This post is a continuation of the previous post on sieve theory, which is an ongoing part of the Polymath8 project. As the previous post was getting somewhat full, we are rolling the thread over to the current post. We also take the opportunity to correct some errors in the treatment of the truncated GPY sieve from this previous post.
As usual, we let be a large asymptotic parameter, and a sufficiently slowly growing function of . Let and be such that holds (see this previous post for a definition of this assertion). We let be a fixed admissible -tuple, let , let be the square-free numbers with prime divisors in , and consider the truncated GPY sieve
where , is the polynomial
and is a fixed smooth function supported on . As discussed in the previous post, we are interested in obtaining an upper bound of the form
as well as a lower bound of the form
for all (where when is prime and otherwise), since this will give the conjecture (i.e. infinitely many prime gaps of size at most ) whenever
It turns out we in fact have precise asymptotics
although the exact formulae for are a little complicated. (The fact that could be computed exactly was already anticipated in Zhang’s paper; see the remark on page 24.) We proceed as in the previous post. Indeed, from the arguments in that post, (2) is equivalent to
and (3) is similarly equivalent to
Here is the number of prime factors of .
We will work for now with (4), as the treatment of (5) is almost identical.
We would now like to replace the truncated interval with the untruncated interval , where . Unfortunately this replacement was not quite done correctly in the previous post, and this will now be corrected here. We first observe that if is any finitely supported function, then by Möbius inversion we have
Note that if and only if we have a factorisation , with and coprime to , and that this factorisation is unique. From this, we see that we may rearrange the previous expression as
Applying this to (4), and relabeling as , we conclude that the left-hand side of (4) is equal to
which may be rearranged as
This is almost the same formula that we had in the previous post, except that the Möbius function of the greatest common divisor of was missing, and also the coprimality condition was not handled properly in the previous post.
We may now eliminate the condition as follows. Suppose that there is a prime that divides both and . The expression
can then be bounded by
which may be factorised as
which by Mertens’ theorem (or the simple pole of at ) is
Summing over all gives a negligible contribution to (6) for the purposes of (4). Thus we may effectively replace (6) by
The inner summation can be treated using Proposition 10 of the previous post. We can then reduce (4) to
where is the function
Note that vanishes if or . In practice, we will work with functions in which has a definite sign (in our normalisations, will be non-positive), making non-negative.
We rewrite the left-hand side of (7) as
We may factor for some with ; in particular, . The previous expression now becomes
Using Mertens’ theorem, we thus conclude an exact formula for , and similarly for :
Proposition 1 (Exact formula) We have
Similarly we have
where and are defined similarly to and by replacing all occurrences of with .
These formulae are unwieldy. However if we make some monotonicity hypotheses, namely that is positive, is negative, and is positive on , then we can get some good estimates on the (which are now non-negative functions) and hence on . Namely, if is negative but increasing then we have
for and , which implies that
for any . A similar argument in fact gives
for any . Iterating this we conclude that
From Cauchy-Schwarz we thus have
Observe from the binomial formula that of the pairs with , of them have even, and of them have odd. We thus have
We have thus established the upper bound
By symmetry we may factorise
The expression is explicitly computable for any given . Following the recent preprint of Pintz, one can get a slightly looser, but cleaner, bound by using the upper bound
In practice we expect the term to dominate, thus we have the heuristic approximation
Now we turn to the estimation of . We have an analogue of (8), namely
But we have an improvment in the lower bound in the case, because in this case we have
From the positive decreasing nature of we see that and so is non-negative and can thus be ignored for the purposes of lower bounds. (There are similar improvements available for higher but this seems to only give negligible improvements and will not be pursued here.) Thus we obtain
Estimating similarly to we conclude that
By (9), (10), we see that the condition (1) is implied by
By Theorem 14 and Lemma 15 of this previous post, we may take the ratio to be arbitrarily close to . We conclude the following theorem.
Theorem 2 Let and be such that holds. Let be an integer, define
and suppose that
As noted earlier, we heuristically have
and is negligible. This constraint is a bit better than the previous condition, in which was not present and was replaced by a quantity roughly of the form .