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Source. T. Tao, discussion comment on the Problem 251 page, 17:17 on 07 Oct 2025 (site clock): "By summation by parts, this is equivalent to the irrationality of ∑npn+1−pn2n\sum_n\frac{p_{n+1}-p_n}{2^n}." The comment states the equivalence in one line; the proof below is written out here and checked; it is elementary and complete.

Statement

Let 2=p1<p2<⋯2=p_1<p_2<\cdots be the primes and gm=pm+1−pmg_m=p_{m+1}-p_m the prime gaps. Both series below converge and

S:=∑n≥1pn2n=2+∑m≥1gm2m,S:=\sum_{n\ge1}\frac{p_n}{2^n}=2+\sum_{m\ge1}\frac{g_m}{2^m},

so SS is irrational if and only if ∑m≥1gm2−m\sum_{m\ge1}g_m2^{-m} is irrational.

Proof

Convergence: by Chebyshev's elementary bound pn≪nlog⁡np_n\ll n\log n, the terms pn2−np_n2^{-n} are O(nlog⁡n 2−n)O(n\log n\,2^{-n}), so ∑npn2−n\sum_np_n2^{-n} converges; the identity below then gives the convergence of the gap series, whose terms are positive.

For every n≥1n\ge1,

pn=2+∑m=1n−1gm,p_n=2+\sum_{m=1}^{n-1}g_m,

the sum being empty for n=1n=1. Multiply by 2−n2^{-n} and sum over n≥1n\ge1. All terms are nonnegative, so the order of summation may be exchanged:

∑n≥1pn2n=2∑n≥112n+∑m≥1gm∑n≥m+112n=2+∑m≥1gm2m,\sum_{n\ge1}\frac{p_n}{2^n} =2\sum_{n\ge1}\frac1{2^n}+\sum_{m\ge1}g_m\sum_{n\ge m+1}\frac1{2^n} =2+\sum_{m\ge1}\frac{g_m}{2^m},

since ∑n≥12−n=1\sum_{n\ge1}2^{-n}=1 and ∑n≥m+12−n=2−m\sum_{n\ge m+1}2^{-n}=2^{-m}. Hence ∑m≥1gm2−m=S−2\sum_{m\ge1}g_m2^{-m}=S-2, and SS is rational exactly when the gap series is rational. □\square

Numerically S=3.67464396601…S=3.67464396601\ldots (OEIS A098990), so the gap series equals 1.67464396601…1.67464396601\ldots.

Variants used by the 2026 manuscripts

For an integer base B≥2B\ge2 the same computation gives ∑n≥1gnB−n=(B−1)∑n≥1pnB−n−p1\sum_{n\ge1}g_nB^{-n}=(B-1)\sum_{n\ge1}p_nB^{-n}-p_1 (Ringer's equation (1)); with zero-based indexing p0=2p_0=2, gn=pn+1−png_n=p_{n+1}-p_n, it reads ∑n≥0pn2−n−1=2+∑n≥0gn2−n−1\sum_{n\ge0}p_n2^{-n-1}=2+\sum_{n\ge0}g_n2^{-n-1} (Cook's form). Land works with the weighted tails Gn=∑j≥0gn+j2−j−1G_n=\sum_{j\ge0}g_{n+j}2^{-j-1}, whose first value is G1=S−2G_1=S-2.

Relation to Problem 251

The identity changes nothing about the problem's status; it shows that the question is about the binary expansion of the dyadic series of prime gaps, where each gap gm≍log⁡mg_m\asymp\log m on average occupies several binary digits and the contributions overlap, so carries couple distant terms. Every 2026 manuscript on the problem starts from this form.

Bears on. #251 (an exact reformulation, not progress).