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 {x} be a large asymptotic parameter, and {w} a sufficiently slowly growing function of {x}. Let {0 < \varpi < 1/4} and {0 < \delta < 1/4+\varpi} be such that {MPZ[\varpi,\delta]} holds (see this previous post for a definition of this assertion). We let {{\mathcal H}} be a fixed admissible {k_0}-tuple, let {I := [w,x^\delta]}, let {{\mathcal S}_I} be the square-free numbers with prime divisors in {I}, and consider the truncated GPY sieve

\displaystyle  \nu(n) := \lambda(n)^2

where

\displaystyle  \lambda(n) := \sum_{d \in {\mathcal S}_I: d|P(n)} \mu(d) g(\frac{\log d}{\log R})

where {R := x^{1/4+\varpi}}, {P} is the polynomial

\displaystyle  P(n) := \prod_{h \in {\mathcal H}} (n+h),

and {g: {\bf R} \rightarrow {\bf R}} is a fixed smooth function supported on {[-1,1]}. As discussed in the previous post, we are interested in obtaining an upper bound of the form

\displaystyle  \sum_{x \leq n \leq 2x} \nu(n) \leq (\alpha+o(1)) (\frac{W}{\phi(W)})^{k_0} \frac{x}{W \log^{k_0} R}

as well as a lower bound of the form

\displaystyle  \sum_{x \leq n \leq 2x} \nu(n) \theta(n+h) \geq (\beta+o(1)) (\frac{W}{\phi(W)})^{k_0} \frac{x}{W \log^{k_0-1} R}

for all {h \in {\mathcal H}} (where {\theta(n) = \log n} when {n} is prime and {\theta(n)=0} otherwise), since this will give the conjecture {DHL[k_0,2]} (i.e. infinitely many prime gaps of size at most {k_0}) whenever

\displaystyle  1+4\varpi > \frac{4\alpha}{k_0 \beta}. \ \ \ \ \ (1)

It turns out we in fact have precise asymptotics

\displaystyle  \sum_{x \leq n \leq 2x} \nu(n) = (\alpha+o(1)) (\frac{W}{\phi(W)})^{k_0} \frac{x}{W \log^{k_0} R} \ \ \ \ \ (2)

and

\displaystyle  \sum_{x \leq n \leq 2x} \nu(n) \theta(n+h) = (\beta+o(1)) (\frac{W}{\phi(W)})^{k_0} \frac{x}{W \log^{k_0} R} \ \ \ \ \ (3)

although the exact formulae for {\alpha,\beta} are a little complicated. (The fact that {\alpha,\beta} 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

\displaystyle  \sum_{d_1,d_2 \in {\mathcal S}_I} \mu(d_1) g(\frac{\log d_1}{\log R}) \mu(d_2) g(\frac{\log d_2}{\log R}) \frac{k_0^{\Omega([d_1,d_2])}}{[d_1,d_2]} \ \ \ \ \ (4)

\displaystyle  = (\alpha + o(1)) (\frac{W}{\phi(W)})^{k_0} \log^{-k_0} R

and (3) is similarly equivalent to

\displaystyle  \sum_{d_1,d_2 \in {\mathcal S}_I} \mu(d_1) g(\frac{\log d_1}{\log R}) \mu(d_2) g(\frac{\log d_2}{\log R}) \frac{(k_0-1)^{\Omega([d_1,d_2])}}{[d_1,d_2]} \ \ \ \ \ (5)

\displaystyle  = (\beta + o(1)) (\frac{W}{\phi(W)})^{k_0-1} \log^{-k_0+1} R.

Here {\Omega(d)} is the number of prime factors of {d}.

We will work for now with (4), as the treatment of (5) is almost identical.

We would now like to replace the truncated interval {I = [w,x^\delta]} with the untruncated interval {I \cup J = [w,\infty)}, where {J = (x^\delta,\infty)}. Unfortunately this replacement was not quite done correctly in the previous post, and this will now be corrected here. We first observe that if {F(d_1,d_2)} is any finitely supported function, then by Möbius inversion we have

\displaystyle  \sum_{d_1,d_2 \in {\mathcal S}_I} F(d_1,d_2) = \sum_{d_1,d_2 \in {\mathcal S}_{I \cup J}} F(d_1,d_2) \sum_{a \in {\mathcal S}_J} \mu(a) 1_{a|[d_1,d_2]}.

