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Statement

Notation as on the Theorem 1 page: $P$, $M(S)$ and accessibility are Definitions 1, 6 and 7 (pp. 193--194), and $M(S)$ is defined for sequences of positive integers.

Theorem 4 (p. 204). Let $S=(s_1,s_2,\ldots)$ and suppose $p/q\in P((M(S))^{-1})$ with $(p,q)=1$. Then

(1) $p/q$ is $(M(S))^{-1}$-accessible, (2) $q$ divides some term of $M(S)$.

No further condition is placed on $S$.

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 4 and its proof on p. 204. 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, and the proof was checked. Nothing here is independently reviewed.

Proof pointer

P. 204. Item (1) holds because every member of $P(T)$ is $T$-accessible. For item (2), a finite sum of reciprocals of terms of $M(S)$ has the form $r/(s_1\cdots s_n)$ for some $r$ and $n$, so $ps_1\cdots s_n=qr$, and $(p,q)=1$ gives $q\mid s_1\cdots s_n$, which is a term of $M(S)$.

Dependencies

None.

Bears on

• Problem 282: only as the necessity half of Theorem 5; it says nothing about the greedy algorithm.