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
Divisor residues and the changing skip set

Expanding each divisor count separates its residue conditions. This is a finite counting identity. It does not prescribe whether an eventual proof should be analytic, combinatorial, or use another method.

Lemma 11.2 (Divisor-residue form of the short-window phase). For all M,L≥1,

ΘL(M) = ∑d≥2 ∑1≤i≤Li≡−M (mod d)2−i = ∑d≥2 2−id(M)⋅1−2−dmd1−2−d,

where id(M)∈[1,d] is the least i≥1 with d∣M+i (so id(M) depends only on Mmodd), md:=#{1≤i≤L:d∣M+i}, and terms with id(M)>L are empty. In particular, for every integer D≥M+L,

ΘL(M) = ∑d=2D ∑i=1L2−i1d∣M+i,

and the right-hand side, with an arbitrary integer x≥0 in place of M, depends on x only through its residues modulo the integers d with 2≤d≤D. The cutoff D has to be fixed before the argument varies: at (M,L)=(1,1) the cutoff is M+L=2, the integers 1 and 3 have the same residue modulo every d with 2≤d≤2, and yet Θ1(1)=1/2 while Θ1(3)=1.

Proof. τ(n)−1=∑d≥2[d∣n]; exchange the order of summation and sum the geometric progression id,id+d,id+2d,… inside [1,L]. No d>M+L divides any M+i with 1≤i≤L, which gives the cutoff form, and each indicator 1d∣x+i depends only on xmodd. ◻

The identity was also checked by exact rational computation at (M,L)∈{1000,10001,123456}×{12,20}.

A finite divisor cutoff makes the averaging question precise. For integers D≥2 and L≥1, define

ΨL,D(x)=∑d=2D∑i=1L2−i1d∣x+i,QD=lcm⁡(1,…,D).

This is the cutoff sum of Lemma 11.2. It is periodic in x with period QD, and ΘL(M)=ΨL,D(M) whenever M≥1 and D≥M+L. Averaging each indicator over a complete period gives

1QD∑x=0QD−1ΨL,D(x)=(1−2−L)∑d=2D1d.

The mean diverges as D→∞, even for fixed L. Thus this family does not converge to a finite integrable observable whose limiting mean could simply be used in an equidistribution argument.

The same issue arises if one uses the profinite integers Z^, whose Haar probability measure is uniform on each finite residue quotient. The finite-cutoff functions are continuous there. Their increasing limit is infinite almost everywhere: the events p∣x+1 are independent over primes by the Chinese remainder theorem, have probabilities 1/p, and occur infinitely often almost surely. Each contributes 1/2 to the term at offset 1. The positive integer inputs M, for which ΘL(M) is finite, do not justify treating this divergent limit as a continuous or integrable function. Only the finite period average above is used in the present argument.

There is a related residue expression in Theorem 5.15, but its index and divisor set differ:

∑d∈Dn{4n2d−1}=∑d∈Dn2−id(2n)1−2−d.

Indeed id(2n)=d−(2nmodd), with value d when the remainder is zero. The argument is 2n, not n, because the quotient scale is 4n=22n. For example, at n=7,d=3 the fractional part is 4/7; using i3(7) would instead give 2/7. This sum uses the selected finite set Dn, whereas ΘL uses all divisors with offsets truncated at L. Their common geometric summand does not make the two sums equal.

To keep the changing support visible, put Sn=∑d∈Skipn⌊4n/(2d−1)⌋. The master identity Δn=K(2n)+Sn and the recurrence for K give, for n≥6 and ℓ≥1,

2−2ℓΔn+ℓ=Δn−Θ2ℓ(2n)+2−2ℓSn+ℓ−Sn.

A phase estimate would have to control this additional term and identify the required reset rows before implying the reset bound. No such implication is proved here. The scales also differ from a logarithmic-window assertion: ℓ quotient transitions use 2ℓ binary steps starting at M=2n. If ℓ is proportional to n, the phase length is comparable to M, not to log⁡M. A theorem for logarithmic windows would need a separate argument connecting its range to the reset condition.

Finite residue expansions, divisor-sum estimates and probabilistic local estimates are possible sources of further information. None is an input already supplied here. An application must state the divisor cutoff, offset length, error at the required precision, and control on the changing skip-set sum, rather than appeal to the name of a method or to equidistribution of a different statistic.

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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