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

Unconditional structure and finite examples

The rank formulas hold at every stated level. The residue tests and Farey interval below instead give conclusions on specified finite sets of shifts, scales or denominators. This distinction matters when applying the later conditions, which require witnesses at unbounded indices.

Theorem 6.1 (Rationality gives an eventual tail period). If $S = a / (2^{c} v)$ with $a \in Z$ , $c \in N$ and $v$ a positive odd integer, then

R_{N + h} - R_{N} \in Z (N \geq c), h = φ (v) > 0.

Euler’s theorem gives $v ∣ 2^{h} - 1$ , and the prefix-tail identity then makes $2^{N} (2^{h} - 1) S$ integral. Thus rationality supplies a positive shift for which all sufficiently late tail differences are integral.

Theorem 6.2 (Rationality forces unbounded carry rank). If $S \in Q$ , there are an integer $v > 0$ and an integer sequence $u$ such that

u (N + 1) = 2 u (N) - v φ (N + 1), u (N) / 2^{N} ⟶ 0,

and, for every $e \geq 0$ , the sequences $n \mapsto u (2^{j} n + r)$ with $1 \leq j \leq e$ and $0 \leq r < 2^{j}$ span a space of $Q$ -dimension at least $2^{e} - 1$ .

The recurrence follows from the scaled tails after clearing a hypothetical denominator of $S$ . The only growth condition imposed here is the displayed limit $u (N) / 2^{N} \to 0$ ; subexponential growth is not assumed. The rank lower bound is a necessary consequence of rationality and yields no contradiction by itself. An upper bound using additional properties of the actual totient coefficients would be needed; the later rational comparison rules out such a bound for arbitrary rational coefficient series.

Theorem 6.3 (Independence of the retained dyadic sections). For every integer $e \geq 0$ , the family

φ (n), φ (2 n), φ (2^{j} n + r) (1 \leq j \leq e, 0 < r < 2^{j}, r odd)

is linearly independent over $Q$ and has $2^{e} + 1$ members. At $e = 0$ this family still contains $φ (2 n)$ ; it is not the full level-zero truncation, which consists only of $φ (n)$ . For $e \geq 1$ it is the basis of the full truncated dyadic family.

The proof separates one section at a time: the Chinese remainder theorem forces suitable prime divisors in the other affine arguments, while Dirichlet’s theorem makes the selected argument prime. This proves independence, not just the displayed cardinality. In particular, the span of all dyadic sections is infinite-dimensional; that conclusion alone says nothing about rationality of their weighted sum.

Proposition 6.4 (Denominator divisibility forces tail integrality). If $S = a / q$ in lowest terms with $q > 0$ and $q ∣ 2^{N} (2^{h} - 1)$ , then $R_{N + h} - R_{N} \in Z$ . Indeed, the prefix identity expresses that difference as $2^{N} (2^{h} - 1) S$ minus an integer. This is the contradiction used by a nonintegrality certificate to exclude the rational value $a / q$ .

The denominator results use the following polynomial, defined for $r \geq 1$ :

P_{r} (X) = \sum_{d ∣ r} μ (d) \frac{r}{d} \sum_{j = 0}^{r / d - 1} X^{d j} .

For example, $P_{p} (X) = p (1 + X + \dots + X^{p - 1}) - 1$ when $p$ is prime. Thus $P_{p} (2) = p (2^{p} - 1) - 1$ is coprime to $2^{p} - 1$ . For general $r \geq 1$ and $H \geq 1$ , write

b_{r} = \frac{P_{r} (2)}{2^{r} - 1}, β_{H} = \frac{b_{rad (H)}}{rad (H)}, r_{t} = rad (H_{t}) .

Here $rad (H)$ is the product of the distinct prime factors of $H$ , and $den (x)$ denotes the positive reduced denominator of $x \in Q$ . These rational numbers are finite rational contributions to the Möbius decomposition; no convergence to $S$ is asserted for them. Also put $P_{t} = {p prime : t / 2 < p \leq t}$ . The hypotheses below specify which factors of $2^{r_{t}} - 1$ remain in the reduced denominator after cancellation.

Proposition 6.5 (A cyclotomic factor remains after cancellation). If $r$ is squarefree, then

gcd (| P_{r} (2) |, | Φ_{r} (2) |) = 1, | Φ_{r} (2) | ∣ den (b_{r}) .

Here $Φ_{r}$ is the $r$ -th cyclotomic polynomial. Proposition 6.40 gives $P_{r} \equiv μ (r) (\mod Φ_{r})$ , and $μ (r) = \pm 1$ because $r$ is squarefree. This proves that the indicated cyclotomic factor cannot cancel from $b_{r}$ .

Proposition 6.6 (Primes in the upper half remain after cancellation). For $t \geq 5$ the whole product

\prod_{p \in P_{t}} (2^{p} - 1) ∣ den (H_{t} β_{H_{t}})

remains after cancellation. More generally, the source gives the same product-divisibility conclusion for a multiplier coprime to that product; it also supplies a sufficient condition on the multiplier’s prime factors. For the multiplier $H_{t} / r_{t}$ , those prime factors are at most $t$ , and the LCM result above follows.

