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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Statement

Lemma (p. 372). Let A>0A>0 be any constant. Each of the two systems

pk−pk−1<pk+1−pk,pk−pk−1<A pk1/2(6)p_k-p_{k-1}<p_{k+1}-p_k,\qquad p_k-p_{k-1}<A\,p_k^{1/2} \qquad (6) pk+1−pk<pk−pk−1,pk+1−pk<A pk1/2(7)p_{k+1}-p_k<p_k-p_{k-1},\qquad p_{k+1}-p_k<A\,p_k^{1/2} \qquad (7)

has infinitely many solutions kk.

Read depth. Claims checked: the statement was read clause by clause on the page image of p. 372 of the print, and the proof on pp. 372--373 was followed. Nothing here is independently reviewed.

Proof pointer

Pp. 372--373; the only input is π(x)>c1x/log⁡x\pi(x)>c_1x/\log x, the paper's (5). For (6): by (5) there are infinitely many mm with pm+1−pm<c2log⁡pmp_{m+1}-p_m<c_2\log p_m, and the least k>mk>m whose next gap exceeds pm+1−pmp_{m+1}-p_m satisfies (6). For (7): if (7) failed for all p>p0p>p_0, then taking such an mm and the first prime prp_r above pm1/2p_m^{1/2}, the gaps from prp_r to pm+1p_{m+1} would be non-decreasing and all below c2log⁡pmc_2\log p_m (the paper's (8)). The paper bounds runs of equal gaps: if pt+1−pt=⋯=pt+s+1−pt+s=dp_{t+1}-p_t=\cdots=p_{t+s+1}-p_{t+s}=d then s≤ds\le d (p. 373), so m−r<(c2log⁡pm)2m-r<(c_2\log p_m)^2 and π(pm)≤pm1/2+(c2log⁡pm)2\pi(p_m)\le p_m^{1/2}+(c_2\log p_m)^2, contradicting (5).

Dependencies

None in the corpus. External input: π(x)>c1x/log⁡x\pi(x)>c_1x/\log x, cited by the paper from Ingham's The distribution of prime numbers.

Source. P. Erdős and P. Turán, On some new questions on the distribution of prime numbers, Bull. Amer. Math. Soc. 54 (1948), 371--378; the edition read is named on the source card. Used in the proof of Theorem 1.

Bears on

  • Problem 6, as context only: (6) and (7) give two consecutive gaps in increasing, and in decreasing, order infinitely often; the problem asks for three consecutive increasing gaps, which the lemma does not give.