Plectis

Long paper

The Binary Totient Series

Erdős #249 By July 2026; revised 1 October 2026 Authorship, AI use and citation

Summary. The dyadic totient sections through level e≥1 have an explicit rational basis of size 2e+1. Truncating a scaled tail difference produces an integer whose residue certifies nonintegrality when its distances from both endpoints of the residue interval exceed an explicit error bound. The record develops the resulting finite denominator exclusions, equivalent residue criteria, and sufficient exponential-sum estimates. Rational comparison sequences identify the limitations of particular size, parity, rank, and finite-prefix arguments. The short paper supplies the all-base basis theorem; here each conditional result carries the arithmetic hypothesis still needed to apply it. Those hypotheses are not established in the form required to prove irrationality of S=∑ φ(n)2−n.

This paper covers a claim-bounded problem-specific reasoning surface for Erdős #249: selected checked premises, finite certificates, 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 #249, which remains open.

In this paper

A bound on the last totient difference

We compare the last totient difference with a centred residue. To use the comparison for irrationality, it would have to hold for arbitrarily large a, with q in the prescribed range.

Definition 9.15. The terminal dominance condition, which is unproved, is the following. Writing H=H(2a),

∀a0 ∃a,q, a0≤a∧8≤a∧2q+2+(a+6)<2⋅2a∧2(H+q+2)≤4q∧
δ2a(2q+2)≤2⋅ua,q.

PowerTwoFlexibleActualTerminalDominanceSupply

Theorem 9.16 (The proved implication). The dominance hypothesis at a single odd rank already excludes integrality of Ωa, and the terminal dominance condition implies S∉Q.

Observation 9.17 (How the sign and size bounds enter the contradiction). Use the hypotheses and notation of Proposition 6.66. Under the assumption z=Ωa∈Z, the exact identity is

2ua,q=δ2a(2q+2)−cH,H,z(2q+1),0<cH,H,z(2q+1)<2H+2q+3.

Terminal dominance δ2a(2q+2)≤2ua,q would contradict the positive lower bound on the carry. The other sufficient inequality, 2ua,q≤δ2a(2q+2)−(2H+2q+3), would contradict its upper bound. Both are valid ways to contradict integrality.

These bounds do not eliminate either approach: they hold under the integrality assumption that the approach is intended to refute. They neither prove nor disprove an unconditional estimate for ua,q. Finite examples lying between the two boundaries do not establish an asymptotic exclusion of either inequality. The identities and conditional bounds are proved in Lean (two_mul_actualOddHalfCenteredLift_eq_terminal_sub_trueCarry, actualLcm_trueEndpointSurvivor_neg, actualLcmTailOrbit_notMem_int_of_actualTerminalCarryCorridorEscape).

Proposition 9.18 (A sufficient extension). Either of the following suffices: (i) the lower-escape branch cofinally, as displayed just above; or (ii) the two-sided magnitude form (item 4 of Proposition 9.14), which asks only H+q+2≤|ua,q| and which the Lean source proves implies corridor escape via a clean sign split (positive branch escapes above the terminal letter, negative branch escapes below the directed bound), without prescribing the sign of the centred representative. Neither branch is supplied by the conditional sign identity itself.

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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