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

One bound on the good indices

Two counting estimates allow us to restrict the cancellation question to $G$ . For fixed $h, s$ , let $L (X)$ be the least $L$ satisfying $h \leq L - s$ and $16 (2 X + h + L + 2) \leq 2^{L}$ , as in Proposition 12.6. With $s = 26$ and $η = 1 / 1000$ , more than $67 X / 100$ indices remain in $G$ for large $X$ . A suitable saving on this set alone would therefore imply irrationality.

Theorem 3.5 (Irrationality from one bound on the good indices).

Suppose that for every $h \geq 1$ and every $A$ there are $X$ , $L$ with $16 (2 X + h + L + 2) \leq 2^{L}$ and a nonempty finite set $T$ of integers in $[A, 2 X)$ with
$\sum_{N \in T} Re E (h, N, L) \leq \frac{9}{10} # T .$
Then $S \notin Q$ (irrational_totient_series_of_support_gap).
Take $s = 26$ , $η = 1 / 1000$ and $L = L (X)$ . Suppose that for every $h \geq 1$ there are arbitrarily large $X$ with
$Re \sum_{N \in G} E (h, N, L (X)) \leq \frac{603}{1000} X .$
Then $S \notin Q$ (irrational_totient_series_of_goodBase_gap).

Proof. (a) At a depth satisfying the size inequality, a nonempty set of indices below $2 X$ whose average of $Re E (h, N, L)$ is at most $9 / 10$ contains an index $N$ with $C (h, N, L)$ (Erdos249257.TotientTailPeriodKiller.exists_certifiedKill_of_first_harmonic_gap_subset). The hypothesis therefore supplies, for every $h \geq 1$ , certificates beyond every threshold, which is Definition 1.2; that condition is equivalent to $S \notin Q$ .

(b) For all large $X$ , fewer than $8 X / 25$ indices in $[X, 2 X)$ are unassigned (Proposition 12.6), and fewer than $X / 100$ assigned indices have $φ (m_{N}) < m_{N} / 1000$ , by the Chebyshev form of Proposition 12.7 given after its proof. These two sets and $G$ partition $[X, 2 X)$ , so $# G > 67 X / 100$ (eventually_card_pivotGoodBases_gt). Since $603 / 1000 = (9 / 10) (67 / 100)$ , each large $X$ in the hypothesis gives $\sum_{N \in G} Re E (h, N, L (X)) < \frac{9}{10} # G$ . For $X \geq A$ the set $T = G$ is nonempty and lies in $[A, 2 X)$ , and $L (X)$ satisfies the size inequality, so (a) applies. ◻

The trivial bound for the sum in (b) is $# G$ , so the hypothesis asks for a fixed saving: along a sequence of blocks, the average of $\cos (2 π D (h, N, L) / 2^{L})$ over $G$ must be at most $603 X / (1000 # G)$ , a number below $9 / 10$ . Equidistributed phases would make the sum $o (X)$ . A rational value $S = a / (2^{c} v)$ with $v$ odd makes the bound fail for $h = φ (v)$ and all large $X$ : every large index then has an integral tail difference, hence no certificate, and an index at a depth satisfying the size condition and having no certificate has $Re E > 9 / 10$ . The hypothesis is sufficient and is not known to be necessary: from irrationality one obtains nonintegral tail differences, and no argument turns them into a fixed saving on $G$ . It is open. No computation in this record bears on it, and the count $# G > 67 X / 100$ holds only for large $X$ .

When the first two inequalities of Definition 3.4 hold with $s = 26$ , $η = 1 / 1000$ and $L = L (X)$ , they add to $Re \sum_{N \in G} E (h, N, L (X)) \leq \frac{57}{100} X$ , so the four-term condition at these parameters implies the hypothesis of (b). At these same parameters, the recorded argument takes the prime number theorem as an input and bounds every group mean $| {\bar{z}}_{m} |$ on $G$ by $1 / 100$ for all sufficiently large $X$ , for each fixed $h$ (fiberMean_le_of_primeNumberTheorem). Since $| w_{N} | = 1$ , this gives $| \sum_{N \in G} w_{N} {\bar{z}}_{m_{N}} | \leq # G / 100 \leq X / 100$ , regardless of the dependence of $w_{N}$ on the prime. It supplies the mean bound, not the mean-subtracted cancellation. Given this mean bound, a bound $\frac{11}{20} X$ on the good-index sum, on arbitrarily large blocks for each $h$ , yields the mean-subtracted inequality and hence irrationality (dtw_of_fiberMean_and_centered, irrational_totient_series_of_primeNumberTheorem_and_goodBase_gap). Theorem 3.5(b) supersedes that route. It assumes neither the prime number theorem nor a bound on the group means, and asks only for $603 X / 1000$ , rather than $11 X / 20$ . The saving is measured against $# G$ , not against the full block size: since $# G > 67 X / 100$ , the good-index average is below $9 / 10$ . There is no need to add the discarded indices back into the sum.

For clarity, the two sufficient estimates have different uses. The centred route controls all four pieces and so bounds the sum on the whole block. The $603 X / 1000$ route uses only the good indices and their cardinality. Bounding the discarded part trivially would give only $(603 / 1000 + 33 / 100) X = 933 X / 1000$ , not $9 X / 10$ ; adding it back would lose the conclusion one was trying to prove.

There is also a useful distinction between eventual and cofinal hypotheses. At $s = 26$ , $η = 1 / 1000$ and the prescribed $L (X)$ , fix $h$ and take the maximum of the thresholds for the two counting bounds and the PNT-input mean bound. Any arbitrarily large sequence of blocks on which the centred bound holds eventually passes this maximum. Thus it supplies common blocks for all four estimates. Merely knowing that several different estimates each hold on some arbitrarily large blocks would not justify that intersection.

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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