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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

For 0<a<N0<a<N let D(a,N)D(a,N) be the least possible value of nkn_k over all expansions

aN=1n1+⋯+1nk,0<n1<⋯<nk,\frac aN=\frac1{n_1}+\cdots+\frac1{n_k},\qquad 0<n_1<\cdots<n_k,

and let D(N)=max⁡0<a<ND(a,N)D(N)=\max_{0<a<N}D(a,N). Theorem 1 (p. 158): "If PP is a prime then D(P)≥P{ ⁣{log⁡2P} ⁣}D(P)\ge P\{\!\{\log_2P\}\!\}, where { ⁣{x} ⁣}=−[−x]\{\!\{x\}\!\}=-[-x] is the least integer not less than xx."

The inequality is not strict as printed. The site's commentary for Problem 305 quotes it as D(p)≫plog⁡pD(p)\gg p\log p.

Source. M. N. Bleicher and P. Erdős, Denominators of Egyptian fractions, J. Number Theory 8 (1976), 157--168; Theorem 1 on printed p. 158 (PDF p. 2), proof on p. 158. The copy read is a scan whose text layer garbles formulas; the statement was read on the page image.

Read depth. Claims checked: the statement and the definition of D(a,N)D(a,N) and D(N)D(N) (p. 158; the Egyptian form, p. 157) were read clause by clause on the page images. The proof was not checked.

Proof pointer

The proof on p. 158 tracks, in an expansion of a/Pa/P with least possible largest denominator, the denominators that are divisible by PP; it is not reconstructed here. The authors add on p. 158: "There is both theoretical and computational evidence to indicate that D(N)/ND(N)/N is maximum when NN is a prime." The closing pages tabulate D(P)D(P) for the primes up to 3737 (p. 165).

Dependencies

None outside the paper.

Bears on

  • Problem 305: the lower bound shows that the exponent 11 of log⁡b\log b in the question D(b)≪b(log⁡b)1+o(1)D(b)\ll b(\log b)^{1+o(1)} cannot be lowered.