Note that {a|[d_1,d_2]} if and only if we have a factorisation {d_1 = a_1 d'_1}, {d_2 = a_2 d'_2} with {[a_1,a_2] = a} and {d'_1 d'_2} coprime to {a_1 a_2}, and that this factorisation is unique. From this, we see that we may rearrange the previous expression as

\displaystyle  \sum_{a_1,a_2 \in {\mathcal S}_J} \mu( [a_1,a_2] ) \sum_{d'_1,d'_2 \in {\mathcal S}_{I \cup J}: (d'_1 d'_2, a_1 a_2) = 1} F( a_1 d'_1, a_2 d'_2 ).

Applying this to (4), and relabeling {d'_1,d'_2} as {d_1,d_2}, we conclude that the left-hand side of (4) is equal to

\displaystyle  \sum_{a_1,a_2 \in {\mathcal S}_J} \mu( [a_1,a_2] ) \sum_{d_1,d_2 \in {\mathcal S}_{I \cup J}: (d_1d_2,a_1a_2)=1}

\displaystyle \mu(a_1d_1) g(\frac{\log a_1d_1}{\log R}) \mu(a_2d_2) g(\frac{\log a_2d_2}{\log R}) \frac{k_0^{\Omega([a_1 d_1,a_2 d_2])}}{[a_1 d_1,a_2 d_2]}

which may be rearranged as

\displaystyle  \sum_{a_1,a_2 \in {\mathcal S}_J} \frac{\mu( (a_1,a_2) ) k_0^{\Omega([a_1,a_2])}}{[a_1,a_2]} \sum_{d_1,d_2\in {\mathcal S}_{I \cup J}: (d_1d_2,a_1a_2)=1} \ \ \ \ \ (6)

\displaystyle  \mu(d_1) g(\frac{\log a_1d_1}{\log R}) \mu(d_2) g(\frac{\log a_1 d_2}{\log R}) \frac{k_0^{\Omega([d_1,d_2])}}{[d_1, d_2]}.

This is almost the same formula that we had in the previous post, except that the Möbius function {\mu((a_1,a_2))} of the greatest common divisor {(a_1,a_2)} of {a_1,a_2} was missing, and also the coprimality condition {(d_1d_2,a_1a_2)=1} was not handled properly in the previous post.

We may now eliminate the condition {(d_1d_2,a_1a_2)=1} as follows. Suppose that there is a prime {p_* \in J} that divides both {d_1d_2} and {a_1a_2}. The expression

\displaystyle  \sum_{a_1,a_2 \in {\mathcal S}_J} \frac{k_0^{\Omega([a_1,a_2])}}{[a_1,a_2]} \sum_{d_1,d_2 \in {\mathcal S}_{I \cup J}: p_* | (d_1d_2,a_1a_2)}

\displaystyle |g(\frac{\log a_1d_1}{\log R})| |g(\frac{\log a_1 d_2}{\log R})| \frac{k_0^{\Omega([d_1,d_2])}}{[d_1, d_2]}

can then be bounded by

\displaystyle  \ll \sum_{a_1,a_2} \sum_{d_1,d_2: p_* | (d_1d_2,a_1a_2)} \frac{k_0^{\Omega([a_1,a_2])}}{[a_1,a_2]} \frac{k_0^{\Omega([d_1,d_2])}}{[d_1, d_2]} (a_1 a_2 d_1 d_2)^{-1/\log R}

which may be factorised as

\displaystyle  \ll \frac{1}{p_*^2} \prod_p (1 + \frac{O(1)}{p^{1+1/\log R}})

which by Mertens’ theorem (or the simple pole of {\zeta(s)} at {s=1}) is

\displaystyle  \ll \frac{\log^{O(1)} R}{p_*^2}.

Summing over all {p_* > x^\varpi} gives a negligible contribution to (6) for the purposes of (4). Thus we may effectively replace (6) by

\displaystyle  \sum_{a_1,a_2 \in {\mathcal S}_J} \frac{\mu( (a_1,a_2) ) k_0^{\Omega([a_1,a_2])}}{[a_1,a_2]} \sum_{d_1,d_2\in {\mathcal S}_{I \cup J}}

\displaystyle \mu(d_1) g(\frac{\log a_1d_1}{\log R}) \mu(d_2) g(\frac{\log a_1 d_2}{\log R}) \frac{k_0^{\Omega([d_1,d_2])}}{[d_1, d_2]}.

