arctan(tan(Pi*bernoulli(2*n)*GAMMA(2*n+1)))*(2*n+1)/Pi= 1 for Prime or zero for non-prime!

In log form:

i*(n+1/2)*log((1-tan(Pi*bernoulli(2*n)*GAMMA(2*n+1)))*(2*n+1)/Pi)/(1+tan(Pi*bernoulli(2*n)*GAMMA(2*n+1)))*(2*n+1)/Pi) )= 1 for Prime or zero for non-prime! ]]>

frac(theta)=arctan(tan(theta)). You see the primes obey this rule for

theta = Pi*bernoulli(2*n)*Gamma(2*n+1), only if 2*n+1 is a prime. The function theta is a consequence of the Riemann Hypothesis: s=1/2+i*tan(theta)/2.

In fact, I show that:

Zeta(s) = sum(-2^nu*bernoulli(nu)*GAMMA(1-s)/(factorial(nu)*GAMMA(-s-nu+2)), nu = 0 .. infinity); and that when the function vanishes, s/(1-s) = (1/2+(1/2*I)*tan(theta))/(1/2-(1/2*I)*tan(theta))= exp(theta) iff RH is true. The connection with primes is obvious, since when s=-2*n and 2*n+1 is a prime, the function takes an exponential form, and von-Staudt Clausen makes it clear the primes can be counted. ]]>

N= sum(i(n+1/2)*log( (1-i*tan(Pi*bernoulli(2*n)*gamma(2*n))/(1-i*tan(Pi*Bernoulli(2*n)*GAMMA(2*n+1))/Pi , n=3..m); which for large n reduces to the Prime Number theorem.

]]>N=sum( arctan(tan(Pi*Bernoulli(2*n)*GAMMA(2*n+1))) *(2*n+1))/Pi, n=3..M), M=m/2 for m even and M =m/2-1/2 for m odd. This formula yields the exact number of primes less than m. It was derived from the Riemann hypothesis and a a solution that I submitted that was never looked at. If this formula is right, then I have solve Riemann’s hypothesis! I invite a review of my paper. ]]>

So for integer programming in bounded dimension when the random subset is of density 1/2, does your argument gives average polynomial running time or even O(poly) running time almost surely. And what precisely the random sampling statement for primes you would need to extend it?

I suppose the next problem to play with is Frobenius coin problem. Ravi Kannan proved that there is a polynomial algorithm to determine the largest amount you canot pay with relatively prime k coins. We want to say that we can also determine the largest integer in a random subset of integers (or the largest prime) that we cannot pay. This does not look automatic from the problem but maybe follows from Kannan’s algorithm.

]]>Applied to prime numbers, the algorithm’s strategy for finding primes in an interval would just reduce to randomly sampling integers in the interval and checking primality. Same goes for what the algorithm would do to find twin primes. Is anything is known about random sampling and twin primes?

]]>A few years ago, I wrote some code in C to calculate the count of primes < n by calculating the various sum of divisor functions and then applying Linnik's identity. I tried to do everything I could to speed it up, but ultimately I never could get it fast enough to be competitive. I think the current version runs in something like O(n^4/5) time (at least from eyeballing it) and O(epsilon) space, but because the various loops for the sums of divisor functions are particularly sped up by aggressive use of wheels, I think its constant time factors are pretty good – or at least for an algorithm with O(epsilon) for memory. It calculates 10^12 in about a second on my reasonably decent laptop with a wheel culling out primes up to 19.

The code is here

http://www.icecreambreakfast.com/math/mckenzie_math.pdf (the code is at the bottom of this pdf, along with descriptions of what I'm doing above it)

if anyone wants to look over it or play with it. I keep trying to find some way to rearrange things in the algorithm to find some clever way to cache… something… to massively speed things up, but so far I've had no real luck. I've tried a lot of things, but I clearly haven't stumbled on the right idea yet. I could list some of those if anyone is interested.

At any rate, if any of you feel like taking a look at the code and seeing if you have any brilliant ideas, I would be overjoyed. While I wasn't looking, this whole topic (counting sums of divisors quickly) has turned into my Ahab-style White Whale.

]]>http://en.wikipedia.org/wiki/Sieve_of_Atkin

This computes the primes up to n in O(n/log log n)

time. The idea is as follows we first use this sieve to find the number of square free numbers up to n^1/4 and save the value for each value less than n^1/4 then for each of these numbers i we find the number of primes in the region n/p^2 is k plus a fractional value then for each prime bigger than n^1/4 we look at those primes for which n/p^2 =k rounded down has the same value then we used the data from the algorithm for lower values to see how man of the values one through k are square free then we have an accurate count of the number of numbers removed by the primes and the whole process should be o(n^1/2) although the memory used is at least O(n^1/2).