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Let me start by paraphrasing Richard Nixon by saying "I am not a crank!"; I am not "working on" Goldbach's Conjecture, this is an idle curiosity along similar lines to this question.

So, as any crank knows, you look at $2m$, and then you start looking at $2m-p$ for primes $p$. Well how about we place $2m$ in an interval where for about $k\approx N/\ln N$ the first $k$ of the $2m-p$ (i.e. $2m-3,\,2m-5,\,2m-7,\dots,\,2m-p_k$) are all composite.

We can do this readily by considering $2m$ in the interval $[N!+2,N!+N]$, say $2m\approx N!+N$.

So some focused questions. The first a cursory Google could not answer for me; the second I doubt Google could answer.

Has anyone seriously looked for counterexamples in such an interval? If not, why not?

I understand that, using this notation, all such intervals up $N=20$ have been accounted for via verification up to $\approx 5\times 10^{18}$.

JP McCarthy
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  • For standard numbers (without your filter on specific ranges), for even numbers $n$ in range $(10^{10}, 10^{11})$ for example, the number of couples $(p,q)$ of prime numbers such that $p+q=n$ is big, greater than $1000$ (I don't remember the estimation... but it is big). And there are simple formulas that give very good estimations of this number of couples. If you focus in range $(N!, N!+N)$, you know that we are safe with prime numbers lower than $N$, (less than $10$ primes numbers in our range), but it still remains many thousands of couples $p,q$ such that $p+q=n$ – Lourrran Nov 26 '24 at 18:05
  • @Lourrran I get you - while it may be true that many small primes exhibit Goldbach, throwing these out doesn't do anything because in general small primes aren't necessary. I think that answers the question (if not, why not?) if you would like to make your comment an answer. – JP McCarthy Nov 26 '24 at 18:15

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Goldbach's conjecture is very 'weak'. It says that for any even number $n$ there exists at least one couple $(p,q)$ of prime numbers, such that $p+q=n$.

I say it is very weak because as far as we can check, we can see that there exist a very big number of couples $(p,q)$ such that $p+q=n$.

We can run small calculs.

For big numbers $n$, in range $(2, n)$, the number of prime numbers can be estimate by $\pi(n)=\frac{n}{ln(n)}$ ; when we combine all couples from this set, we obtain $\frac{\pi(n)^2}{2}$ couples. For all these couples, the sum $p+q$ is less than $2n$.

We have $\frac{\pi(n)^2}{2}$ couples, that are 'dispached' in less than $n$ boxes.

This dispatching is not uniform, but we have in average $\frac{n}{ln(n)^2}$ couples in each box.

For numbers $r$ like $10^{20}$, this rough estimation says that we should have around $2 \times 10^{16}$ couples $(p,q)$ that give $p+q=r$.

If by any process, for some specific $r$, we can reduce this estimation by $10$, or $20$, it remains many many couples.

Lourrran
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