The inner summation can be treated using Proposition 10 of the previous post. We can then reduce (4) to

\displaystyle  \sum_{a_1,a_2 \in {\mathcal S}_J} \frac{\mu( (a_1,a_2) ) k_0^{\Omega([a_1,a_2])}}{[a_1,a_2]} G_{k_0}( \frac{\log a_1}{\log R}, \frac{\log a_2}{\log R} ) = \alpha+o(1) \ \ \ \ \ (7)

where {G_{k_0}} is the function

\displaystyle  G_{k_0}(t_1,t_2) := \int_0^1 g^{(k_0)}(t+t_1) g^{(k_0)}(t+t_2) \frac{t^{k_0-1}}{(k_0-1)!}\ dt.

Note that {G} vanishes if {t_1 \geq 1} or {t_2 \geq 1}. In practice, we will work with functions {g} in which {g^{(k_0)}} has a definite sign (in our normalisations, {g^{(k_0)}} will be non-positive), making {G_{k_0}} non-negative.

We rewrite the left-hand side of (7) as

\displaystyle  \sum_{a \in {\mathcal S}_J} \frac{k_0^{\Omega(a)}}{a} \sum_{a_1,a_2: [a_1,a_2] = a} \mu((a_1,a_2)) G_{k_0}( \frac{\log a_1}{\log R}, \frac{\log a_2}{\log R} ).

We may factor {a = p_1 \ldots p_n} for some {x^\delta < p_1 < \ldots < p_n} with {p_1 \ldots p_n \leq R}; in particular, {n < \frac{1 + 4\varpi}{4\delta}}. The previous expression now becomes

\displaystyle  \sum_{0 \leq n < \frac{1+4\varpi}{4\delta}} k_0^n \sum_{x^\delta < p_1 < \ldots < p_n} \frac{1}{p_1 \ldots p_n}

\displaystyle  \sum_{\{1,\ldots,n\} = S \cup T} (-1)^{|S \cap T|} G_{k_0}( \sum_{i \in S} \frac{\log p_i}{\log R}, \sum_{j \in T} \frac{\log p_j}{\log R} ).

Using Mertens’ theorem, we thus conclude an exact formula for {\alpha}, and similarly for {\beta}:

Proposition 1 (Exact formula) We have

\displaystyle  \alpha = \sum_{0 \leq n < \frac{1+4\varpi}{4\delta}} k_0^n \int_{\frac{4\delta}{1+4\varpi} < t_1 < \ldots < t_n} G_{k_0,n}(t_1,\ldots,t_n) \frac{dt_1 \ldots dt_n}{t_1 \ldots t_n}

where

\displaystyle  G_{k_0,n}(t_1,\ldots,t_n) := \sum_{\{1,\ldots,n\} = S \cup T} (-1)^{|S \cap T|} G_{k_0}( \sum_{i \in S} t_i, \sum_{j \in T} t_j ).

Similarly we have

\displaystyle  \beta = \sum_{0 \leq n < \frac{1+4\varpi}{4\delta}} (k_0-1)^n \int_{\frac{4\delta}{1+4\varpi} < t_1 < \ldots < t_n} G_{k_0-1,n}(t_1,\ldots,t_n) \frac{dt_1 \ldots dt_n}{t_1 \ldots t_n}

where {G_{k_0-1}} and {G_{k_0-1,n}} are defined similarly to {G_{k_0}} and {G_{k_0,n}} by replacing all occurrences of {k_0} with {k_0-1}.

These formulae are unwieldy. However if we make some monotonicity hypotheses, namely that {g^{(k_0-1)}} is positive, {g^{(k_0)}} is negative, and {g^{(k_0+1)}} is positive on {[0,1)}, then we can get some good estimates on the {G_{k_0}, G_{k_0-1}} (which are now non-negative functions) and hence on {\alpha,\beta}. Namely, if {g^{(k_0)}} is negative but increasing then we have

