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Source. Theorem 2, p. 1 of the arXiv PDF; proof pp. 4--5. Read in the text layer and checked on the rendered pages.

Statement

Let $b\ge2$ be an integer that is not a proper power, and let $g:\mathbb{N}\to\mathbb{R}$ be continuous and nondecreasing with $g(n+1)/g(n)\to1$. Let $f:\mathbb{N}\to\mathbb{N}$ be nondecreasing with $f(n+1)/f(n)\sim g(n)$, and let $\alpha$ be the real number whose base-$b$ expansion is $0.f(1)f(2)f(3)\ldots$, the digits of $f(1)$ followed by those of $f(2)$, and so on. If $\alpha$ is rational, then $g(n)$ tends to a limit $c$ that is a power of $b$, and $f(n+1)-cf(n)$ is bounded.

The paper notes that rational $\alpha$ do occur: $b=10$, $f(n)=(10^n-1)/9$, $g(n)\to10$, $\alpha=1/9$. For $f(n)=a^n$ with an integer $a\ge2$ the result is Mahler's for $b=10$ and Bundschuh's for arbitrary $b$, including proper powers.

Proof structure (pp. 4--5)

If $\alpha$ is rational its digit sequence is eventually periodic with some period $p$; the fractional parts of $\log_bf(n)$ then have at most $p$ limit points, hence those of $\log_bg(n)$ have finitely many, and since $\log_bg(n+1)-\log_bg(n)\to0$ the sequence $g(n)$ converges to some $c$. If $\log c/\log b$ is rational, a periodicity argument gives $f(n+1)=cf(n)+O(1)$ and shows $c$ rational and a rational power of $b$, hence a power of $b$ as $b$ is not a proper power. If $\log c/\log b$ is irrational, equidistribution of $\{\theta n/p\}$ for irrational $\theta$ produces infinitely many $n$ for which $f(n)$ is an initial digit segment of $f(n+1)$, forcing $f(n+1)=f(n)b^k+O(b^k)$ and $c$ a power of $b$, a contradiction.

Relation to problem 251

Problem 251 asks whether $\sum p_n/2^n$ is irrational. Theorem 2 concerns the number whose base-$b$ digits are the concatenated digits of $f(n)$; a weighted sum $\sum f(n)/b^n$ is not of that form, because the base-$b$ digits of $f(n)$ (about $\log_bn$ of them for $f(n)=p_n$) overlap when placed at position $n$, with carries. For $f(n)=p_n$ the concatenation is the Copeland–Erdős number, not the series of problem 251. The theorem is therefore related to problem 251 by analogy only and records no progress on it; it is filed because this card's earlier digest singled it out.

Bears on. #251 (a mention that explains why the theorem does not apply).