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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Divisor terms and their complement

At $H = 6$ and $s = 2$ , the difference $φ (2 H + s) - φ (H + s) = φ (14) - φ (8)$ is $2$ , whereas the terms with divisor index $d ∣ H$ sum to $3$ . The other indices contribute $- 1$ . The following decomposition keeps both contributions. It holds for every positive integer $H$ , not only a least common multiple; an estimate is needed before its second sum can be discarded.

Proposition 6.131 (The sum over divisor indices). For integers $H > 0$ and $s \geq 0$ , the part of the Möbius expansion indexed by divisors of $H$ is

\sum_{\begin{matrix} d ∣ H \\ d ∣ s \end{matrix}} μ (d) \frac{H}{d} = H \frac{φ (gcd (H, s))}{gcd (H, s)} .

This is the standard identity $\sum_{d ∣ g} μ (d) / d = φ (g) / g$ at $g = gcd (H, s) > 0$ . The formal statement writes the same value after extracting $H / rad (H)$ , where $rad (H) = \prod_{p ∣ H} p$ .

Proposition 6.132 (The divisor sum and its complement). For integers $H > 0$ and $s \geq 0$ ,

\begin{aligned} φ (2 H + s) - φ (H + s) & = H \frac{φ (gcd (H, s))}{gcd (H, s)} \\ + \sum_{\begin{matrix} 1 \leq d \leq 2 H + s \\ d ∤ H \end{matrix}} μ (d) (\frac{2 H + s}{d} 1_{d ∣ 2 H + s} - \frac{H + s}{d} 1_{d ∣ H + s}) . \end{aligned}

Here $1$ is the indicator of the stated divisibility condition. The identity follows by expanding both totients as $φ (n) = \sum_{d ∣ n} μ (d) n / d$ and separating the indices $d ∣ H$ . Such an index divides either endpoint precisely when it divides $s$ , and its difference is then $μ (d) H / d$ .

Theorem 6.133 (Doubling the full totient difference). For nonnegative integers $H, r$ with $H$ even,

φ (4 H + 2 r) - φ (2 H + 2 r) = {\begin{cases} 2 (φ (2 H + r) - φ (H + r)), & r even, \\ φ (2 H + r) - φ (H + r), & r odd . \end{cases}

This applies to the full difference, not just its divisor contribution. It follows from $φ (2 n) = 2 φ (n)$ for even $n$ and $φ (2 n) = φ (n)$ for odd $n$ .

Observation 6.134 (Squarefree support of the complementary terms). A nonzero summand in the complementary sum must have a squarefree index $d$ dividing exactly one of $2 H + s$ and $H + s$ . Indeed, otherwise $μ (d) = 0$ or both indicators vanish. Both endpoints cannot be divisible by $d$ : their difference is $H$ , and $d ∤ H$ . Erdos249257.DiagonalFreshLossBridge.squarefree_of_foreignChannelPhaseTerm_ne_zero Erdos249257.DiagonalFreshLossBridge.foreign_endpoint_support_disjoint The restriction identifies which indices can contribute. It does not control the sign or magnitude of their sum.

Proposition 6.135 (A doubling identity for two portions of the sum). Let $H > 0$ , $s \geq 0$ and $d \geq 1$ , with $d ∤ H$ and $d ∣ H + s$ . The $d$ -summands in the complementary sums at offsets $s$ and $2 s$ are, respectively,

- μ (d) \frac{H + s}{d} and 2 μ (d) \frac{H + s}{d} .

For the second expression, $d ∣ 2 H + 2 s$ and $d ∤ H + 2 s$ ; otherwise $d$ would divide their difference $H$ . Thus the second contribution is $- 2$ times the first, including the zero case $μ (d) = 0$ . The two terms have different binary weights in a window sum, so this identity alone is not a cancellation of their weighted contributions.

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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