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> There are an infinite number of primes and of weak primes (and possibly of strong primes too, though that's still only conjectured)

Given that a strong prime is < (p[n-1] + p[n+1])/2, and that an infinite number of prime pairs with prime gap < 400 exist (twin prime conjecture and derivatives), and that the average prime gap is ln(p[n]), does it then not also follow that there is an infinite number of strong primes?

Yes, nice! Here's a more rigorous way of proving it:

A strong prime is precisely one for which the prime gap preceding it is strictly larger than the one following it. There are arbitrarily large prime gaps (for example none of the n numbers following n! are prime, since n!+m divides by m whenever m≤n). But it has recently been proven that there are infinitely many prime gaps with length less than or equal to 246. So given any strong prime we can find a prime gap after it which is longer than 246, and then a prime gap after that whose length is less than or equal to 246. Somewhere in between those two prime gaps there must be a prime at which the gap size decreases; a strong prime larger than our original strong prime. Hence there are infinitely many strong primes.

n!+1 can be prime, as primes can be divisible by 1. None of the n-1 numbers after n!+1 are prime.

Good point! But n-1 still gets arbitrarily large so the proof still works.

I don't think that we need to know the average prime gap for this.

Assume that there are only a finite number of strong primes.

Then after some point for all triplets of consecutive primes p1, p2, p3, we must have p2-p1 <= p3-p2. In other words, the gap between consecutive primes must be non-decreasing after some point.

But since there are arbitrarily large runs of non-primes (e.g., [n!+2, n!+n] is a run of n-1 consecutive non-primes), non-decreasing gaps contradicts the theorem that there are an infinite number of gaps < 400.

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