I have a conjecture related to the strong Goldbach conjecture and the Goldbach function. It is that: for any $g(E)$, there are a finite number of even numbers which can be expressed as a sum of two primes in precisely that many ways. More formally, I'm introducing a new function $h(g(E))$, which is defined as the number of evens which yield that $g(E)$. My conjecture is that, for any finite $n$, $h(n)$ is finite.

The strong Goldbach conjecture is simply that $h(0) = 2$ (only 2 and 4 cannot be expressed as the sum of odd primes).

My conjecture seems to me very likely to be true. The question then arises: what does the function $h(x)$ look like. Are there any good approximations?

See also:



  • $\begingroup$ See oeis.org/A229492 for the first few (conjectured) values. $\endgroup$ – Barry Cipra Jul 20 '16 at 23:32
  • $\begingroup$ As a follow-up, if you follow links in my previous comment, you get to a (conjectured) graph at oeis.org/A000974/graph $\endgroup$ – Barry Cipra Jul 20 '16 at 23:42
  • $\begingroup$ '$h(n)$ is finite for every $n$' is much stronger than "Goldbach conjecture is true for $E$ large enough" (the negation of "Goldbach conjecture is true for $E$ large enough" is that $h(0)$ is infinite). And the best approximation you can find for all those prime counting functions (and related) is the simplest : mark $2k+1$ as prime with a probability $\frac{1}{\ln 2k+1}$ $\endgroup$ – reuns Jul 21 '16 at 0:06
  • $\begingroup$ @user1952009 Actually, that's subtly wrong: the '$n$ is prime with probability $\frac1{\ln n}$' holds only when $n$ is effectively random (i.e., when you know nothing about it). By walking only through the odd numbers you increase that probability by a factor of 2. (Consider that if you mark no even numbers as prime and mark every $2k+1$ prime with probability $\frac1{\ln (2k+1)}$, you'll only mark (approximately) $\frac{n}{2\ln n}$ numbers $\lt n$, half as many as you should.) $\endgroup$ – Steven Stadnicki Jul 21 '16 at 0:16
  • $\begingroup$ Thanks Barry. Interesting graph. $\endgroup$ – Thomas Delaney Jul 26 '16 at 9:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.