeed true provided r is a prime. It fails to work when r is a composite. For example, N=75 works fine producing F(75)= (710-1)/(714-1)= 0.00041649. But the function F(127)=(1214- 1)/(1220-1)= 0.3348979767x10-6 fails to agree with the computer result of F(127)= 0.41862x10-7 based on the definition of F(N) given at the beginning of this note. Thus we can say the above analytic result for F(rn)works provided we also have that F(r)=1. May 2014 .
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