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

Signed moments and parity

A signed determinant can serve in rational approximation only with both a nonzero numerator and a small error. Here parity proves the first requirement for the particular finite signed sum.

Lemma 10.3 (Nonvanishing from a unique largest denominator exponent). Let $I$ be a finite set, let $u_{i} \in Z$ and $e_{i} \in N$ for $i \in I$ , and suppose that $m \in I$ satisfies $e_{i} < e_{m}$ for every $i \neq m$ . If $u_{m}$ is odd, then

2^{e_{m}} \sum_{i \in I} \frac{u_{i}}{2^{e_{i}}} = \sum_{i \in I} u_{i} 2^{e_{m} - e_{i}} \equiv 1 (\mod 2) .

In particular, the sum is nonzero.

Every term except $u_{m}$ is even, which proves the assertion. For example, $1 / 2 - 1 / 8 = 3 / 8$ has odd cleared numerator. Uniqueness is necessary for this argument: two odd contributions with the same largest exponent can cancel.

Lean also proves a separate determinant expansion from the Leibniz formula (det_mul_rectangular) and defines the truncated signed moments $\sum_{N = 1}^{Y} a_{N} 2^{- N (k + 1)}$ and their Hankel determinants (truncatedMoment, hankelDet). The parity lemma is not derived from that determinant identity. A determinant application must first identify a sum to which the parity hypotheses apply.

For the totient series, one would still need suitable configurations at arbitrarily large scales and a denominator-weighted error tending to zero. Nonvanishing alone provides neither. No such family is supplied by these two algebraic statements; compare the separate rank discussion in §10.8. In another base, the analogous argument would require divisibility by a prime factor of that base rather than an unchanged appeal to parity.

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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