---

Source. J. Koizumi, Irrationality of the reciprocal sum of doubly exponential sequences, arXiv:2504.05933v1 (8 April 2025); Remark 22 on p. 13, Example 11 on p. 8. Published as Integers 26 (2026), paper A28, where they are Remark 4 (p. 16) and Example 1 (p. 10). The editions are identified on the source card.

Read depth. Claims checked: the remark and Example 11 were read clause by clause on the page images of both editions. The transcendence of $c(m)$ is the paper's citation of Dubickas and was not checked here; nothing here is independently reviewed.

Statement

For a positive integer $m$, Example 11 (p. 8) defines $s_1(m)=m+1$ and $s_{n+1}(m)=s_n(m)^2-s_n(m)+1$, the pseudo-greedy expansion of $1/m$, and cites the constant $c(m)>1$ with $s_n(m)\approx c(m)^{2^n}$ (ratio tending to $1$), shown irrational by Wagner and Ziegler and transcendental by Dubickas.

Remark 22 (p. 13). Suppose Question 5 has an affirmative answer, equivalently that Conjecture 6 holds. Then the exceptional set of Theorem 4 is

$$(1,\infty)\setminus\mathcal I=\{c(m)^{2^{-N}}: N\ge0,\ m>0\},$$

so every real algebraic $\alpha>1$ lies in $\mathcal I$; in particular $2^{2^n}$ would be a Type 2 irrationality sequence.

Proof pointer

Page 13. If $a_n\approx\alpha^{2^n}$ has rational reciprocal sum, the assumed answer makes the sequence eventually follow $a_{n+1}=a_n^2-a_n+1$, so $a_{N+n}=s_n(m)$ for some $N\ge0$ and $m>0$, and $\alpha=c(m)^{2^{-N}}$. The transcendence of $c(m)$ makes every such $\alpha$ transcendental.

Dependencies

Theorem 16 and Theorem 4 of the same paper; A. Dubickas, Ramanujan J. 57 (2022), 569--581, for the transcendence of $c(m)$, as the paper cites it (not held here).

Bears on

• Problem 263: a conditional implication. An affirmative answer to the question of Problem 243 would answer the first question of Problem 263 affirmatively; the remark proves nothing unconditionally, and says nothing on the second question.
• Problem 243: the remark records a consequence of an affirmative answer, not progress on the question.