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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
General identities for perturbed greedy recurrences

Theorem 6.124 (A general perturbed greedy recurrence). Let the old integer values be separated by at least g≥1, and let the update be t(x)=4s(x)+p(x) with 0≤p(x)≤B<g. Suppose x− and x+ are the adjacent old values on either side of capacity C. Theorem 6.62 gives the largest admissible updated value at capacity 4C+g and the exact three-branch remainder after testing the extra weight 2g+4.

The underlying structure in Definition 6.61 assumes only B<3g; that suffices for order preservation, but the cited maximality application has the additional hypothesis B<g. The counterexample after Theorem 6.62 shows what can fail without it. For a maximum over both choices of the extra weight, rather than the stated two-stage rule, one also needs a separation argument; 4g−B≥2g+4 suffices, as shown after that theorem.

Theorem 6.125 (Spacing of compatible reverse-carry words). Consider two integer carry recurrences bi(m)+2ui(m)=ai(m)+ui(m+1), i=1,2. If a1(k)=a2(k) and the output bits at k are 1 and 0, then the carry difference at k+1 is odd. If the coefficients and bits agree for the following L positions, the difference at k+L+1 is 2L times that odd integer. Consequently absolute terminal bounds B1,B2 give 2L≤B1+B2. Lemma 6.46 states the exact identity and proves it by subtraction. The linked formal statements encode these agreements; they do not supply terminal bounds for an unrelated digit system.

Proposition 6.126 (Changing one support bit at a doubled rank). Let two supports agree except that the second includes N+1 while the first does not. At the argument 2(N+1) their divisor counts differ by exactly 1, because N+1 divides 2(N+1). More generally the difference at a positive argument m is 1N+1∣m. The linked statements apply this identity to the finite supports in their hypotheses. No conclusion about a different coefficient sequence follows without identifying its own support change.

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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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