\displaystyle  -g^{(k_0)}(t+t_1) \leq -g^{(k_0)}(\frac{t}{1-t_1})

for {0 \leq t_1 < 1} and {t \in [0,1]}, which implies that

\displaystyle  G_{k_0}(t_1,t_1) \leq (1-t_1)_+^{k_0} G_{k_0}(0,0)

for any {t_1 \geq 0}. A similar argument in fact gives

\displaystyle  G_{k_0}(t_1+t_2,t_1+t_2) \leq (1-t_1)_+^{k_0} G_{k_0}(t_2,t_2)

for any {t_1,t_2 \geq 0}. Iterating this we conclude that

\displaystyle  G_{k_0}(\sum_{i \in S} t_i, \sum_{i \in S} t_i) \leq (\prod_{i \in S} (1-t_i)_+^{k_0}) G_{k_0}(0,0)

and similarly

\displaystyle  G_{k_0}(\sum_{i \in T} t_i, \sum_{i \in T} t_i) \leq (\prod_{i \in T} (1-t_i)_+^{k_0}) G_{k_0}(0,0).

From Cauchy-Schwarz we thus have

\displaystyle  G_{k_0}( \sum_{i \in S} t_i, \sum_{i \in T} t_i ) \leq (\prod_{i=1}^n (1 - t_i)_+^{k_0/2}) G_{k_0}(0,0).

Observe from the binomial formula that of the {3^n} pairs {(S,T)} with {S \cup T = \{1,\ldots,n\}}, {\frac{3^n+1}{2}} of them have {|S \cap T|} even, and {\frac{3^n-1}{2}} of them have {|S \cap T|} odd. We thus have

\displaystyle  -\frac{3^n-1}{2} (\prod_{i=1}^n (1 - t_i)_+^{k_0/2}) G_{k_0}(0,0) \leq G_{k_0,n}(t_1,\ldots,t_n) \ \ \ \ \ (8)

\displaystyle  \leq \frac{3^n+1}{2} (\prod_{i=1}^n (1 - t_i)_+^{k_0/2}) G_{k_0}(0,0).

We have thus established the upper bound

\displaystyle  \alpha \leq G_{k_0}(0,0) (1 + \kappa) \ \ \ \ \ (9)

where

\displaystyle  \kappa := \sum_{1 \leq n < \frac{1+4\varpi}{4\delta}} \frac{3^n+1}{2} k_0^n \int_{\frac{4\delta}{1+4\varpi} < t_1 < \ldots < t_n} (\prod_{i=1}^n (1 - t_i)_+^{k_0/2}) \frac{dt_1 \ldots dt_n}{t_1 \ldots t_n}.

By symmetry we may factorise

\displaystyle  \kappa := \sum_{1 \leq n < \frac{1+4\varpi}{4\delta}} \frac{3^n+1}{2} \frac{k_0^n}{n!} ( \int_{\frac{4\delta}{1+4\varpi} < t \leq 1} (1-t)^{k_0/2}\ \frac{dt}{t})^n.

The expression {\kappa} is explicitly computable for any given {\varpi,\delta,k_0}. Following the recent preprint of Pintz, one can get a slightly looser, but cleaner, bound by using the upper bound

\displaystyle  1-t \leq \exp(-t)

and so

\displaystyle  \kappa \leq \sum_{1 \leq n < \frac{1+4\varpi}{4\delta}} \frac{3^n+1}{2} \frac{k_0^n}{n!} (\int_{4\delta/(1+4\varpi)}^\infty \exp( - \frac{k_0}{2} t )\ \frac{dt}{t})^n.

Note that

\displaystyle  \int_{4\delta/(1+4\varpi)}^\infty \exp( - \frac{k_0}{2} t )\ \frac{dt}{t} = \int_1^\infty \exp( - \frac{2k_0 \delta}{1+4\varpi} t)\ \frac{dt}{t}

\displaystyle < \int_1^\infty \exp( - \frac{2k_0 \delta}{1+4\varpi} t)\ dt

\displaystyle  = \frac{1+4\varpi}{2k_0\delta} \exp( - \frac{2k_0 \delta}{1+4\varpi} )

and hence

\displaystyle \kappa \leq \tilde \kappa

where

\displaystyle  \tilde \kappa := \sum_{1 \leq n < \frac{1+4\varpi}{4\delta}} \frac{1}{n!} \frac{3^n+1}{2} (\frac{1+4\varpi}{2\delta} \exp( - \frac{2k_0 \delta}{1+4\varpi} ))^n.

