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
Bounds for the remaining binary positions

Definition 6.92 (The sharper binary bound). Let c≥4, D⊆{2,…,c−1}, and XD(2)<1/2. The sharper bound is the condition

S(D,1,2c−2)<2c−2.

It permits a binary completion using ranks c+1,…,2c−2, with rank c omitted, as explained below.

localBinarySuffix_two_mul_sub_two_lt_criticalCapacity_iff

Theorem 6.93 (Equivalent forms of the sharper bound). Under the hypotheses of Definition 6.92:

S(D,1,2c−2)<2c−2⟺2(2c−2)−1≤Q(D∪{c}, 2c−2).

The threshold on the right is one greater than the exact-row target 22c−3−1. Thus this is a test for omitting rank c, not for including it in the completed row.

Proof. Put M=2c−2. Since XD(2)<1/2, one has Q(D,M)<2M−1, and the integer remainder is S(D,1,M)=2M−1−Q(D,M)−1≥0 without truncation. The geometric quotient identity gives q(M,d)=2M−d for c≤d≤M, since 2d>M. In particular q(M,c)=2c−2, so

S(D,1,M)<2c−2⟺Q(D∪{c},M)≥2M−1.

For the completion, use the ranks c+1,…,M instead. Their quotient weights are 2c−3,…,1 and represent every integer from 0 to 2c−2−1. A subset of them therefore has quotient sum S(D,1,M) and completes D to the exact target 2M−1−1 while omitting c. These finite identities prove the equivalence and explain the sharper bound. They do not assert that a completed row stays below one half in real value. ◻

For example, D={2,3} and c=5 give M=8, Q(D,8)=121 and remainder 6<8. Inserting rank 5 adds 8, giving 129 rather than the exact target 127. The correct completion adds ranks 6 and 7, whose quotients are 4 and 2.

Theorem 6.94 (An unconditional bound with one extra bit). Unconditionally, for every c≥4 and every below-half core D⊆[2,c) with deficit <w(c):

S(D,1,2c−2)<2c−1.

The bound holds uniformly in c, but is twice the threshold needed in Theorem 6.93 to apply Theorem 6.75. Before using |D|≤c−2≤2c−2, the proof gives the sharper additive estimate S(D,1,2c−2)<2c−2+|D|. Thus the critical-capacity inequality would follow by excluding the integer band [2c−2,2c−2+c−3], which contains c−2 integers. This is the same band shape as Theorem 6.79’s dyadic-band condition. Its width is linear in c, whereas the square-root reset condition discussed in Section 12.5 has width 2(r+5)/2 in a different parameter r. Comparing these widths alone proves no implication between the two hypotheses.

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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