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Statement

Setting (p. 107). $s_1<s_2<\cdots$ are the squarefree numbers, and $g_t(x)$ is the number of $s_i<x$ with $s_{i+1}-s_i=t$.

Lemma 2 (p. 107), quoted: "There exists an absolute constant $c_{17}$ so that"

$$\sum_{l>t}g_l(x)<c_{17}\,\frac{x}{t^2(\log t)^2}.$$

The print states no range for $t$ or $x$; the bound is meaningful for $t\ge2$, where $\log t>0$.

Source. P. Erdős, Some problems and results in elementary number theory, Publ. Math. Debrecen 2 (1951), 103--109, doi:10.5486/pmd.1951.2.2.04: Lemma 2 on p. 107, its proof on pp. 107--108. The edition read is identified on the source card.

Read depth. Claims checked: the statement was read clause by clause on the printed page, and the proof was read through but not checked step by step. Nothing here is independently reviewed.

Proof pointer

Pages 107--108. A gap $s_{i+1}-s_i=r>t$ contains at least $r/16$ integers divisible by the square of a prime $P>t\log t/100$: by (24), with $\sum1/p^2<3/4$ and Chebyshev's bound for $\pi(t\log t/100)$, at most $7r/8$ of the $r-1$ non-squarefree integers inside the gap are divisible by $P^2$ for a smaller prime $P$. Summing over gaps longer than $t$ ((25)) and counting the integers up to $x$ divisible by such $P^2$ ((26)) gives (27), $\sum(s_{i+1}-s_i)<16c_{18}\,x/(t(\log t)^2)$ over those gaps, and dividing by $t$ bounds their number.

Dependencies

Chebyshev's bound for the prime counting function and the convergence of $\sum1/p^2$.

Bears on

• Problem 145 and Problem 489: the lemma is the tail bound in the proof of (23) for α = 2, whose page states the relation; on its own it bounds how many long gaps there are and decides neither problem.