Two-prime running-lcm series

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For Problem 269, the case of two primes follows from Bugeaud and Laurent's Theorem 1.1 in [[../library/irrationality/bugeaud_2022_transcendence_continued_fraction_hecke_mahler/_index|their 2022 Hecke–Mahler paper]]. The deduction below is author-recorded and imports that theorem as a source result. Its exact statement and series normalization have been inspected in the primary TeX; its proof has not been independently examined. This is a partial result and supplies no conclusion for three or more primes.

Let $P$ be a finite set of primes, including the smooth integer $1$ in the increasing enumeration. At a smooth cutoff $m$, the running lcm equals

$$D(m)=\prod_{p\in P}p^{\lfloor\log_p m\rfloor}.$$

Every earlier smooth integer divides this product. Conversely each maximal prime power in the product occurs by the cutoff, so every displayed factor is required. In particular, each smooth integer contributes its own term, including cutoffs at which the lcm does not change.

Now let $P=\{p,q\}$ with $p<q$, and put

$$\theta=\frac{\log p}{\log q},\qquad x=p^{-1},\qquad y=q^{-1}.$$

Here $0<\theta<1$ is irrational: a rational slope would equate positive powers of the two different primes. At $m=p^i q^j$ one has

$$D(m)=p^{i+\lfloor j/\theta\rfloor} q^{j+\lfloor i\theta\rfloor}.$$

Therefore the full series factors as

$$S_P=AB,\qquad A=\sum_{i\ge0}x^i y^{\lfloor i\theta\rfloor},\qquad B=\sum_{j\ge0}y^j x^{\lfloor j/\theta\rfloor}.$$

All terms are positive and the two factors are bounded by convergent geometric series, so the factorization preserves both convergence and multiplicity. Define the source's Hecke–Mahler series at intercept zero by

$$F_\theta(x,y)=\sum_{i\ge1}\sum_{j=1}^{\lfloor i\theta\rfloor}x^iy^j.$$

Summing the inner geometric progression gives

$$F_\theta(x,y)=\frac{y}{1-y}\left(\frac1{1-x}-A\right).$$

Because the slope is irrational, the positive lattice points partition into $j<i\theta$ and $i<j/\theta$. The corresponding double series sum to $xy/((1-x)(1-y))$. Applying the same geometric identity to the second region and rearranging yields

$$y(1-x)A+x(1-y)B=x+y-xy.$$

Thus $B$ is an affine function of $A$ with nonzero rational slope, and $AB$ is a nonconstant quadratic polynomial in $A$ with rational coefficients. If $AB$ were algebraic, this polynomial relation would make $A$ algebraic over the algebraic numbers, hence algebraic over $\mathbb Q$. Consequently transcendence of $F_\theta(x,y)$ implies transcendence of $A$ and then of $S_P$. For $p=2,q=3$, the relation simplifies to $A+2B=4$.

Bugeaud–Laurent, arXiv:2203.12901v1, Section 1, Theorem 1.1, treats $F_{\theta,\rho}(\beta,\alpha)$ with irrational $\theta\in(0,1)$, $\rho\in[0,1)$, and nonzero algebraic $\alpha,\beta$. Its hypotheses for this value are $\beta\ne1$, $|\beta|<1$ and $|\beta\alpha^\theta|<1$. Setting

$$\rho=0,\qquad\beta=1/p,\qquad\alpha=1/q$$

meets all of them, since $\beta\alpha^\theta=p^{-2}<1$. The theorem therefore gives the transcendence needed in the preceding argument. No assumption that $\theta$ is algebraic, has bounded partial quotients, or has unbounded partial quotients is required by the numbered theorem.

The public account of Problem 269 credits Steve Fan with the two-prime factorization and conclusion in a forum post of 26 June 2026. The post is quoted and read on its claim page, and no novelty is asserted here. The stated Hecke–Mahler source is from 2022, so this is a source-supported partial deduction identified in a later problem discussion, not a new theorem for all finite prime sets.

For three or more primes, the prime exponents contain floors of sums of several logarithmic terms. The displayed separation into two one-variable series no longer follows. A further application needs an exact higher-dimensional identity and a transcendence theorem with matching hypotheses; the two-prime result alone supplies neither. That is the next source-fit question, with failure to obtain such an identity the stopping point for this route.