bibkey: hanna2025a380708 authors: Paul D. Hanna year: 2025 title: “OEIS A380708, g.f. satisfying A(x) = 1 + xabs(1/A(x))^2” doi: null url: https://oeis.org/A380708 claim: “G.f. A(x) satisfies A(x) = 1 + xabs( 1/A(x) )^2. Conjecture: for n > 0, a(n) is odd iff n = 2*A003714(k) + 1 for k >= 0, where A003714 are the fibbinary numbers.” strata_touched:

• D5/S1/Recurrence/Residue/AbsoluteReciprocalSquareFibbinaryParity license: citation-only triage: anchor

OEIS A380708

Paul D. Hanna’s entry, dated February 8, 2025, specifies the generating equation and parity conjecture quoted above. Absolute value is applied coefficientwise to the reciprocal series before squaring. The constant coefficient is one. The fibbinary numbers in A003714 are the nonnegative integers with no adjacent ones in their binary expansion.

The module constructs the unique integer solution by degree contraction. Modulo two, absolute value disappears. If F is the reduced series and B its reciprocal, then F = 1 + xB^2 and B = 1 + xB^3. The latter equation identifies B with the reduced ternary series from StripThreeTernaryCatalanParity. Its degree-f coefficient is binomial(3f,f) modulo two. Frobenius gives F = 1 + xB(x^2). Lucas descent and the binary-digit recursions show that binomial(3*f,f) is odd exactly when f is fibbinary, proving the conjecture.

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• URL: https://oeis.org/A380708