Recent content by e2theipi2026

Re: Equivalent Goldbach proof impossible queston. No, this is equivalent to proving Goldbach Conjecture, no one is going to solve it. Just a joke. You can delete if you wish.

Prove that the number of unordered partitions of an even number 2n into 2 composites is greater than the number of unordered partitions of an odd number 2n+1 into 2 composites for n>1 and n\ne p prime.

f(n) was defined at the beginning of the post. If you mean more detail, it is counting the number of "smaller" twin composites \le n. So, it is counting all composite k\le n such that k+2 is also composite. It would seem what I have shown is that the Twin Prime Conjecture is equivalent to...

Only counting once. Twin prime counting function is counting "smaller" twin primes and twin composite version is counting "smaller" twin composites. As an example for the composite version: given 22,23,24,25,26,27 it would count every even number and include 25, but it would not count 27 because...

In that movie they say he calculated it painstakingly by hand. Could be a different movie I'm thinking of, but that is what was stated.

I can prove the twin prime counting function has this form: \pi_2(n)=f(n)+\pi(n)+\pi(n+2)-n-1, where \pi_2(n) is the twin prime counting function, f(n) is the number of twin composites less than or equal to n and \pi(n) is the prime counting function. At n=p_n, this becomes \pi_2(p_n) =...

Helo. Welcome.

I must say, I am baffled at this question. If you have the opportunity to attend college take it! I get what you are saying about self teaching, but if you don't have a degree, your prospects for a career in your field of interest will be essentially non-existent. What is to stop you from...

I can prove the twin prime counting function has this form: \pi_2(n)=f(n)+\pi(n)+\pi(n+2)-n-1, where \pi_2(n) is the twin prime counting function, f(n) is the number of twin composites less than or equal to n and \pi(n) is the prime counting function. At n=p_n, this becomes \pi_2(p_n) =...