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Statement
Setting (p. 5). Let be a positive integer and a positive real number. By a result of Panaitopol (the paper's reference [13]) there are positive reals and positive reals and with
displays (5.1) and (5.2). Let be the least positive integer such that for every .
Proposition 5.1 (p. 5, quoted). "Let be a positive integer and be positive real numbers with . Then for all real numbers with and
"
Proof pointer
P. 5. The threshold on and give , which with (5.1) bounds below by over the denominator of (5.2) taken at , display (5.3). Since and , the same denominator at is at least , so Dusart's bound for gives the matching lower bound for , displays (5.4) and (5.5). Adding these and comparing with (5.2) at gives the inequality.
Read depth
Claims checked: the setting and the statement 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
- Displays (5.1) and (5.2), cited from L. Panaitopol, Nieuw Arch. Wiskd. (5) 1 (2000), 55--56.
- for , 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: for real the proposition proves the problem's inequality once , and is beyond the stated thresholds, for any admissible choice of , , and expansion constants. With , together with Theorem 1.1 for smaller arguments, it yields Theorem 1.3. Pairs with are not covered, so it does not decide the problem.