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Claim. Theorem 1.3 of Terence Tao and Joni Teräväinen, Quantitative correlations and some problems on prime factors of consecutive integers, arXiv:2512.01739 (v1 2025-12-01, v2 2026-04-25), states that

$$\sum_{n\ge1}\frac{\omega(n)}{2^n}=\sum_p\frac{1}{2^p-1}$$

is irrational, the sum on the right running over the primes. This is the case of Problem 257 in which $A$ is the set of primes, answered yes; the paper's introduction names it as the primes case of this problem, and it is the question of Problem 69, where the same theorem is the accepted full claim on its claim page. The identity comes from $\omega(n)=\sum_{p\mid n}1$ and the geometric series $\sum_{m\ge1}2^{-pm}$. The authors write that a similar argument can also establish the case of the prime powers, leaving its details to the reader, so that case is not covered here. The main tool, as the source card tao_2025_quantitative_correlations_problems_prime_factors_consecutive records, is a quantitative two-point correlation estimate for multiplicative functions (their Theorem 3.1) combined with probabilistic and circle-method arguments; this outline is a reading aid, not proof coverage.

Covers. The support $A$ equal to the set of primes. Not covered: the prime powers, which the paper only sketches, and every other infinite support.

Acceptance. Reviewed: Thomas Bloom, the site's curator, labels Problem 69, whose statement is exactly this instance, PROVED (page last edited 2026-04-15) and credits the unconditional proof to Tao and Teräväinen in its remarks; the remarks of Problem 257, which the site labels OPEN, record the same credit for the primes case. No journal publication of the preprint is recorded (the arXiv record lists no journal reference), so the claim carries no refereed evidence. The formal-conjectures file for Problem 69 (the record link, pinned to its commit of 2026-10-06) tags its specialization of Problem 257 to the primes research solved and records no formal proof. This corpus has not reproved the theorem and awards no tier of its own.

Depends on. Nothing in this wiki.