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Statement

C(r)C(r) is "the maximal length C(r)C(r) of a sequence of consecutive integers each divisible by one of rr arbitrarily chosen primes" (p. 225), and C0(r)C_0(r) "the maximal length of the sequence of consecutive integers each divisible by one of the first rr primes" (p. 226).

Corollary (p. 226, quoted). "We have C(r)≪r2log⁡2rC(r)\ll r^2\log^2r."

It follows the Theorem on the same page: for an absolute c>0c>0 and arbitrary primes q1,…,qrq_1,\ldots,q_r, r>1r>1, each interval of length c∏i≤r(1−1/qi)−1r2log⁡rc\prod_{i\le r}(1-1/q_i)^{-1}r^2\log r contains at least r2r^2 integers coprime to q1⋯qrq_1\cdots q_r. The paper prints no step between the two; the step is that (1−1/q)−1(1-1/q)^{-1} decreases in qq, so the product is at most ∏p≤pr(1−1/p)−1≪log⁡pr≪log⁡r\prod_{p\le p_r}(1-1/p)^{-1}\ll\log p_r\ll\log r by Mertens's theorem, and an interval of length ≪r2log⁡2r\ll r^2\log^2r then contains an integer coprime to QQ, so no run of consecutive integers each divisible by one of the qiq_i is that long (a filing remark, not the paper's text). The page also records the context: "Then the results of [4] imply C0(r)≪r2exp⁡(log⁡r)13/14C_0(r)\ll r^2\exp(\log r)^{13/14} while in [3] it is proved (1) C0(r)≪r2log⁡2rC_0(r)\ll r^2\log^2r. Jacobsthal asked whether C(r)=C0(r)C(r)=C_0(r) and whether C(r)≪r2C(r)\ll r^2. The aim of this paper is to prove (1) for C(r)C(r)." The introduction (p. 225) has the weaker C(r)<c(ε)r2+εC(r)<c(\varepsilon)r^{2+\varepsilon} from the Jurkat--Richert sieving limit and remarks that "by the sieve method the exponent 2 cannot be reduced". The note added in proof (p. 230) records Vaughan's C(r)≪r2log⁡4rC(r)\ll r^2\log^4r, derived from [3].

In the problems' notation. Three one-line steps made here and named as such; the paper states none of them.

  • Problem 970's h(k)h(k) is the least mm such that, for each nn with at most kk distinct prime factors, any mm consecutive integers contain one coprime to nn, that is h(k)=max⁡{j(n):ω(n)≤k}h(k)=\max\{j(n):\omega(n)\le k\} with jj Jacobsthal's function. A run of consecutive integers each sharing a factor with nn is a run each divisible by one of the ω(n)\omega(n) primes of nn, so max⁡{j(n):ω(n)=r}=C(r)+1\max\{j(n):\omega(n)=r\}=C(r)+1, and CC is nondecreasing (a run for rr primes is a run for those primes and one more), so h(k)=C(k)+1h(k)=C(k)+1. Erdős's 1965 lecture writes the same max⁡g(n)=C(r)+1\max g(n)=C(r)+1 over ν(n)≤r\nu(n)\le r. The Corollary is therefore h(k)≪(klog⁡k)2h(k)\ll(k\log k)^2, the site's statement.
  • Problem 687's Y(x)Y(x), the longest initial interval covered by one residue class per prime p≤xp\le x, is j(P(x))−1=C0(π(x))j(P(x))-1=C_0(\pi(x)), the longest run of consecutive integers each divisible by a prime p≤xp\le x ([FGKMT18] display (1.3)). Since C0(r)≤C(r)C_0(r)\le C(r), Y(x)≤C(π(x))≪π(x)2log⁡2π(x)≪x2Y(x)\le C(\pi(x))\ll\pi(x)^2\log^2\pi(x)\ll x^2, using π(x)≪x/log⁡x\pi(x)\ll x/\log x and log⁡π(x)≤log⁡x\log\pi(x)\le\log x. This is the "Y(x)≪x2Y(x)\ll x^2, which comes from Iwaniec's work [26]" of [FGKMT18] p. 4; the paper itself credits the primorial bound (1) to its [3] and proves the general bound here.
  • Problem 929's S(k)S(k) is the least xx with Y(x)≥kY(x)\ge k. If Y(x)≤Cx2Y(x)\le Cx^2 and Y(x)≥kY(x)\ge k then x≥(k/C)1/2x\ge(k/C)^{1/2}, so S(k)≫k1/2S(k)\gg k^{1/2}; this is Erdős's "Iwaniec's result B(n)>cnB(n)>c\sqrt n" of 1979, and it is stronger than the site's S(k)>k1/2−o(1)S(k)>k^{1/2-o(1)} from Rosser's sieve.

Source. H. Iwaniec, On the problem of Jacobsthal, Demonstratio Math. 11 (1978), no. 1, 225--231; the Corollary, the Theorem, the definition of C0(r)C_0(r), display (1) and Jacobsthal's questions on printed p. 226 (PDF p. 2 of the publisher's scan), the definition of C(r)C(r) and the Jurkat--Richert bound on p. 225 (PDF p. 1), the note added in proof on p. 230 (PDF p. 6), read on the page images (the text layer garbles the displays). The edition read is identified in the source digest.

Read depth. Claims checked: the Corollary, the Theorem, both definitions, display (1) and the questions were read clause by clause on the page images on 2026-09-22. The proof of the Theorem (pp. 228--230) was followed at the level of its displays and not checked; the step from the Theorem to the Corollary is the filing remark above. Nothing here is independently reviewed.

Proof pointer

Page 226: the Corollary is stated directly after the Theorem, with no printed argument; the Mertens step above supplies it. The Theorem is proved in § 3 (pp. 228--230) by the shifted sieve of § 2 with the linear-sieve weights and the two estimates (5) and (6) quoted from the author's 1971 paper; see the theorem page.

Dependencies

The Theorem (p. 226) and Mertens's theorem for the product over the first rr primes. The translation to Y(x)Y(x) uses Chebyshev's bound π(x)≪x/log⁡x\pi(x)\ll x/\log x and the identity Y(x)=j(P(x))−1Y(x)=j(P(x))-1 of Lemma 1.1 with (1.3).

Bears on

  • Problem 970: the upper bound h(k)≪(klog⁡k)2h(k)\ll(k\log k)^2 that the site attributes to the paper, now read at its source; the displayed question h(k)≪k2h(k)\ll k^2 is Jacobsthal's second question as p. 226 reports it, and the paper leaves it open.
  • Problem 687: the upper bound Y(x)≪x2Y(x)\ll x^2, through Y(x)=C0(π(x))≤C(π(x))Y(x)=C_0(\pi(x))\le C(\pi(x)); the paper does not state the bound in this form, and it settles neither displayed question there.
  • Problem 929: the lower bound S(k)≫k1/2S(k)\gg k^{1/2} by inversion of Y(x)≪x2Y(x)\ll x^2, the strongest lower bound on that page.