The Binary Totient Series

Proposition 6.61 (An equivalent diagonal nonintegrality condition). The following statements are equivalent:

This is an exact reformulation of irrationality, not a weaker theorem about a finite set of scales. The later sufficient conditions seek a finite inequality that establishes the right-hand side.

Proposition 6.62 (A sufficient short-window condition). Suppose that for every $a_0\in \mathbb{N}$ there are $a\ge a_0$ and $L<2\cdot 2^a$ with $\mathcal{C}(H_{2^a},H_{2^a},L)$. Then $S\notin \mathbb{Q}$. Writing out the certificate, the required inequality is

It places the residue away from both endpoints of the modulus interval. The examples at $a=4,6$ establish the bounded statement in Proposition 6.67, not the condition for every threshold. Pointwise completeness allows a sufficiently large depth but does not supply a depth below $2\cdot 2^a$.

Proposition 6.63 (A sufficient approximation condition). Use the prescribed index $q_a$ and the error ${\epsilon }_{a,q_a}$ defined above. Suppose

Then $S\notin \mathbb{Q}$. The estimate $|{\mathrm{\Omega }}_a-{\rho }_{a,q_a}|<{\epsilon }_{a,q_a}$ implies $|{\rho }_{a,q_a}-z|>1/32$ for every integer $z$, and hence the finite residue separation from which the proof derives irrationality. This is a fixed separation condition along an unbounded family of LCM scales. It is not an irrationality-measure theorem for arbitrary rational approximants, and nonintegrality alone does not give the stated margin. The depth is prescribed, not an additional free witness.

Proposition 6.64 (A sufficient upper-endpoint separation). Let $a,J,K,m\in \mathbb{N}$, $a\ge 8$, and $H=H(2^a)$. Assume

Define the finite condition

Then ${\mathcal{G}}_a(J,K,m)$ implies $R_{2H+J}-R_{H+J}\notin \mathbb{Z}$. In particular, irrationality follows if for every $a_0$ there are $a\ge max(a_0,8)$ and $K,m$ satisfying $K+(a+6)<2\cdot 2^a$ and ${\mathcal{G}}_a(0,K,m)$.

The sign theorem forces an integral tail to give a residue in the upper endpoint interval. This one-sided test excludes that interval; it does not require a lower residue bound. Both the sign-range hypothesis and the strict modulus bound are essential premises of this implication. The required unbounded family is not established.

Proposition 6.65 (Relations among the sufficient conditions). The sufficient conditions do not form a single linear chain. The proved implications are

There is also a separate sufficient argument: terminal dominance and flexible centred magnitude each imply the two-sided exclusion in Proposition 6.66, which implies diagonal nonintegrality. These statements refer to the quantified versions, with the bounds on $a,q$ and the depth retained. The detailed conditions and sources appear in Proposition 9.14. No reverse implication between terminal dominance and flexible magnitude is asserted, and none of the required unbounded families is proved here.

Proposition 6.66 (An exact endpoint formula). Let $a,q\in \mathbb{N}$, $a\ge 8$, and $H=H(2^a)$. Assume

If $z\in \mathbb{Z}$ and $z={\mathrm{\Omega }}_a$, then

Consequently either of the inequalities

is sufficient for ${\mathrm{\Omega }}_a\notin \mathbb{Z}$. Under integrality the recurrence equals a positive tail difference strictly smaller than $2H+2q+3$, and the identity rules out both displayed inequalities. The identity is exact; the resulting nonintegrality criterion is only sufficient. Its converse is not asserted.

Proposition 6.67 (A sufficient extension of the examples through exponent 6). The supplied finite result is

The witnesses at $(a,L)=(4,23)$ and $(6,93)$ give this bounded family. A witness with $a\ge 7$ would extend the verified range, but any finite extension would still leave the assertion for arbitrarily large thresholds in Proposition 6.62 unproved.

Proposition 6.68 (A sufficient extremal-order condition). Let $H\ge 1$ and $j\ge 0$. If

then

If instead the middle value is strictly larger than both outer values, the displayed second difference is negative. In each case the claim follows by adding the two strict differences. No lower bound on their sizes is needed.

For $H=H_{2^a}$, this reduces nonvanishing of one second difference to an ordering of three totient values. For a prescribed $j$, the remaining question is whether either strict ordering occurs for arbitrarily large $a$. The elementary implication does not establish those occurrences.

Proposition 6.69 (A directed certificate condition). For fixed $h,N\in \mathbb{N}$, a depth $L\in \mathbb{N}$ satisfying

exists if and only if $R_{N+h}-R_N\notin \mathbb{Z}$. The endpoint inequalities are non-strict.

