Metadata

slug: oeis-a393173-hanna-power-of-two-parity
bibkey: hanna2026a393173
doi: null
url: https://oeis.org/A393173
triage: theorem
motivation_gids:
  - D5/S1/Recurrence/Invariants/HannaA393173PowerTwoParity

Hanna’s power-of-two parity conjecture for OEIS A393173

Problem

Paul D. Hanna’s OEIS A393173 states:

G.f. A(x) satisfies [x^n] 1/(A(x) - n^2*x) = 0 for n >= 1.

Conjecture: for n > 0, a(n) is odd iff n is a power of 2.

The offset-zero data specify $a (0) = 1$ . Preregistration #18824 classifies the published conjecture as tier 1. The source is documented in the Library note.

Statement echo

Names below belong to D5.S1.Recurrence.Invariants.HannaA393173PowerTwoParity.

Source clause	Lean counterpart	Fidelity
Offset zero, $a (0) = 1$	`IsGF` constant coefficient	Literal normalization.
$[x^{n}] (A - n^{2} x)^{- 1} = 0$ for $n \geq 1$	`IsGF` inverse coefficient, natural index cast into $Q$	Literal source expression.
Positive indices	`claim`, $n > 0$	Zero is excluded.
$a (n)$ is odd iff $n = 2^{k}$	`claim`, integer coefficient witness and both directions of the equivalence	Exponent $k$ is natural.
Existence and integrality implicit in the sequence	`claim`, an existing solution and an integer witness for every coefficient of every solution	Explicit strengthened clauses from the preregistration; these prevent vacuity.

D5/S1/Recurrence/Invariants/HannaA393173PowerTwoParity.result proves the complete preregistered proposition. The normalized rational inverse-diagonal solution exists, its positive coefficients are integers, and their odd support is exactly the powers of two. Its admission basis is open-problem-resolution (#18824; Proved), generality is I, and utility is none. The argument is symbolic at all natural degrees.

Gap

The conjecture is a universal statement about the full formal series. A finite prefix does not establish it. A normalized integral lift and uniqueness of the coefficient recursion connect reduction modulo two to every rational solution.

The escape audit #19804 is unfinished for the only public theorem, the full Lean name D5.S1.Recurrence.Invariants.HannaA393173PowerTwoParity.result. The enrolled existential and iff readouts do not alone supply a readout of its whole conjunction over the infinite power-series domain. A faithful dependent family reconstruction, observational dependence and intervention/sensitivity proofs for this exact telescope have not been supplied. No validated registration is claimed.

Route

Let $S = \sum_{j \geq 0} x^{2^{j}}$ over $F_{2}$ . The frozen CubicFifteenSubstitutionParity.thue_series_equation supplies $S = x + S^{2}$ . For normalized series, changing only the coefficient of degree $d$ changes the inverse’s degree- $d$ coefficient by its negative. Thus the correction that adds the offending inverse coefficient is a strict coefficient contraction. The frozen PowerSeriesNormalizedContraction.exists_unique_fixed supplies its integral fixed point and uniqueness. The frozen QuotientThetaCompositionModFour.map_inverse transports the equation through coefficient ring maps.

Put $P = 1 + S$ . Since $P (S / x) = 1$ , its inverse is $S / x$ . At positive even $n$ , $n^{2} = 0$ modulo two and $[x^{n}] P^{- 1} = [x^{n + 1}] S = 0$ : the odd index $n + 1 \geq 3$ is not a power of two. At odd $n$ , $P - x = P^{2}$ in characteristic two, so its inverse is a square and has zero odd-degree coefficients by CubicFifteenSubstitutionParity.prime_power_subst. Hence $P$ satisfies every reduced equation. Uniqueness identifies the reduction of the integral fixed point with $P$ . Rational uniqueness and the integer-cast parity equivalence establish result.

Declaration classification

Every authored declaration has proof_shape: bind-only after expanding the same-module helpers to the frozen owners and pinned Mathlib. Each helper is consumed; escape_witness: none. This classification concerns the proof steps; the published open-problem settlement supplies the module’s admission basis.