In practice we expect the {n=1} term to dominate, thus we have the heuristic approximation

\displaystyle  \kappa \lessapprox \frac{1+4\varpi}{\delta} \exp( - \frac{2k_0 \delta}{1+4\varpi} ).

Now we turn to the estimation of {\beta}. We have an analogue of (8), namely

\displaystyle  -\frac{3^n-1}{2} (\prod_{i=1}^n (1-t_i)^{(k_0-1)/2}) G_{k_0-1}(0,0) \leq G_{k_0-1,n}(t_1,\ldots,t_n)

\displaystyle  \leq \frac{3^n+1}{2} (\prod_{i=1}^n (1-t_i)^{(k_0-1)/2}) G_{k_0-1}(0,0).

But we have an improvment in the lower bound in the {n=1} case, because in this case we have

\displaystyle  G_{k_0-1,n}(t) = G_{k_0-1}(t,0) + G_{k_0-1}(0,t) - G_{k_0-1}(t,t).

From the positive decreasing nature of {g^{(k_0-1)}} we see that {G_{k_0-1}(t,t) \leq G_{k_0-1}(t,0)} and so {G_{k_0-1,n}(t)} is non-negative and can thus be ignored for the purposes of lower bounds. (There are similar improvements available for higher {n} but this seems to only give negligible improvements and will not be pursued here.) Thus we obtain

\displaystyle  \beta \geq G_{k_0-1}(0,0) (1-\kappa') \ \ \ \ \ (10)

where

\displaystyle  \kappa' := \sum_{2 \leq n < \frac{1+4\varpi}{4\delta}} \frac{3^n-1}{2} \frac{(k_0-1)^n}{n!}

\displaystyle (\int_{4\delta/(1+4\varpi)}^1 (1-t)^{(k_0-1)/2}\ \frac{dt}{t})^n.

Estimating {\kappa'} similarly to {\kappa} we conclude that

\displaystyle  \kappa' \leq \tilde \kappa'

where

\displaystyle  \tilde \kappa' := \sum_{2 \leq n < \frac{1+4\varpi}{4\delta}} \frac{1}{n!} \frac{3^n-1}{2} (\frac{1+4\varpi}{2\delta} \exp( - \frac{2(k_0-1) \delta}{1+4\varpi} ))^n.

By (9), (10), we see that the condition (1) is implied by

\displaystyle  (1+4\varpi) (1-\kappa') > \frac{4G_{k_0}(0,0)}{k_0 G_{k_0-1}(0,0)} (1+\kappa).

By Theorem 14 and Lemma 15 of this previous post, we may take the ratio {\frac{4G_{k_0}(0,0)}{k_0 G_{k_0-1}(0,0)}} to be arbitrarily close to {\frac{j_{k_0-2}^2}{k_0(k_0-1)}}. We conclude the following theorem.

Theorem 2 Let {0 < \varpi < 1/4} and {0 < \delta < 1/4 + \varpi} be such that {MPZ[\varpi,\delta]} holds. Let {k_0 \geq 2} be an integer, define

\displaystyle  \kappa := \sum_{1 \leq n < \frac{1+4\varpi}{4\delta}} \frac{3^n+1}{2} \frac{k_0^n}{n!} (\int_{4\delta/(1+4\varpi)}^1 (1-t)^{k_0/2}\ \frac{dt}{t})^n

and

\displaystyle  \kappa' := \sum_{2 \leq n < \frac{1+4\varpi}{4\delta}} \frac{3^n-1}{2} \frac{(k_0-1)^n}{n!}

\displaystyle (\int_{4\delta/(1+4\varpi)}^1 (1-t)^{(k_0-1)/2}\ \frac{dt}{t})^n

and suppose that

\displaystyle  (1+4\varpi) (1-\kappa') > \frac{j_{k_0-2}^2}{k_0(k_0-1)} (1+\kappa).

Then {DHL[k_0,2]} holds.

As noted earlier, we heuristically have

\displaystyle  \tilde \kappa \approx \frac{1+4\varpi}{\delta} \exp( - \frac{2k_0 \delta}{1+4\varpi} )

and {\tilde \kappa'} is negligible. This constraint is a bit better than the previous condition, in which {\tilde \kappa'} was not present and {\tilde \kappa} was replaced by a quantity roughly of the form {2 \log(2) k_0 \exp( - \frac{2k_0 \delta}{1+4\varpi})}.