The restriction $t \geq 5$ belongs to the sufficient prime-factor condition, not just to the LCM example. At $t = 3$ , one has $r_{t} = 6$ , $b_{6} = 298 / 63$ and $P_{3} = {2, 3}$ . The multiplier $9$ has only prime factors at most $3$ , but $den (9 b_{6}) = 7$ is not divisible by $(2^{2} - 1) (2^{3} - 1) = 21$ . Thus the prime-factor condition alone does not imply survival of the product at every scale. This is a denominator statement about a finite rational contribution, not a nonintegrality assertion about a real tail.

Proposition 6.7 (A lower bound for the reduced denominator). For every integer $t \geq 5$ ,

2^{⌊ t / 2 ⌋} \leq \prod_{p \in P_{t}} (2^{p} - 1) \leq den (H_{t} β_{H_{t}}) .

Bertrand’s postulate supplies a prime in $P_{t}$ ; its factor is at least $2^{⌊ t / 2 ⌋}$ . Proposition 6.6 then gives the second inequality. The complementary term in the real tail decomposition can still cancel this rational contribution, so denominator growth alone does not prove irrationality of $S$ .

Proposition 6.8 (The exact reduced denominator). At every integer scale $t \geq 0$ ,

den (H_{t} β_{H_{t}}) = \frac{2^{r_{t}} - 1}{gcd (2^{r_{t}} - 1, \frac{H_{t}}{r_{t}} \prod_{\begin{matrix} p ∣ r_{t} \\ p odd prime \end{matrix}} (p^{2} - 1))} .

The product is $1$ when there is no odd prime divisor. This gives the exact cancellation, including the multiplier $H_{t} / r_{t}$ , rather than merely a lower bound for the denominator. For example, $H_{5} = 60$ and $r_{5} = 30$ , and the denominator is $(2^{30} - 1) / 3$ .

Proposition 6.9 (Nonvanishing of a signed dyadic sum). Let $I$ be finite, let $u_{i} \in Z$ and $e_{i} \in N$ , and suppose that $m \in I$ is the unique index with maximal exponent $e_{m}$ . 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) .

Indeed, every summand other than $u_{m}$ is even. The sum is therefore nonzero. In the determinant application, the rectangular Cauchy–Binet formula produces such finite signed sums from truncated moments. The result tests a specified configuration; it does not show that a suitable maximal-exponent term exists at arbitrarily large scales.

Proposition 6.10 (Small certificates). For each integer $1 \leq h \leq 8$ , the finite test $C (h, 12, 16)$ holds. Each discrepancy uses two 16-term windows. Across all eight shifts, only the 24 distinct totient values at $13 \leq n \leq 36$ are needed, since the windows overlap. The source verifies the eight integer residue inequalities by exact computation.

Consequently, if $a / b$ is a reduced fraction with $b > 0$ and $b ∣ 2^{12} (2^{h} - 1)$ for at least one $1 \leq h \leq 8$ , then $S \neq a / b$ . This follows from Propositions 6.18 and 6.4.

Proposition 6.11 (A common certificate for sixteen shifts). For each integer $1 \leq h \leq 16$ , one has $C (h, 14, 9)$ . The common basepoint is $14$ and the depth is $9$ ; these parameters have different roles. The two windows for each shift use only the 25 distinct totient values at $15 \leq n \leq 39$ across the whole family. Thus $S \neq a / b$ for every reduced fraction with $b > 0$ such that $b ∣ 2^{14} (2^{h} - 1)$ for at least one $1 \leq h \leq 16$ .

Proposition 6.12 (Historical diagonal examples and the complete band through 82). Let $H_{t} = lcm (1, \dots, t)$ and let

P (t) :⟺ \exists L, C (H_{t}, H_{t}, L) .

The checked certificate table proves $P (t)$ at 28 explicit indices, ending at $t = 64$ . For the initial indices

t = 1, 2, 3, 4, 5, 7, 8, 9, 11, 13, 16, 17, \dots,

the corresponding depths begin $6, 5, 7, 7, 9, 14, 15, 14, 21, 22, 23, 26, \dots$ . Each entry is a finite kernel computation on explicit totient values. The factorisations use checked prime-power blocks and Lucas primality certificates. Thus the table proves a finite list of instances of $P (t)$ ; it does not prove that $P (t)$ holds for infinitely many $t$ .

The historical 28 examples are now a strict subset of the aggregate theorem $\forall t \leq 82, P (t)$ . That theorem closes the finite interval without holes but supplies neither $P (83)$ nor a cofinal family.

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.

Cite and contact. Cite this paper by its title, author and date above, with its PDF; cite the earlier sources it uses for a mathematical result. For software, use the release citation and give the commit used. Contact Will with questions or corrections.

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