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Source. Algorithm 3.1 and the verification after it, preprint p. 6; the abstract and the introduction, pp. 1--2. Read on the rendered pages.

Statement

Algorithm 3.1 (p. 6) takes a monotonically non-decreasing sequence of positive integers $\{b_n\}_{n=1}^{\infty}$ such that $T=\sum_{n=1}^{\infty}b_n2^{-n}$ converges, and a real number $S\in(\frac T3,T]$. It sets $S_1=S$ and $T_N=\sum_{n=N}^{\infty}b_n2^{-(n-N)}$, and defines, for $n=1,2,\ldots$,

$$a_n=\begin{cases} 2 & \text{if } \frac13T_n<S_n\le\frac12T_n,\\ 3 & \text{if } \frac14T_n<S_n\le\frac13T_n,\\ 4 & \text{if } \frac16T_n<S_n\le\frac14T_n, \end{cases} \qquad S_{n+1}=a_nS_n-b_n .$$

The verification on the same page shows that every $a_n$ is defined and lies in $\{2,3,4\}$ and that

$$\sum_{n=1}^{\infty}\frac{b_n}{a_1\ldots a_n}=S .$$

So every real number in $(T/3,T]$, rational or not, is the value of such a series with all denominators in $\{2,3,4\}$.

Why it works (p. 6)

Since $T_1=2T$, the starting value lies in $(\frac16T_1,\frac12T_1]$. The choice of $a_n$ puts $a_nS_n$ in $(\frac23T_n,T_n]$. Since $T_n=b_n+\frac12T_{n+1}$ and the monotonicity of $b_n$ gives $T_{n+1}\ge2b_{n+1}\ge2b_n$, the new value $S_{n+1}$ falls again in $(\frac16T_{n+1},\frac12T_{n+1}]$. The partial sums differ from $S$ by $S_{N+1}/(a_1\cdots a_N)$, which is at most the tail $\sum_{n>N}b_n2^{-n}$ of $T$, and so tends to $0$.

In the paper

The abstract announces the result for "every monotonic sequence $\{b_n\}_{n=1}^{\infty}$" with a rational value of the series, and the introduction (p. 2) for any monotonic sequence of positive integers with "a wanted value in a prescribed interval"; the algorithm as printed treats non-decreasing sequences with $T$ convergent. The introduction uses it to show that "Some growth condition on $\{a_n\}_{n=1}^{\infty}$ and $\{b_n\}_{n=1}^{\infty}$ is needed" for criteria of the kind the paper proves, in which the sum is irrational unless $b_n/(a_n-1)$ is eventually constant. The Remark before it (p. 6) gives the case $b_n=1$: uncountably many bounded sequences $a_n$ with $\sum1/(a_1\cdots a_n)=7/11$.

Relation to problem 251

With $b_n=p_n$, the $n$-th prime, the series $T$ is the constant $\sum p_n/2^n$ of problem 251, and the case $S=T$ of the algorithm returns $a_n=2$ for every $n$. Every other number of $(T/3,T]$, among them infinitely many rationals, is $\sum p_n/(a_1\cdots a_n)$ for some sequence with all $a_n\in\{2,3,4\}$; such a sequence is bounded and is not covered by the growth conditions of Theorem 5.1 or Theorem 6.1. The algorithm says nothing about whether $T$ itself is rational.

Bears on. #251 (context: the problem's constant is the right end of the interval; every other value in it, rationals included, is a sum of $p_n/(a_1\cdots a_n)$ with all $a_n$ in $\{2,3,4\}$; nothing about the constant itself).