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If p is a prime number, and k is an even integer, what is the probability p+k is a prime number?

According to my simulations p+108 is prime twice as often as p+344

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There are asymptotics for $k=2$, though you'll have to translate that result to your setting. It'd be very nice to know the general case. What do your simulations indicate? My own crude simulations indicate heavy clusters for these probabilities: 0.12, 0.25, 0.13. –  lhf Sep 29 '11 at 1:43
@lhf: The general case is due to Beteman, Horn, and Stemmler: en.wikipedia.org/wiki/Bateman%E2%80%93Horn_conjecture –  Charles Sep 29 '11 at 7:07
Also see Dickson's conjecture. (Note typo above: first name should be Bateman.) –  Charles Sep 30 '11 at 14:33
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2 Answers

up vote 12 down vote accepted

Note that if $p \equiv 1 \mod 3$, $p + 344$ is divisible by 3 and so must be composite. On the other hand, $p + 108 \equiv p \mod 3$. Thus primes $p$ with $p+344$ prime can occur in only one residue class mod 3, but those with $p+108$ prime can occur in two residue classes mod 3. On the other hand, $p + 344 \equiv p \mod 43$. So I would expect that primes $p + 108$ would occur $(2/1) (41/42) = 41/21$ times as often as $p + 344$.

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If and only if you have a proof that probability for the prime fall int one on the following groups are equal: p+108, p+344 (the eqality of a number to not being a prime for the p+344 and p+108 for the given prime p). In general, you must have a proof that probability for the p+108 to be prime is 41/21 greater then probability p+344 to be a prime. –  Artur Mustafin Sep 30 '11 at 9:28
I did not say I had a proof. I was careful to use the wording "I would expect...". –  Robert Israel Oct 2 '11 at 8:59
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Technically, the answer to your question is zero. If you fix $k$ and take the number of primes $p$ less than $N$ such that $p+k$ is prime and divide it by the number of primes less than $N$ and take the limit as $N$ goes to infinity, you get zero.

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Of course. Nevertheless, I'd interpret the question as being about asymptotics. –  Robert Israel Oct 2 '11 at 9:11
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