Declaration	`proof_shape`	`escape_witness`	Direct consumers
`IsGF`	bind-only	none	`claim`, `isGF_iff_fixed`, `result`
`claim`	bind-only	none	`result`
`inverse_top`	bind-only	none	`step_contract`
`binary_quadratic`	bind-only	none	`binary_inverse`, `candidate_fixed`
`binary_parity`	bind-only	none	`result`
`step`	bind-only	none	`candidate_fixed`, `fixed_iff`, `fixed_map`, `fixed_unique`, `integral_mod_two`, `isGF_iff_fixed`, `normalized_solution`, `result`, `step_contract`, `step_zero`
`denominator_zero`	bind-only	none	`fixed_map`, `isGF_iff_fixed`, `step_contract`
`step_zero`	bind-only	none	`normalized_solution`
`step_contract`	bind-only	none	`normalized_solution`
`fixed_iff`	bind-only	none	`candidate_fixed`, `fixed_map`, `isGF_iff_fixed`
`fixed_map`	bind-only	none	`integral_mod_two`, `result`
`isGF_iff_fixed`	bind-only	none	`result`
`binary_inverse`	bind-only	none	`candidate_fixed`
`candidate_fixed`	bind-only	none	`integral_mod_two`
`normalized_solution`	bind-only	none	`fixed_unique`, `result`
`fixed_unique`	bind-only	none	`integral_mod_two`, `result`
`integral_mod_two`	bind-only	none	`result`
`result`	bind-only	none	External named settlement

Falsifier

A positive degree whose integral coefficient violates the power-of-two parity equivalence would contradict result. A rational normalized series satisfying the source equations without the stated integral coefficients would also contradict it. The separate modulo-four claim has a different conclusion and is outside this theorem.

Evidence

The canonical module and its Blueprint show the full proposition. The public surface consists of IsGF, claim, and the sole theorem result; all proof helpers are private. The axiom closure of every public declaration is contained in {propext, Classical.choice, Quot.sound}.

The numerical experiment is oeis-hanna-parity-family-research at 9dfe5bdc210cac281a148e7180239ddb21b0b34b. Command: python3 check.py in that directory; exit 0. The check.py SHA-256 is ec87bb1637d6a946d4e5fb43f078bc623ce327257ef53b9305c05c564ca154b5. For A393173 the first 12 terms agree with the recorded OEIS data; coefficients through degree 40 are integral and parity agrees for $1 \leq n \leq 40$ . The same program reports those finite checks for A382122 and A393850, and tests an alternate A382122 sign branch. These are finite computations, independent of the unbounded theorem.

Triage

What the settlement shows

A393173 — proved, kernel-checked: result establishes existence, integrality and parity for every normalized rational solution. The decisive reduced series is $1 + S$ ; the even case uses $S / x$ and the odd case uses Frobenius. Uniqueness — proved in prose: two normalized fixed points agree below degree zero; strict coefficient contraction extends agreement below degree $d$ to agreement below degree $d + 1$ . Induction gives coefficientwise equality. The private isGF_iff_fixed identifies precisely these fixed points with the source equation. Uniqueness is not a separate public theorem.
A382122 — proved in prose below: every normalized rational solution of its absolute-value equation is integral and has the same positive-degree parity. This statement does not assume uniqueness or positivity of the inverse coefficients. Its separate preregistration is #18822; it is not a Lean settlement delivered by this module.
A393850 — proved in prose below: the integral Lagrange-root recurrence reduces to $B = 1 + x$ , its root $G$ to $x + x^{2}$ , its reversion to $S$ , and its series $A$ to $1 + S$ . This is a mathematical consequence of the coefficient recurrence and the classical Lagrange coefficient formula, not a new general Lagrange or derivative theorem in this module. Its separate preregistration is #18823.
Modulo four and neighbouring entries — open: the source’s separate conjecture $a (n) \equiv 2 (mod 4)$ iff $n /2$ is a sum of two distinct powers of two is not settled here. A393170, whose diagonal has $n$ in place of $n^{2}$ , has no settlement claim in this dossier. No conclusion about another source result is inferred without its own hypotheses and proof.

