The Binary Totient Series

Will Cook

Long paper

The Binary Totient Series

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

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Summary. The dyadic totient sections through level $e \geq 1$ have an explicit rational basis of size $2^{e} + 1$ . 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.

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In this paper

A two-adic congruence that does not give a certificate

Theorem 9.34. The theorem is unconditional and holds for every $K$ , $H$ , $B$ . For every $K \geq 2$ , every $H > K$ , and every bound $B$ , there are primes $p > B$ with a length- $(K - 1)$ zero prefix and a terminal half-modulus, giving

D (H, p - K, K) \equiv 2^{K - 1} (\mod 2^{K}) .

Under eventual integrality this transfers to an integer $z$ with $(z : R) = R (p + H) - R_{p}$ and $z \equiv 2^{K - 1} (\mod 2^{K})$ .

Remark 9.35 (Why the central residue still fails the test). This places the residue at the centre of the residue interval: $2^{K - 1}$ is as far as possible from $0$ modulo $2^{K}$ . Taking $N = p - K$ , $h = H$ , $L = K$ gives radius $N + h + L + 2 = p + H + 2$ , so $C$ requires $2^{K - 1} > p + H + 2$ . But $z \equiv 2^{K - 1} (\mod 2^{K})$ forces $| z | \geq 2^{K - 1}$ , while the directed tail bound forces $| z | < p + H + 2$ ; and the construction’s own congruence $p \equiv 1 + 2^{K - 1} (\mod 2^{K})$ (equivalently $v_{2} (p - 1) = K - 1$ ) forces $p \geq 1 + 2^{K - 1}$ . So $p + H + 2 > 2^{K - 1}$ always, and no choice of $p$ rescues it. The quantifier order is inverted: the theorem gives, for fixed $K$ , cofinally many large $p$ ; the certificate needs a $p$ small relative to $2^{K - 1}$ , impossible for a single totient difference since $| φ (N + H) - φ (N) | < N + H$ . The depth is not the obstacle: the construction already works uniformly for every $K$ , combining by the Chinese remainder theorem one congruence for $p + H$ and two for each of the $K - 1$ preceding positions (Proposition 6.92).

Proposition 9.36 (A sufficient accumulated-residue condition). Suppose that for every integer $h \geq 1$ and every $N_{0} \in N$ there are $N \geq N_{0}$ and $L \geq 1$ such that

D (h, N, L) \equiv 2^{L - 1} (\mod 2^{L}), N + h + L + 2 < 2^{L - 1} .

Then $S \notin Q$ : these two inequalities imply $C (h, N, L)$ , and the full quantified certificate condition applies. The half-modulus condition here is imposed on the accumulated weighted sum $D (h, N, L)$ , not on a single totient difference. An unbounded set of triples with no control of the shift does not supply the displayed quantifiers. The required family remains unproved.

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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