bibkey: hanna2022a355872 authors: Paul D. Hanna year: 2022 title: “OEIS A355872, g.f. satisfying x = Sum_{n=-oo..+oo} (-x)^(n^2) * A(x)^((n-1)^2)” doi: null url: https://oeis.org/A355872 claim: “G.f. A(x) satisfies: x = Sum_{n=-oo..+oo} (-x)^(n^2) * A(x)^((n-1)^2). Conjecture: a(n) == 2 (mod 4) for n >= 1. Conjecture: a(2n-1) == 2 (mod 8) for n >= 1. Conjecture: a(2n) == 6 (mod 8) for n >= 1.” strata_touched:

• D5/S1/Recurrence/Parity/BilateralThetaReversionModSixteen license: citation-only triage: anchor

OEIS A355872

Paul D. Hanna’s entry, dated August 9, 2022, specifies the bilateral generating equation and the three coefficient conjectures quoted above. The integer series has zero constant coefficient. Its coefficient function is c(m) = [x^m]A(x); the sequence is indexed by a(n) = c(4*n-3) for n >= 1.

The formal equation uses the coefficientwise finite window of integer indices -N <= j <= N at degree N. This is the exact meaning of the bilateral sum: the summand has order at least j^2 + (j-1)^2, and every index outside the window has j^2 > N. No bilateral infinite-sum operation on formal series is needed.

The module constructs the unique solution by degree contraction, proves that c(m) = 0 unless m mod 4 = 1, and identifies its reduction modulo 16 with 2*x/(1+x^4). Hence a(n) is congruent to 2*(-1)^(n-1) modulo 16, which implies all three conjectures. The quotient here denotes a formal unit inverse.

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• URL: https://oeis.org/A355872