Identities by Zeta Function Wrap

MAIN We write out the terms of the series for the Rieman zeta function into a grid of width k, wrapping down onto the next line. By drawing k + 1 lines through the terms, which also wrap, all of the terms are accounted for. Using this relationships can be drawn upon by using known values of the ζ(s). I.e. for integer s. These results are for example {12^2}+\psi ^{(1)}\left({5}\right)+\psi ^{(1)}\left({5}\right)+\psi ^{(1)}\left({5}\right)+\psi ^{(1)}\left({5}\right)=4 \pi ^2 with ψn the polygamma function. GRID First we write the terms of the zeta function in a width k grid, for example for k = 4 {1^2} & {2^s} & {3^s} & {4^s} \\ {5^2} & {6^s} & {7^s} & {8^s} \\ {9^2} & {10^s} & {11^s} & {12^s} \\ {13^2} & {14^s} & {15^s} & {16^s} we may draw a line through the 1, 6, 11, 16, ... terms, and also lines through the 2, 7, 12, ..., the 3, 8, 13, ..., the 4, 9, 14, ... and the 5, 10, 15, ... terms. Then these five sequences describe all of the terms. Then the sum of the terms (which is the zeta function itself), is equal to the five sequences summed. This is obvious and arbitrary. However, we can also define the next terms to sum in the grid based on the old terms. We may write {1^s} + {6^s} + {11^s} + \cdots = {1^s} + {2^s} + \left( {6^s} - {2^s} \right) + {3^s} + \left( {11^s} - {3^s} \right) + \cdots which is actually equal to ζ(s) plus an additional sum, we can call the additional sum A. If we do this for the other 4 sequences, we get more sums B, C, D, E, and find them to all be polylog expressions. Equationg the whole lot gives the expressions \zeta(s) = 5\zeta(s)-4-{2^s} - {3^s} - {4^s} + A + B + C + D + E where explicitly A=^\infty (nk+n+1)^{-s} - (n+1)^{-s}\\ B=^\infty (nk+n+2)^{-s} - (n+2)^{-s}\\ C=^\infty (nk+n+3)^{-s} - (n+3)^{-s}\\ D=^\infty (nk+n+4)^{-s} - (n+4)^{-s}\\ E=^\infty (nk+n+5)^{-s} - (n+5)^{-s} with k = 4. However, the selection of k was arbitrary, we can do this for any k and write \zeta(s) = (k+1)\zeta(s) -^{k-1} {(m+1)^s} + ^{k+1} S_n where then S_n=^\infty (mk+m+n)^{-s} - (m+n)^{-s}\\ IDENTITIES {54}+\psi ^{(2)}\left({5}\right)+\psi ^{(2)}\left({5}\right)+\psi ^{(2)}\left({5}\right)+\psi ^{(2)}\left({5}\right) -125 \left( \psi ^{(2)}(2)+ \psi ^{(2)}(3)+ \psi ^{(2)}(4)+ \psi ^{(2)}(5) \right)=752 \zeta (3) \psi ^{(1)}\left({3}\right)+\psi ^{(1)}\left({3}\right)={3}-{4} -\psi ^{(2)}\left({7}\right)-\psi ^{(2)}\left({7}\right)-\psi ^{(2)}\left({7}\right)-\psi ^{(2)}\left({7}\right)-\psi ^{(2)}\left({7}\right)-\psi ^{(2)}\left({7}\right)+343 \psi ^{(2)}(2)+343 \psi ^{(2)}(3)+343 \psi ^{(2)}(4)+343 \psi ^{(2)}(5)+343 \psi ^{(2)}(6)+343 \psi ^{(2)}(7)=686 \left({36000}-{343}\right) which may be written 7^3^7\psi ^{(2)}(n) - ^6\psi ^{(2)}\left({7}\right) ={2^4 3^2 5^3} - 3432 \zeta(3) also have ^1\psi ^{(2)}(n)= {1} - 2\zeta(3) \\ ^2\psi ^{(2)}(n)= {1} - 4\zeta(3) \\ ^3\psi ^{(2)}(n)= {2^2} - 6\zeta(3) \\ ^4\psi ^{(2)}(n)= {2\cdot3^3}-8 \zeta (3) \\ ^5\psi ^{(2)}(n)= {2^5\cdot3^3}-10 \zeta (3) \\ ^6\psi ^{(2)}(n)= {2^4\cdot3^2\cdot5^3} -12 \zeta(3) \\ ^7\psi ^{(2)}(n)= {2^5\cdot3^2\cdot5^3} - 14 \zeta(3) \\ ^8\psi ^{(2)}(n)= {2^3\cdot3^2\cdot5^3\cdot7^3}-16 \zeta(3) \\ ^9\psi ^{(2)}(n)= {2^8\cdot3^2\cdot5^3\cdot7^3}-18 \zeta(3) \\ ^{10}\psi ^{(2)}(n)={800150400}-20 \zeta (3)\\ ^{11}\psi ^{(2)}(n)={8001504000}-22 \zeta (3)\\ ^{12}\psi ^{(2)}(n)={2^6×3^5×5^3×7^3×11^3}-24 \zeta (3) \\ It would appear that there is a product of powers of primes as the denominator for this constant! This creates a very convoluted prime test. We have the famous \psi^{(n)}(z+1)=\psi^{(n)}(z) + (-1)^nn!z^{-n-1} and therefore \psi^{(2)}(z+1)=\psi^{(2)}(z) + {z^3} using that \psi^{(2)}(1)=-2\zeta(3) we can then write ^{z+1}\psi^{(2)}(n)=^z\psi^{(2)}(1) + {z^3} we need some way of separating the numerator and denominator. For example we may use , {l, 1, 50}]]} and we see the higher prime factors growing (with the exception of glitches for low values). The benefit of this is that if we know a number p is prime. Then we could work out the next prime immediately by looking at (sensible) differences in the value of the polylog until the denominator becomes zero mod p, then we know the new prime and we can repeat the process. Example. Say I know 389 is prime, and no other information. If we work out the denominator D₃₈₉, it is a very large number. (This is the main problem, we need a modular extension to all of this). But it is quick to verify D₃₈₉ mod 389 = 0, we then calculate the next D389 + k’s for k for candidate primes (digit sum rules, odd numbers etc.), checking for each whether D389 + k mod k = 0. ... (unfortunately this is not strong enough to test primality).