bibkey: hanna2025a381365 authors: Paul D. Hanna year: 2025 title: “OEIS A381365, scaled bilateral product generating function” doi: null url: https://oeis.org/A381365 claim: “G.f. A(x) satisfies 1/3 = Sum_{n=-oo..+oo} x^nA(x)^n * (A(x)^n + 2x)^(2n-1) * (x^n + 2A(x))^(2*n-1). Conjecture: for n > 0, a(n) == 6 (mod 9).” strata_touched:

• D5/S1/Recurrence/Bilateral/ScaledBilateralProductModNine license: citation-only triage: anchor

OEIS A381365

Paul D. Hanna’s entry, dated February 21, 2025, specifies the bilateral generating equation and the coefficient conjecture quoted above. The offset is 0,2 and the constant coefficient is one. This is the parameter c = 2 instance of the family with both inner exponents equal to c*n-1.

For a negative index n = -k, factoring the Laurent powers gives x^(c*k^2) * A^(c*k^2) * (1+2*x*A^k)^(-c*k-1) * (1+2*x^k*A)^(-c*k-1). For c >= 1, positive and negative terms have order at least k. Thus degree N is determined by the finite window -N <= n <= N. The zero-index term is (1+2*x)^(-1)*(1+2*A)^(-1) over the rationals.

The module proves the existence and uniqueness of a normalized integer solution to this literal equation. Opposite-index terms have equal reductions modulo two, so their sum has even coefficients. Writing the nonzero remainder as 2*J gives the integral contracting iteration A = 1 - 3*x*G + 3*(1+2*A)*J, where G = (1+2*x)^(-1). This first gives A = 1+3*B, then A - (1-3*x*G) = 9*(1+2*B)*J. Since the coefficient of G at m is (-2)^m, every positive coefficient of A has remainder six modulo nine. The general result applies to this entry.

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• URL: https://oeis.org/A381365