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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Claim. The single Theorem of P. Erdős, On arithmetical properties of Lambert series, J. Indian Math. Soc. (N.S.) 12 (1948), 63--66, received 8 July 1948, states that for every integer tt with ∣t∣>1|t|>1 both f(1/t)f(1/t) and g(1/t)g(1/t) are irrational, where

f(x)=∑n≥1xn1−xn,g(x)=∑n≥1xn1−xnsin⁡nπ2.f(x)=\sum_{n\ge1}\frac{x^n}{1-x^n},\qquad g(x)=\sum_{n\ge1}\frac{x^n}{1-x^n}\sin\frac{n\pi}{2}.

Since f(1/t)=∑n≥11/(tn−1)f(1/t)=\sum_{n\ge1}1/(t^n-1), the theorem at t=2t=2 is the case A=NA=\mathbb N of Problem 257, answered yes; the site's remarks state this case through the identity ∑n≥11/(2n−1)=∑n≥1τ(n)/2n\sum_{n\ge1}1/(2^n-1)=\sum_{n\ge1}\tau(n)/2^n, with τ(n)\tau(n) the number of divisors of nn. For d≥1d\ge1 and A=dNA=d\mathbb N, the positive multiples of dd, the series is ∑k≥11/(2dk−1)=f(1/2d)\sum_{k\ge1}1/(2^{dk}-1)=f(1/2^d), so the theorem at t=2dt=2^d answers these instances yes as well. The proof is written for f(1/t)f(1/t): with k=[(log⁡n)1/10]k=[(\log n)^{1/10}] and the consecutive primes above (log⁡n)2(\log n)^2, the system of simultaneous congruences (2) produces a block of consecutive integers rr whose divisor counts d(r)d(r) are divisible by high powers of tt, so that the base-tt expansion of ∑rd(r)/tr\sum_rd(r)/t^r contains at least k/2k/2 consecutive zeros for arbitrarily large kk and never terminates; the case of negative tt is said to follow by the same method, and the proof for g(1/t)g(1/t) by this method and Chowla's. The closing remark lists the analogous series for ϕ(n)\phi(n), for the sum of divisors and for the number of prime factors as presenting difficulties. The source card is erdos_1948_arithmetical_properties_lambert_series. Erdős's 1968 theorem for pairwise coprime supports with convergent reciprocal sum is the accepted partial claim on its own page.

Covers. The support A=NA=\mathbb N and, through the theorem at the base t=2dt=2^d, every set of multiples A=dNA=d\mathbb N with d≥1d\ge1. Not covered: every other infinite support; the other settled classes are listed on the problem page.

Acceptance. Refereed: the Journal of the Indian Mathematical Society, new series, volume 12 (1948), pp. 63--66, received 8 July 1948 as the paper's header prints. The site labels the problem OPEN, and its remark crediting the paper with the case A=NA=\mathbb N is commentary on an open problem, not acceptance, so no reviewed evidence is listed. The proof is not checked here.

Formalization. A Lean 4 theorem in Will Cook's plectis-erdos repository, erdos_257_variants_tsum_top at the lines of the formalization link, states that ∑n≥1τ(n)/2n\sum_{n\ge1}\tau(n)/2^n is irrational, the form of the catalog's variant; the same file derives it, and an every-integer-base form erdos_1049_variants_geq_2_integer, from the repository's theorem for ∑n≥11/(bn−1)\sum_{n\ge1}1/(b^n-1) at every integer base b≥2b\ge2, which its header describes as Erdős's 1948 theorem, so the declaration is a link on this page and not an independent claim. The formal-conjectures catalog (the record link, pinned to its commit of 2026-09-23) tags its variant erdos_257.variants.tsum_top, the irrationality of ∑nτ(n)/2n\sum_n\tau(n)/2^n, research solved with that declaration as its formal proof, while its main statement erdos_257 stays research open. This corpus has not built the development or audited its statement against the theorem, so the claim carries no formalized evidence.

Depends on. Nothing in this wiki.