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Statement

Lemma 5 (printed p. 383). For each $N>0$, however large, there is $x_0(N)$ with the following property. Given integers $x\ge x_0$, $y$ (of any size and sign) and $a>0$, some arithmetic progression $b+a,b+2a,\ldots,b+ta$ with common difference $a$ has (a) length $t\ge N\log x$, (b) first term $b+a$ in the interval $y<b+a\le y+x$ of length $x$, and (c) every term divisible by some prime $p\le(\log x)/N$.

The authors' remarks (p. 383): the lemma "is merely an extension of the Westzynthius--Erdös--Rankin result ([17], [1], [12]) that $p_{n+1}-p_n$ sometimes exceeds 'any constant' times $\log p_n$ (to obtain this last, set $a=1$, $y=x$)"; the condition $p\le(\log x)/N$ means the terms have "rather small factors"; and "the crucial point involves getting the first term $b+a$ to fall between $y$ and $y+x$."

Source. D. Hensley and I. Richards, Primes in intervals, Acta Arith. 25 (1973/74), 375--391; Lemma 5 and its proof on printed pp. 383--384 (PDF pp. 5--6 of the retained scan), read on the page images.

Read depth. Claims checked: the statement and the remarks were read clause by clause on the page image. The proof (pp. 383--384) was read for its structure; nothing is independently reviewed.

Proof pointer

Combine Lemma 4 (given the sieving bound $T=T(t)$ of Lemma 3, for every $a>0$ some $b$ makes every term of $b+a,\ldots,b+ta$ divisible by a prime $p\le T$, by the Chinese remainder theorem) with Lemma 3 ($T(t)=o(t)$): since $N$ is fixed, $t\ge N\log x$ and $p\le(\log x)/N$ hold together once $t$ and $x$ are large. For (b), the first term may be moved by any multiple of $\prod_{p\le(\log x)/N}p$, which the prime number theorem for $\psi$ makes about $x^{1/N}$, much smaller than $x$, so $b+a$ can be placed in any interval of length $x$.

Dependencies

Lemma 3 (Mertens's theorem and a two-range hard sieve), Lemma 4 (the Chinese remainder theorem), the prime number theorem for $\psi(x)$.

Bears on

• Problem 1204: the lemma is what the Polymath paper cites ("It follows from Lemma 5 of [45] that one can take $m=o(k/\log k)$", p. 78) for the second-order bound (150) on the diameter of the narrowest admissible $k$-tuple, the problem's $A(k)$.