A382122: absolute values and all solution branches

Write $U = A^{- 1}$ and let $C = 1 + x C^{2}$ be the Catalan series. The equation is $\sum_{m \geq 0} x^{m} ∣ U^{m} ∣ = C$ , with absolute values taken coefficientwise. At degree $n + 1$ , the $m = 1$ term is $∣ u_{n} ∣$ ; the $m = 0$ term is zero and all $m \geq 2$ terms depend only on $u_{0}, \dots, u_{n - 1}$ . Starting from $u_{0} = 1$ , induction therefore shows $∣ u_{n} ∣$ is an integer, so $u_{n}$ is an integer. The inverse of a normalized integral series is integral, hence so is $A$ . For every integer $z$ , $∣ z ∣ \equiv z (mod 2)$ , regardless of sign. Reducing the whole equation gives $(1 - x \overset{ˉ}{U})^{- 1} = \overset{ˉ}{C}$ . But $C (1 - x C) = 1$ , so cancellation yields $\overset{ˉ}{U} = \overset{ˉ}{C}$ . Consequently $\overset{ˉ}{A} = \overset{ˉ}{C}^{- 1} = 1 + x \overset{ˉ}{C} = 1 + S$ : $x \overset{ˉ}{C}$ is the zero-constant solution of $T = x + T^{2}$ , whose coefficients are uniquely determined. Existence follows by taking $A = C^{- 1}$ over $Q$ : every coefficient of $C^{m}$ is nonnegative, so the absolute-value equation is the geometric identity $(1 - x C)^{- 1} = C$ . This proves the assertions for every solution without selecting a sign branch.

A393850: the integral root recurrence

For a normalized rational $A$ , let $G = x A (G)$ . The classical formal Lagrange coefficient identity $[x^{m}] G^{k} = (k / m) [t^{m - k}] A (t)^{m}$ converts the source equation $[t^{n}] A^{n + 1} = n (n + 1) [t^{n - 1}] A^{n + 1}$ into $[x^{m}] G = (2 m) [x^{m}] G^{2}$ for $m \geq 2$ . Writing $G = x B$ gives $b_{0} = 1$ and $b_{r} = (2 r + 1) [x^{r - 1}] B^{2}$ for $r \geq 1$ . This recurrence depends only on lower coefficients, so it constructs a unique integral $B$ . The series $G = x B$ has linear coefficient one; its reversion $R$ is integral, as are $R / x$ and $A = (R / x)^{- 1}$ . Reversing the Lagrange coefficient identity shows this construction satisfies the source equation. The coefficient of $x$ in $A$ is one. Every rational solution gives the same $B$ and hence the same $A$ .

Modulo two, $B = 1 + x$ satisfies the recurrence: its square is $1 + x^{2}$ ; $r = 1$ gives one, $r = 3$ gives $(2 4) = 6 = 0$ , and all other positive coefficients are zero. Coefficient uniqueness identifies the reduction of $B$ with this candidate. Thus $\overset{ˉ}{G} = x + x^{2}$ , $\overset{ˉ}{R} = S$ , and $\overset{ˉ}{A} = x / S = 1 + S$ . The positive-degree parity conclusion follows. This argument is a prose consequence; this dossier makes no claim of a canonical Lean proof of the sibling statement.

ASSUMED-UNVERIFIED

Literature completeness is ASSUMED-UNVERIFIED. The preregistration records reading the OEIS entry and attributes the revision-history and A393170 screen to the research producer. The supplied screen found no published proof; a search miss is not a completeness theorem or a claim of original discovery. The independent bounded reuse queries include heartbeat-limited queries and do not establish exhaustive coverage of the Lean ecosystem. The escape audit remains unfinished under #19804.

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