*[Corrected, thanks – T.]*

Let us informally call a “divisor sum” some function of the form where are some coefficients and are restricted to not be too large (how large depends on what polytope we are using and what distribution hypothesis we would like to assume; I ignore these issues here). We normalise these sums so that is comparable to 1 (this is not our usual normalisation, being more of a “probability theoretic” normalisation, but never mind this). Our basic sieve-theoretic strategy is to obtain upper and lower bounds for quantities of the form

(*)

and hope to parlay this into a proof that at least one of the pairs (n,n+2), (n+2,n+6) are simultaneously prime, giving a proof of . (We are of course also interested in auxiliary quantities such as , but we can’t control these directly; the only way we can proceed within sieve theory is to upper bound this quantity by something like for some other divisor sum , so that it becomes of the form in (*). [Basically, even GEH only allows one to asymptotically control sums with at most one copy of the von Mangoldt function in them; asymptotics for sums such as , which basically count twin primes, are out of reach even on GEH, with the only option being to get upper or lower bounds by some combination of sums (*) for which asymptotics are available, together with the non-negativity properties of sieves.]

Now, the Mobius pseudorandomness conjectures predict that the sieve is asymptotically orthogonal to the Mobius functions , thus for instance . (This is because we expect the Mobius function to exhibit significant cancellation on even very short arithmetic progressions.) The same pseudorandomness heuristic also predicts is asymptotically orthogonal to products .

What about ? This function is no longer orthogonal to , since of course is when n is prime, but from Mobius pseudorandomness we still expect this function to be orthogonal to higher products , because the constraint that n is prime can kill at most one of the two or more Mobius factors that are present.

In particular, if one introduces the weight function

(ignoring the case when n+2 is not squarefree, which can be sieved out by the W-trick (or else we replace Mobius with the Liouville function)), we see that a and 1 are indistinguishable for the purposes of correlations (*), thus for instance

Also, observe that the weight sequence is non-negative; it equals 4 when has the opposite Mobius function to both and , and is zero otherwise. Because of this, any sieve-theoretic manipulation we perform on the sums (*) to produce lots of tuples with a desired property as weighted by the constant function , will also produce lots of tuples with the same property as weighted by the function . For instance, if we can get a lower bound for by some clever combination of asymptotics for (*) (and exploiting non-negativity), we should also be able to get the same lower bound for . In particular, we would find an n in the support of a(n) for which either n,n+2 are both prime, or n+2,n+6 are both prime. But this contradicts the support of a, which forces n+2 to have a Mobius function of opposing sign to both n and n+6. So we cannot use sieve theory to force the event “n,n+2 both prime or n+2,n+6 both prime”, which indicates that we cannot establish through a sieve-theoretic analysis of the tuple n,n+2,n+6.

EDIT: The intuition here is that sieve theory cannot prevent the two “conspiracies” and from simultaneously occurring for (almost) all n for which n, n+2, n+6 are almost prime. (Of course, formally two applications of shifted versions of would imply , contradicting what I just said, but the space of n for which n,n+2,n+6 are almost prime is not shift-invariant, and one cannot force incompatibility this way. Alternatively, one can consider conspiracies in which for “almost all sufficiently large n with , which I believe cannot be ruled out by any easy “local” argument, and would mean that prime gaps less than 6 occur extremely infrequently.)

]]>For Q2, the heuristic I’ve been using is that the probability that a given member is prime should be , and that the probability that two of the are prime is bounded above by (the 2 being there for parity reasons). So if one starts with and deletes the three pairs which are separated by distance 16 or 24, we would get a prime pair separated by 8 if

which is in fact not attainable. The corresponding calculation for k=3 gives the condition

which requires , which is also blocked by parity.

]]>Question 1: Suppose that we achieved good enough numerics to pull off the trick of passing from to . Would similar methods allow us to look at the admissible 3-tuple and allow us to conclude that there are infinitely many prime gaps of size exactly 6? I know that Maynard’s sieve treats all admissible -tuples the same, but I wasn’t sure if the same is true of the trick of forcing the endpoints not simultaneously prime.

This would seem to overcome the parity problem for the pair , so would seem to preclude the ability to pass from to .

Question 2: We easily got , using a linear polynomial on a trivial decomposition, so there is a lot of room for improvement. What sort of bound would we need to push this to in order to play the game of forcing not simultaneously prime?

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