Proposition 6.70 (A sufficient condition at the prescribed mod-four pulses). Suppose that for every positive integer $h$ and every $B\in \mathbb{N}$ there are a prime $p>B$ and $K\in \mathbb{N}$ such that

and every integer $z$ with $|z|\le p+4h+1$ and $z\equiv 2(\mathrm{mod}4)$ has an index $0\le i\le K$ satisfying

Then $S\notin \mathbb{Q}$. Under a hypothetical eventual period $h$, the proved congruence argument at a sufficiently large such prime forces $R_{p+4h}-R_p$ to be an integer in the tested residue class. The finite test excludes every candidate in that class, a contradiction.

The mod-four congruence is supplied by Proposition 6.91; exclusion at the same prime is not. Restricting to $2(\mathrm{mod}4)$ retains one residue class out of four, not necessarily exactly one quarter of the candidates in a finite interval. The quantifier over every bound $B$ cannot be replaced by one fixed prime for each $h$.

Proposition 6.71 (The full quantified condition).

This is exactly the quantifier structure $\mathrm{\forall }h\ge 1\mathrm{\forall }{N}_0\mathrm{\exists }N\ge N_0\mathrm{\exists }L\mathcal{C}(h,N,L)$. The quantified condition is not supplied here. The equivalence specifies exactly what a residue-based proof would need beyond the finite examples.

Proposition 6.72 (One diagonal parameter suffices). The following condition is equivalent to irrationality:

For a hypothetical rational value, let $h_0$ and $N_0$ be its eventual tail period and starting index. Taking $t\ge max(h_0,N_0)$ ensures $h_0\mid H(t)$ and $H(t)\ge N_0$. The corresponding diagonal difference is then integral, contradicting a certificate at that scale. Conversely, irrationality and pointwise completeness provide a witness at every prescribed $t$. The reduction uses one scale parameter; it neither establishes the certificate condition nor makes one fixed scale sufficient for all rational values.

Proposition 6.73 (A sufficient condition on the LCM grid).

One residue certificate anywhere on the two-multiplier LCM cone, at arbitrarily large $t$, suffices; the diagonal choice $q=m=1$ recovers Proposition 6.72.

Proposition 6.74 (A finite-grid certificate condition). Let $H,L\in \mathbb{N}$ and let $Q\subseteq {\mathbb{N}}_{>0}$ be finite and nonempty. Define $A_q=\sum _{j=1}^L\phi (qH+j)2^{L-j}$ and $B_q=qH+L+2$. If $B_q<2^L$ for every $q\in Q$ and

then $R_{q_{j}H}-R_{q_{i}H}\notin \mathbb{Z}$ for some $q_i,q_j\in Q$. Such a test at $H=H_t$ for arbitrarily large $t$ implies $S\notin \mathbb{Q}$. Neither $|Q|\ge 3$ nor compatibility of every pair is a hypothesis. The proof, by choosing a minimum tail in the finite set, is given in Theorem 6.56.

Proposition 6.75 (Unbounded Farey bounds imply irrationality). The Farey-gap denominator bound at window $K$ is currently $\sim 7.96\times {10}^{34}$ at $K=240$ (Prop. 6.99). If

(that is, the proved exclusion bounds are arbitrarily large), then #249 follows via the denominator-exclusion implication, without assuming the certificate supply in Proposition 6.71. No unboundedness claim is proved or attempted in the record.

Proposition 6.76 (The lower bound and a false proposed upper bound). By Prop. 6.98, the canonical dyadic totient-kernel family is unconditionally $(2^e+1)$-dimensional at every level $e\ge 1$. Rationality of $S$ forces an associated carry orbit with $\mathbb{Q}$-rank $\ge 2^e-1$ at every level (Prop. 6.90). An upper bound independent of $e$ for these same carry ranks would contradict the lower bound. More generally, a bound $g(e)$ with $g(e)<2^e-1$ at some level $e\ge 1$ would suffice, provided it applies to the actual carry under the hypothetical rationality of $S$. A bound that merely grows with $e$ need not contradict anything. The generic assertion for arbitrary rational coefficient series is false: the $5/4$ comparison sequence in Section 10.8 has an integer carry satisfying the growth condition and the same rank lower bound. The incompatible integer identity and the finite-shift counterexample (Observation 6.115) remain separate counterexample results. None gives a rank upper bound from rationality alone. The Lean proof covers the rank lower bound and counterexample results, not the counterfactual upper bound.

Proposition 6.77 (Soundness of the finite carry test). For fixed $h,N,K\in \mathbb{N}$, use the integer recurrences $c_{h,N,z}$ defined above. Test each of the $2(N+h+1)+1$ candidates $z\in \mathbb{Z}$ with $|z|\le N+h+1$. If every candidate has some $i\le K$ for which

then $R_{N+h}-R_N\notin \mathbb{Z}$. An integral tail difference would give one of the initial candidates and would remain in the open strip $|c_{h,N,z}(i)|<N+i+h+2$ at every step. Equality at a boundary already excludes a candidate. This is the soundness implication for a specified finite test; no bound for a successful search depth is asserted.