Plectis

Long paper

Reciprocal Mersenne Subseries

Summary. A weighted summability criterion proves irrationality of reciprocal Mersenne subseries: if h is the part of a supported on a finite nonempty prime set P, every infinite A with ∑ h(a)/(a(bh(a)−1)) finite has irrational ∑ (ba−1)−1. The base-two condition gives the conclusion at every integer base for every infinite subset, and permits an example with divergent reciprocal sum, while excluding full support and all odd exponents. The proof averages over a finite part of the support and over dyadic observation lengths. Erdős's weaker reciprocal-summable theorem is proved in full, and a common window extends both arguments to unions of weighted and positive-cover supports. Finite-denominator and achievement-set calculations formulate tests for the targets 1/2 and 1/21; those tests do not decide either target or the universal question. Prime support at base 2 and squarefree support at power-of-two bases are attributed as existing work.

This paper covers a claim-bounded problem-specific reasoning surface for Erdős #257: selected checked premises, finite computations, closed routes, surviving obligations, and their exact evidence bands.

It does not establish the validity of claims tagged Lean, which belongs to the cited kernel-checked source, or a solution to Erdős #257, whose universal and half-value questions remain open.

In this paper

Introduction

For an integer b≥2, the series XA(b)=∑a∈A(ba−1)−1 is irrational for several classes of infinite supports A. Erdős proved the full-support case and later stated a criterion assuming only convergence of ∑a∈A1/a; the periodic-support results belong to Luca and Tachiya. We begin with these antecedents and prove two hereditary criteria which also admit supports of divergent reciprocal sum. The proofs subtract an explicitly chosen integer from (bN−1)XA(b), leaving a positive difference Δb,A(N). If XA(b)=p/q with q>0, then qΔb,A(N)∈Z, so Δb,A(N)≥1/q. We rule out rationality by making this difference arbitrarily small. In a finite average, the incomplete residue periods produce an error without the factor 1/a needed for summation over exponents. Averaging again over dyadic lengths restores this factor. The decisive point for the union theorem is stronger than proving the two criteria separately: both errors are controlled on the same finite distribution, and one index is chosen only after they have been added. Keeping the modulus dependence explicit makes that common distribution available to the cover estimate.

Part I gives the support theorems and the proofs followed by the short paper, Irrationality criteria for Lambert subseries. The subsequent parts concern a different question: whether a prescribed rational number belongs to the Mersenne achievement set. The finite identities in Sections 5–8 describe the integer quotients and their relation to the real greedy sequence. Sections 9–12 state the additional uniform or unbounded-depth assertions needed to apply them. Notation specific to a recurrence is introduced with that recurrence.

Section 13 returns to the support criteria. A logarithmic divisor cost cannot uniformly replace the fractional-cover cost when sampling on multiples of a prescribed modulus, even over complete periods. We prove the counterexample and an initial-interval estimate that survives it.

None of these arguments decides the universal problem, half-membership or 1/21-membership. The conditional results specify sufficient hypotheses or equivalent membership conditions, and the computations cover only the stated finite ranges. Each counterexample concerns the particular estimate under discussion.

Notation used throughout

For A⊆N>0 and an integer b≥2, put

XA(b)=∑a∈A1ba−1,cA(n)=#{a∈A:a∣n}(n≥1).

At base two write wn=(2n−1)−1, Rn=∑j>nwj and A={XA(2):A⊆N>0}. For x≥0 the real greedy remainder is defined by r0(x)=x and

rn(x)=rn−1(x)−{wn,wn≤rn−1(x),0,wn>rn−1(x).

The greedy support Gx consists of the selected indices; G=G1/2. For the half target index 1 is skipped. Thus a recurrence begun at index 2 with remainder 1/2 is the same process, with its index shifted.

The finite quotient arguments use

q(M,d)=⌊2M2d−1⌋,Q(D,M)=∑d∈Dq(M,d).

For M≥k their nonnegative binary remainder and signed next-step expression are

S(D,k,M)=max{2M−k−Q(D,M)−1,0},H(D,k,n)=2S(D,k,n−1)+1−cD(n)(n>k).

The maximum matters: subtraction in the corresponding natural-number source definition is truncated at zero. Statements that remove the maximum establish nonnegativity first.

The integer carry for a support omitting 1 is

ihc⁡(A,N)=2N−∑j=2N+12N+1−jcA(j),CA(N)=ihc⁡(A,N)−1.

Some finite-depth statements use KA(m)=ihc⁡(A,m−1). The recurring bound B(m)=2⌊m⌋+4 is integral; bounds written with m rather than its floor are real-valued bounds. Quantities specific to one argument are defined in that argument. Some formal sources allow a support to be a subset of N, including zero. In all reciprocal sums in this paper only positive exponents contribute: for such a support, XA(b) means XA∩N>0(b). The expression 1/(b0−1) is never evaluated.

About this paper

Authorship and AI use. Will Cook built and directed the research infrastructure and maintains the public release. He reviewed claims when he could. AI agents did most of the research and drafting. Cook did not independently verify every claim.

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