## Tuesday, January 31, 2012

### Connecting Positive (Even) and Negative (Odd) Integers of Zeta Function

I have discovered for myself a very interesting relationship which - based on initial empirical evidence - I would conjecture, universally holds.

The first part can be stated in the following manner.

If s (representing the dimensional power) of the Riemann Zeta Function is a positive even integer and if the denominator, i.e. D1 of ζ(1 - s) is divisible by s + 1 then, s + 1 is prime.

Furthermore in this case the denominator of ζ(s) i.e. D2 will be divisible also by s + 1.

Then if ai (from i = 1 to k) represent all the factors of D1, D1 will be divisible by ai + 1 (from 1 to k) whenever ai + 1 is a prime.
Furthermore in such cases D1 will have no other prime factors (though multiples of some existing factors may occur).

Also in many cases when D1 is divided singly by all of the prime factors (ai + 1), frequently the result = s.
Also the probability that s will result seems to be related to the degree to which s is of a very composite nature (i.e. with many factors). I carried out calculations for all negative odd integers up to 100 to find that half of the results (after division of denominator by the relevant prime factors) = s!

Likewise D2 will be divisible by all prime numbers from 3 to s (and only these primes) though in the case where s represents a power of 2, D2 will be divisible by all primes from 2 to s + 1 (and only these primes).

Even when D2 is not divisible by s + 1, or in the cases where s is an odd integer (> 1), D2 will be divisible by all prime numbers from 2 to the largest prime immediately prior to s + 1!

So to illustrate!

s = 10 is an even integer.

Therefore if the denominator of ζ(- 9) is divisible by 11, then 11 is prime.

The relevant denominator = 132 (in absolute terms) and 132 is indeed divisible by 11.

Therefore 11 is prime.

This also means that the denominator of ζ(10) will be divisible by 11.

And D2 = 93555 which again is divisible by 11.

The factors of 10 are 1, 2, 5 and 10.
When we add 1 to each factor we get 2, 3, 6 and 11.

So the prime numbers here are 2, 3 and 11.

Therefore 132 is divisible by 2, 3 and 11 (and only these primes).

The remainder that results on division by these prime numbers = 2 (≠ s).

So 132 is not an especially composite number (with 3 prime factors).

Also 93555 is thereby divisible by 3, 5, 7 and 11 (and only these primes).

So hidden in the structure of the rational denominator values of the Riemann Zeta Function, for negative odd integer values, is a simple test to determine whether a number is prime! Though of course it does not represent a practical way of finding prime numbers, it remains however an extremely interesting fact!

And it is no less interesting that for positive even integer values of s (regardless of how large is the value of s) that when s + 1 is prime, the denominator of ζ(s) will be divisible by all primes from 3 to s + 1 (and only those primes). Again where s is a square of 2 the denominator of ζ(s) is divisible by all primes from 2 to s + 1 (and only these primes).

It probably would help to illustrate with refernce to a couple of more examples.

s = 30 is an even integer.

Therefore if the denominator of ζ( - 29) is divisible by 31, then 31 is prime.

The relevant denominator = 85932 (in absolute terms) and 85932 is indeed divisible by 31.

Therefore 31 is prime.

This also means that the denominator of ζ(30) will be divisible by 31.

And D2 = 5660878804669082674070015625 is indeed divisible by 31.

The factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30.
When we add 1 to each factor we get 2, 3, 4, 6, 7, 11, 16 and 31.

So the prime numbers here are 2, 3, 7, 11 and 31.

Therefore 85932 is divisible by 2, 3, 7, 11 and 31 (and only these primes).

The remainder that results on division by these prime numbers = 6 (≠ s).

Perhaps it is a liitle surprising here that s = 30 is not the remainder, as 85932 is quite a composite number (though with no prime factors from 13 to 29 inclusive)!

The denominator of D2 (5660878804669082674070015625) will however in this case be divisible by all prime factors from 3 to 31 (inclusive) and only these factors.

In fact 5660878804669082674070015625 = 3^15.5^6.7^5.11^3.13^2.17.19.23.29.31

When however s = 32 the denominator of ζ-(31) is not now divisible by s + 1 = 33

i.e. the absolute value of denominator of ζ-(31) = 16320 (which is not divisible by 33). So 33 therefore is not prime!

In fact not only can we directly confirm whether a number is prime (through division of the relevant denominator) of the Zeta Function (as described above) we can also confirm directly that a number is not prime (as illustrated in this case whenever the denominator is not divisible by the number in question).

However because 32 is a square of 2, this implies that the denominator of ζ(32) will in this case be divisible by every prime number from 2 to 31 inclusive and only these prime factors (though of course these prime factors may repeat).

And this denominator

= 64290220571022341207266406250 = 2.3^15.5^8.7^4.11^2.13^2.17^2.19.23.29.31