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Statement

Printed p. 126. Let g(k)g(k) be the least, over all blocks of kk consecutive integers each greater than kk, of the number of members of the block having a prime factor greater than kk; the paper defines it as "the smallest integer so that among kk consecutive integers each greater than kk there are at least g(k)g(k) of them having prime factors greater than kk", and notes that the Sylvester--Schur theorem is the statement g(k)≥1g(k)\ge1.

Theorem 2.

g(k)=(1+o(1))klog⁡k.g(k)=(1+o(1))\frac{k}{\log k}.

A remark on printed p. 128 records that k=10k=10, n=12n=12 shows g(k)g(k) can be smaller than π(2k)−π(k)\pi(2k)-\pi(k).

Source. P. Erdős, On consecutive integers, Nieuw Arch. Wisk. (3) 3 (1955), 124--128; the definition of g(k)g(k) and Theorem 2 on printed p. 126, the proof on pp. 126--127, the remark on p. 128.

Read depth. Claims checked: the definition and the statement were read clause by clause on the page images. The proof was read for the sketch below; it is not verified.

Proof pointer

The upper bound: the block k+1,…,2kk+1,\ldots,2k contains π(2k)−π(k)=(1+o(1))k/log⁡k\pi(2k)-\pi(k)=(1+o(1))k/\log k primes, and its other members have no prime factor above kk. For the lower bound it suffices to show that for n≥kn\ge k the block n+1,…,n+kn+1,\ldots,n+k (the paper's display (9)) contains (1+o(1))k/log⁡k(1+o(1))k/\log k members with a prime factor greater than kk. The paper splits into three ranges (p. 127):

  • k≤n≤2kk\le n\le2k: the prime number theorem gives π(n+k)−π(n)=(1+o(1))k/log⁡k\pi(n+k)-\pi(n)=(1+o(1))k/\log k primes in the block.
  • 2k<n≤k3/22k<n\le k^{3/2}: by the Hoheisel--Ingham estimate (*) the block holds at least (1+o(1))k/(32log⁡k)(1+o(1))k/(\tfrac32\log k) primes, and, as n>2kn>2k, at least (1+o(1))k/(2⋅32log⁡k)(1+o(1))k/(2\cdot\tfrac32\log k) further members of the form 2p2p with p>kp>k prime; the two counts sum to (1+o(1))k/log⁡k(1+o(1))k/\log k.
  • n>k3/2n>k^{3/2}: for kk larger than some k0k_0 at least k/6k/6 members have a prime factor greater than kk, by the lemma on prime powers exactly dividing a binomial coefficient and the inequality (6) from the proof of Theorem 1, applied to (n+kk)\binom{n+k}{k}.

Dependencies

The prime number theorem; the Hoheisel--Ingham estimate (*) for primes in short intervals (cited to Ingham, Quart. J. Math. 8 (1937), 255--266); the paper's lemma (p. 126) that pα ∥ (u+tt)p^\alpha\,\|\,\binom{u+t}{t} implies pα≤u+tp^\alpha\le u+t, from Legendre's formula.

Bears on

No problem page of this corpus. The theorem counts the members of a block with a large prime factor, where Problem 961 asks for the least block length guaranteeing one; the problem page does not cite it.