---

Claim. Let $P=\{p,q\}$ with primes $p<q$, and let $S$ be the sum of Problem 269 for this $P$, with every term retained, including those at which the running lcm repeats. Then $S$ is transcendental, and in particular irrational. Steve Fan posted the proof in the problem's thread on 26 June 2026. With $\beta=\log q/\log p$, which is irrational, the running lcm at $a_n=p^iq^j$ is $p^{i+\lfloor j\beta\rfloor}q^{j+\lfloor i/\beta\rfloor}$, so $S=S_1S_2$ with

$$S_1=\sum_{i\ge0}p^{-i}q^{-\lfloor i/\beta\rfloor},\qquad S_2=\sum_{j\ge0}p^{-\lfloor j\beta\rfloor}q^{-j}.$$

For the Hecke–Mahler series $F_\theta(z_1,z_2)=\sum_{i\ge1}z_1^i\sum_{1\le j\le i\theta}z_2^j$, the post shows that $S_2=1+1/(q-1)-(p-1)F_\beta(1/q,1/p)$ and, from the reflection identity $F_\theta(z_1,z_2)+F_{1/\theta}(z_2,z_1)=z_1z_2/((1-z_1)(1-z_2))$, that $(p-1)S_1+(q-1)S_2=p+q-1$. So $S$ is a quadratic polynomial with rational coefficients in $S_2$. The post takes the transcendence of $F_\beta(1/q,1/p)$ from the classical theory of Hecke–Mahler series, naming no theorem, and concludes that $S_2$ and then $S$ are transcendental.

Submission note. Posted to the erdosproblems.com forum by Steve Fan (account Steve_Fan) on 26 June 2026:

Here is a short proof of the case $|P|=2$, which actually shows that $S$ is transcendental. Let $P =\{p_1<p_2<...<p_k\}$. Then each $a_n$ is of the form $p_1^{i_1}\cdots p_k^{i_k}$, where $i_1,...,i_k\ge0$, and

$$> \text{lcm}[a_1,...,a_n]=\prod_{j=1}^{k}p_j^{\big\lfloor \frac{\log a_n}{\log p_j}\big\rfloor}. >$$

Hence

$$> S=\sum_{i_1,...,i_k\ge0}\prod_{j=1}^{k}p_j^{-\big\lfloor \frac{\log p_1^{i_1}\cdots p_k^{i_k}}{\log p_j}\big\rfloor}. >$$

Specializing to the case $P=\{p<q\}$ and setting $\beta=\log q/\log p>1$ which is irrational, we have

$$> S=\sum_{i,j\ge0}p^{-i-\lfloor j\beta\rfloor}q^{-j-\lfloor i\beta^{-1}\rfloor}=S_1S_2, >$$

where \begin{align*} S_1&:=\sum_{i\ge0}p^{-i}q^{-\lfloor i\beta^{-1}\rfloor},\ S_2&:=\sum_{j\ge0}p^{-\lfloor j\beta\rfloor}q^{-j}. \end{align*} We can relate $S_1,S_2$ to the Hecke--Mahler series

$$> F_{\theta}(z_1,z_2):=\sum_{i_1\ge1}z_1^{i_1}\sum_{1\le i_2\le i_1\theta}z_2^{i_2}, >$$

where $|z_1|,|z_2|<1$ and $\theta>0$. When $\theta\ge1$, we have

$$> F_{\theta}(z_1,z_2)=\frac{z_2}{1-z_2}\left(\frac{z_1}{1-z_1}-\sum_{i_1\ge1}z_1^{i_1}z_2^{\lfloor i_1\theta\rfloor}\right), >$$

and so

$$> S_2=\frac{1}{q-1}-(p-1)F_{\beta}(1/q,1/p)+1. >$$

Moreover, for any $\theta>0$,

$$> F_{\theta}(z_1,z_2)+F_{\theta^{-1}}(z_2,z_1)=\sum_{i_1,i_2\ge1}z_1^{i_1}z_2^{i_2}=\frac{z_1z_2}{(1-z_1)(1-z_2)}. >$$

This yields the nice identity

$$> (p-1)S_1+(q-1)S_2=p+q-1, >$$

which allows one to write $S$ as a quadratic polynomial with rational coefficients in $S_2$. We know by the classical theory of Hecke--Mahler series that $F_{\beta}(1/q,1/p)$ is transcendental. Therefore, $S_2$ is also transcendental, and so is $S$.

This argument does not seem to generalize immediately to $|P|\ge3$, since $S$ does not factor nicely in the first place.

Covers. Every $P$ of exactly two primes: the sum is transcendental, so the answer is yes for those sets. Sets of three or more primes are not covered; the post itself says that the argument does not seem to generalize immediately to $|P|\ge3$, since the sum does not factor there.

Claimant and attribution. The post appears under the account Steve_Fan, whose display name is Steve Fan, and mentions no AI assistance. The same day the account old-bielefelder replied that the argument works for any two coprime integers in place of the two primes, and Fan agreed in a further post, suggesting more generally sets of nonzero algebraic numbers whose logarithms are linearly independent over $\mathbb Q$, ordered by their norms; that suggestion carries no proof and lies outside this problem. Will Cook's post of 11 September 2026 in the same thread (display name williamwkcook) states that the two-prime factorization, the Hecke–Mahler reduction and the transcendence conclusion are Fan's. Cook's write-up, Running least common multiples, reproduces the calculation with that credit and claims no priority for it.

The transcendence input. The post's slope $\beta=\log q/\log p$ exceeds $1$, while Bugeaud and Laurent's Theorem 1.1 (its card) treats slopes in $(0,1)$. By the post's reflection identity, $F_\beta(1/q,1/p)$ is a rational number minus $F_{1/\beta}(1/p,1/q)$, the value of slope $\log p/\log q\in(0,1)$ whose fit to that theorem's hypotheses the two-prime research page checks. Cook's write-up attributes the same input to Loxton and van der Poorten, in Bugeaud and Laurent's formulation.

Relation to the corpus's research. The research page records Plasma AI's author-recorded deduction of the same conclusion, through the factor $S_1$ (its $A$) rather than $S_2$, with Bugeaud and Laurent's theorem named; its affine relation between its two factors is the post's identity $(p-1)S_1+(q-1)S_2=p+q-1$ divided by $pq$. It asserts no novelty, and no claim card or other claim page covers the two-prime case.

Depends on. No page of this wiki. The argument takes the transcendence of $F_\beta(1/q,1/p)$ from the classical theory of Hecke–Mahler series, an outside input the post does not cite further.

Standing. Claimed. The erdosproblems.com page for the problem, last edited 28 December 2025, does not mention the post, and its proof-claims list was empty when read on 2026-10-18; Cook's credit in the thread is attribution, not review. There is no refereed publication and no Lean proof checked here. Cook's write-up reports a formal proof of the deduction that assumes the cited value theorem; this corpus has not built it. Plasma AI's research page is the project's own deduction and is not outside acceptance.