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Source. Theorem 3, p. 58, with its proof on p. 63, of P. Erdős, On some applications of Brun's method, Acta Univ. Szeged. Sect. Sci. Math. 13 (1949), 57--63, as identified on the source card.

Statement

Write p1<p2<⋯p_1<p_2<\cdots for the primes in increasing order, as the paper does (p. 58).

Theorem 3 (p. 58). Let c5c_5 be any constant and nn sufficiently large. Then there are a constant c6=c6(c5)c_6=c_6(c_5) and primes pk<pk+1<⋯<pk+r<np_k<p_{k+1}<\cdots<p_{k+r}<n, with r=[c6log⁡n]r=[c_6\log n], such that

pk+i+1−pk+i>c5,i=0,1,…,r−1.p_{k+i+1}-p_{k+i}>c_5,\qquad i=0,1,\ldots,r-1.

The print calls these "[c6log⁡n][c_6\log n] primes", although the list pk,…,pk+rp_k,\ldots,p_{k+r} has r+1=[c6log⁡n]+1r+1=[c_6\log n]+1 members; the display concerns the rr gaps between them. The print places the constant c6c_6 after nn, but it depends only on c5c_5, as the notation c6(c5)c_6(c_5) says and the proof shows.

The paper presents the theorem (p. 58) as a sharpening of Sierpiński's result that lim sup⁡min⁡(pn+1−pn, pn−pn−1)=∞\limsup\min(p_{n+1}-p_n,\,p_n-p_{n-1})=\infty, that is, that infinitely many primes are isolated on both sides.

Proof pointer

Page 63. By Schnirelmann's sieve bound, the number of mm with pm+1−pm≤c5p_{m+1}-p_m\le c_5 and pm≤np_m\le n is less than a constant times n/(log⁡n)2n/(\log n)^2, while π(n)\pi(n) exceeds a constant times n/log⁡nn/\log n; so the gaps of size at most c5c_5 are too few to break every run of [c6log⁡n]+1[c_6\log n]+1 consecutive primes below nn when c6c_6 is small in terms of c5c_5. The paper says this gives the theorem immediately.

Read depth

Claims checked: the statement was read clause by clause on the printed p. 58 and the proof on p. 63. Nothing here is independently reviewed.

Bears on

  • Problem 238: the problem fixes c1,c2>0c_1,c_2>0 and asks whether every sufficiently large xx has more than c1log⁡xc_1\log x consecutive primes ≤x\le x with all pairwise differences greater than c2c_2. Theorem 3 with c5=c2c_5=c_2 and n=xn=x gives [c6log⁡x]+1>c6log⁡x[c_6\log x]+1>c_6\log x consecutive primes below xx whose successive gaps, and hence (as the primes increase) all pairwise differences, exceed c2c_2; this answers the question yes for every pair with c1≤c6(c2)c_1\le c_6(c_2). The paper gives no value of c6(c2)c_6(c_2), and the theorem says nothing about larger c1c_1.