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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

Theorem 1.4 (p. 2, quoted). "Let c1=2(1−log⁡2)=0.6137…c_1=2(1-\log2)=0.6137\ldots. Then we have π(m+n)≤π(m)+π(n)\pi(m+n)\le\pi(m)+\pi(n) for all integers m≥n≥2m\ge n\ge2 satisfying m+n≤1020m+n\le10^{20} and

n≥2m(1−2c1log⁡m+c1).n\ge2\sqrt m\left(1-\frac{2c_1}{\log m+c_1}\right).

"

The range is bounded: it concerns only pairs with m+n≤1020m+n\le10^{20}.

Proof pointer

Section 6, pp. 6--7. The input is Dusart's Proposition 6.1 (p. 6): π(x)≤li⁡(x)\pi(x)\le\operatorname{li}(x) for real 2≤x≤10202\le x\le10^{20}, and li⁡(x)−2x/log⁡x≤π(x)\operatorname{li}(x)-2\sqrt x/\log x\le\pi(x) for real 1 090 877≤x≤10201\,090\,877\le x\le10^{20}. By Theorems 1.1 and 1.3 only nn below mmin⁡{1/1950,c0/log⁡2m}m\min\{1/1950,c_0/\log^2m\} remains, and Proposition 2.4 disposes of m≤39 687 876 365m\le39\,687\,876\,365. For larger mm, the mean value theorem and Proposition 6.1 give π(m+n)≤π(m)+2m/log⁡m+n/log⁡m\pi(m+n)\le\pi(m)+2\sqrt m/\log m+n/\log m, and Dusart's bound π(t)≥t/(log⁡t−1)\pi(t)\ge t/(\log t-1) for t≥5393t\ge5393 turns this into the comparison (6.4). The paper then checks (6.4) in four ranges of nn, from n≥mlog⁡m/log⁡log⁡mn\ge\sqrt m\log m/\log\log m down to the stated lower bound.

Read depth

Claims checked: the statement and Proposition 6.1 were read clause by clause on the pages of the copy named on the source card. The proof was read but not checked. Nothing here is independently reviewed.

Dependencies

  • Proposition 6.1 (p. 6), cited from P. Dusart, Ramanujan J. 47 (2018), 141--154, Lemma 2.2.
  • Theorem 1.1, Theorem 1.3 and Proposition 2.4.
  • π(t)≥t/(log⁡t−1)\pi(t)\ge t/(\log t-1) for t≥5393t\ge5393, cited from P. Dusart, C. R. Math. Acad. Sci. Soc. R. Can. 21 (1999), 53--59, p. 55.

Source. Christian Axler, "Some Results on a Conjecture of Hardy and Littlewood," arXiv:1909.12625v2 (2019), the edition read for the source card.

Bears on

  • Problem 855: the theorem proves the problem's inequality for integers X≥Y≥2X\ge Y\ge2 with X+Y≤1020X+Y\le10^{20} and Y≥2X(1−2c1/(log⁡X+c1))Y\ge2\sqrt X(1-2c_1/(\log X+c_1)). Every pair it covers is bounded, so it cannot decide a statement about all large XX and YY.