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This question is inspired by the announcement of a proof that "fake twin" primes, i.e. pairs of consecutive primes differing by at most K, are -in infinite number- where K is a fixed integer which can be taken = 70,000,000.

It is not known OTOH whether the above would hold where K would be = 2 instead (the twin prime conjecture).

Anyway, my question may be (hopefully is) much simpler : it's been known after Viggo Brun that the sum of reciprocals of actual "twin primes" is finite (whether the number of terms in this sum is finite or infinite). Does the sum of reciprocals of the "fake twins" as defined above (for K=70 million) also converge ? Does it follow from an easy extension of Brun's method ?

I should make it clear I am no expert, just curious... I do not have accesss to a proof of Brun's original theorem, even a sketchy one.

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    $\begingroup$ Far from a proof, but the first Hardy-Littlewood Conjecture concerns the general case of k-tuples, but for pairs of primes separated by distance k (not necessarily consecutive primes) in general the more prime factors k has the more numerous the pairs are. See mathworld.wolfram.com/k-TupleConjecture.html $\endgroup$
    – Foo Barrigno
    May 20, 2013 at 17:05

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Your instinct is correct: the same proof that establishes the convergence of the sum of the reciprocals of the actual twin primes also establishes the convergence of the sum of the reciprocals of "$K$-twin primes" for any value of $K$.

To give a tiny amount of detail: sieve methods (which Brun pioneered) can show that the number of primes $p\le x$ for which $p+K$ is also prime is at most $c(K) x/\log^2 x$, for some constant $c(K)$ depending on $K$. This is enough to show that the sum of the reciprocals of such primes $p$ converges (consider the contribution of such primes between $2^n$ and $2^{n+1}$, then sum over $n$).

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  • $\begingroup$ Dr Martin, many thanks ! I suspect I'd have to enroll onto stackexhange here and award you marks or votes of sort, but being my first visit I'm clueless still. $\endgroup$
    – NimbUs
    May 20, 2013 at 20:56
  • $\begingroup$ No worries. Enroll if you feel like hanging around more :) $\endgroup$
    – Greg Martin
    May 20, 2013 at 23:28
  • $\begingroup$ I think my registration has succeeded, tentatively marking your illuminative answer as "accepted" now. Thanks again... $\endgroup$
    – NimbUs
    May 21, 2013 at 10:21

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