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Source. J. Koizumi, Irrationality of the reciprocal sum of doubly exponential sequences, arXiv:2504.05933v1 (8 April 2025); Conjecture 6 and the computer check on p. 3, Definition 7 and Lemma 8 on p. 7, Definition 9 on p. 8, the comparison with the odd greedy expansion on p. 9, Remark 17 on p. 11. Published as Integers 26 (2026), paper A28, where they are Conjecture 1 (p. 4), Definition 1 and Lemma 1 (p. 9), Definition 2 (p. 9), the comparison on p. 11 and Remark 1 (p. 13). The editions are identified on the source card.

Read depth. Claims checked: the conjecture, the two definitions, Lemma 8, Lemma 13 and Remark 17 were read clause by clause on the page images of both editions. The computer check is the author's report and was not repeated here.

Definitions

The paper writes $\lfloor x\rceil=\lfloor x+1/2\rfloor$ for the integer closest to $x$ (p. 3).

Definition 7 (p. 7). The pseudo-greedy expansion of a positive real $r$ is the sequence of positive integers

$$a_n=\left\lfloor\Bigl(r-\sum_{k=1}^{n-1}\frac1{a_k}\Bigr)^{-1}+1\right\rceil , \qquad n\ge1 .$$

By Lemma 8 (p. 7), $r=\sum_{n\ge1}1/a_n$.

Definition 9 (p. 8). Its remainder sequence is $x_n=r-\sum_{k<n}1/a_k$, so that $a_n=\lfloor x_n^{-1}+1\rceil$, and its gap sequence is $\varepsilon_n=x_n^{-1}+1-a_n$.

So $\varepsilon_n$ is the signed distance from $x_n^{-1}$ to the nearest integer, with $-1/2\le\varepsilon_n<1/2$ (p. 10). For $r=1$ the expansion is Sylvester's sequence, with $\varepsilon_n=0$ throughout (Example 11, p. 8).

Statement

Conjecture 6 (p. 3). "Let $r$ be a positive rational number and $(\varepsilon_n)_{n=1}^\infty$ be the gap sequence of the pseudo-greedy expansion of $r$. If $\lim_{n\to\infty}\varepsilon_n=0$, then $\varepsilon_n=0$ holds for $n\gg0$." The paper defines "for $n\gg0$" as holding for all $n\ge n_0$ for some positive integer $n_0$ (p. 3).

The author expects the conclusion even without the hypothesis $\varepsilon_n\to0$, and reports a computer check of the conjecture for $r=p/q$ with $0<p\le q\le10^5$ (p. 3). Once some $\varepsilon_n=0$, every later gap is $0$ (Lemma 13, p. 9).

Heuristic (Remark 17, p. 11). Writing $x_n=c_n/d_n$ with $d_n=qa_1\cdots a_{n-1}$ and $\varepsilon_n=e_n/c_n$ (Lemma 15, pp. 9--10), the positive integers $c_n$ satisfy $c_{n+1}/c_n=1-e_n/c_n$ with $-1/2\le e_n/c_n<1/2$. Modelling the ratio as uniform on $[1/2,3/2)$ gives a negative expected logarithm, $\tfrac32\log3-\log2-1<0$, so $c_n$ is expected to shrink and $e_n=0$ to occur. The paper offers this as a heuristic, not a proof.

Dependencies

None; the conjecture is open. Its equivalence with Erdős and Graham's question is Theorem 16, and its known special cases are Proposition 19 and Corollary 20.

Bears on

• Problem 243: by Theorem 16, the conjecture holds exactly when the problem's question has an affirmative answer. The computer check is evidence, not a case of the problem.
• Problem 282: the paper remarks (p. 9) that the conjecture "resembles the termination problem of the odd greedy expansion", which it records (p. 7) as open, citing Guy's problem book. No implication either way is stated or proved.