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Statement

Notation as on the Theorem 1 page: PP, completeness, M(S)M(S) and accessibility are Definitions 1, 2, 6 and 7 (pp. 193--194), and pp, qq are positive integers by the convention of §3 (p. 196).

Theorem 5 (p. 205). Let S=(s1,s2,…)S=(s_1,s_2,\ldots) be a sequence of positive integers such that

(1) M(S)M(S) is complete, (2) sn+1/sns_{n+1}/s_n is bounded.

Then, for (p,q)=1(p,q)=1, p/q∈P((M(S))−1)p/q\in P((M(S))^{-1}) if and only if

(3) p/qp/q is (M(S))−1(M(S))^{-1}-accessible, (4) qq divides some term of M(S)M(S).

The paper calls this the main result of the paper (p. 205). In the remark after it (p. 205) it says that no example is known showing that condition (2) cannot be omitted, and its example shows that condition (1) cannot be omitted.

Source. R. L. Graham, On finite sums of unit fractions, Proc. London Math. Soc. (3) 14 (1964), no. 2, 193--207, doi:10.1112/plms/s3-14.2.193; Theorem 5 and the remark after it on p. 205. The edition read is named on the source card.

Read depth. Claims checked: the statement was read clause by clause on the page image of the print; the proofs of its ingredients were read as their pages record. Nothing here is independently reviewed.

Proof pointer

P. 205: immediate from Theorems 3 and 4. Sufficiency is Theorem 1 without its condition (2) that sns_n be unbounded. Theorem 2 (p. 204) covers bounded SS with infinitely many terms other than 11: with k>1k>1 a value taken infinitely often, the sequence S∗=(s1,k,s2,k,k2,s3,k,k2,k3,s4,…)S^*=(s_1,k,s_2,k,k^2,s_3,k,k^2,k^3,s_4,\ldots) has M(S∗)=M(S)M(S^*)=M(S), unbounded terms and bounded ratios. The remark before Theorem 3 (p. 204) notes that if only finitely many sn≠1s_n\ne1 then M(S)M(S) is finite and so not complete. Necessity is Theorem 4.

Dependencies

Theorem 1, Theorems 2 and 3 (p. 204) and Theorem 4 of the same paper.

Bears on

  • Problem 282: the paper states in §4, without proof, applications of Theorem 5 that include an arithmetic-progression criterion containing the odd-denominator case. Theorem 5 concerns which rationals have a representation; it says nothing about the greedy algorithm or its termination.