Parity of the A380558 Square-Reversion Series

Abstract

The normalized A380558 series has odd coefficients exactly at twice the stated binary-prefix indices.

Paul D. Hanna’s A380558 entry defines A by A(x-A(x))=x^2/(1-x^2) and labels its parity characterization a conjecture. The source also gives A=B^2/(1-B^2), where B(x-A(x))=x. The earlier project formalization of A380678 supplies B; this result identifies A with the unique normalized integral solution before proving parity.

A and B are integral formal power series, X is the indeterminate, and all indices and exponents are natural. Rational expressions mean formal inverses of denominators with constant coefficient one. The notation coeff(n,A) extracts degree n. The uniqueness clause quantifies over every integral series F whose constant and linear coefficients vanish; the parity clause covers every natural n.

Definition 1.1 (The source generating series).

$$A=\frac{(B{)}^2}{1-((B{)}^2)}$$

Formalization. D5/S1/Recurrence/Parity/A380558.generatingSeries (✓ std3).

Citation. Paul D. Hanna (2025). OEIS A380558, g.f. satisfying A(x - A(x)) = x^2/(1 - x^2). URL: https://oeis.org/A380558.

Commentary.

This is the transform in A380558’s formula field, applied to the previously constructed integral reversion series B. The denominator 1-B^2 is a unit because B has zero constant term.

Theorem 1.2 (Source identity, uniqueness, and complete parity support).

$$\begin{array}{r}((\mathrm{constantCoeff}\left(A\right)=0)\wedge (\mathrm{coeff}\left(1,A\right)=0))\wedge \\ (\mathrm{subst}\left(A,X-(A)\right)=\frac{(X{)}^2}{1-((X{)}^2)})\wedge \\ (\forall F:\mathrm{PowerSeries}\left(\mathbb{Z}\right),(((\mathrm{constantCoeff}\left(F\right)=0)\wedge (\mathrm{coeff}\left(1,F\right)=0))\wedge (\mathrm{subst}\left(F,X-(F)\right)=\frac{(X{)}^2}{1-((X{)}^2)}))⟹(F=A))\wedge \\ \forall n:\mathbb{N},\mathrm{Odd}\left(\mathrm{coeff}\left(n,A\right)\right)⟺(n=2\vee \exists m,j:\mathbb{N},(n=(2)\cdot (m))\wedge (((3)\cdot ((2{)}^j)\le m)\wedge (m<(4)\cdot ((2{)}^j))))\end{array}$$

Proof. Machine-checked in Lean as D5/S1/Recurrence/Parity/A380558.result (✓ std3). ∎

Resolves. Problems/oeis-a380558-parity (proved) by D5/S1/Recurrence/Parity/A380558.result.

Source. Repository-derived.

Acknowledgement. Paul D. Hanna (2025). OEIS A380558, g.f. satisfying A(x - A(x)) = x^2/(1 - x^2). URL: https://oeis.org/A380558.

Commentary.

Compositional inversion proves that the transformed series satisfies the source equation and is its only normalized integral solution. Modulo two, write L for the previously proved lacunary reduction of B and T=L/(1-L). Algebra gives T=X+X^2+(1+X)T^2. Frobenius and strong induction show that the coefficient of T is one exactly at 1 and in the dyadic half-open intervals [32^j,42^j). The reduction of A is T^2, which doubles those degrees. The exceptional value 1 in A004760 gives n=2; odd degrees and the upper endpoints are excluded. This settles the full bidirectional A380558 comment with its natural zero extension at degrees zero and one.

References

• Truth anchor: D5/S1/Recurrence/Parity/A380558.generatingSeries
• Truth anchor: D5/S1/Recurrence/Parity/A380558.result
• Dependency: D5/S1/Recurrence/Parity/QuadraticSquareReversionDyadicSupportParity