bibkey: hanna2024a379204 authors: Paul D. Hanna year: 2024 title: “OEIS A379204, g.f. satisfying 1/x = Sum_{n in Z} A(x)^n (A(x)^n + 4)^(n+1)” doi: null url: https://oeis.org/A379204 claim: “G.f. A(x) satisfies 1/x = Sum_{n=-oo..+oo} A(x)^n * (A(x)^n + 4)^(n+1). Conjecture: a(n) is even for n > 1. Conjecture: a(n) == 2 (mod 4) iff n = (k-1)^2 + 1 for some k > 1.” strata_touched:
- D5/S1/Recurrence/Parity/BilateralQuarticPowerThetaModFour license: citation-only triage: anchor
OEIS A379204
Paul D. Hanna’s entry, dated December 20, 2024, specifies the bilateral
generating equation and the two coefficient conjectures quoted above. The
normalization is A(0)=0, a(1)=1, and a(n)=[x^n]A(x).
For index n=-r, with r>=2, the summand is read as
A^((r-1)^2-1) * invOfUnit(1+4*A^r,1)^(r-1). The index -1 is the Laurent
term A^(-1). After multiplication by X*A, this term contributes the isolated
X in A=X+X*A*sum(bilateralTerm(A,n)), and bilateralTerm(A,-1) is zero.
The sum at degree N uses exactly the finite window -N-2<=n<=N.
The order bound proves that every omitted summand has zero coefficient at N;
thus no bilateral infinite-sum operation on formal power series is needed.
The module constructs the unique zero-constant solution by degree contraction.
Modulo four, multiplying each summand by A and pairing indices m and -m-2
gives two copies of A^((m+1)^2). The resulting equation implies A=X modulo
two. Doubling lifts equality modulo two to equality modulo four, yielding
A=X*thetaSeries modulo four. Hence the remainder is two exactly at the
positive-square shifts, and this classification implies evenness for n>1.
The companion remainder-zero classification includes the condition n>1,
since a(1)=1.
Verified locator
- URL: https://oeis.org/A379204
Freeze prerequisite
D5/S1/Recurrence/Parity/ThetaSelfCompositionModFour- statement_id:
sha256:0aec24647d5df63513f2c164c40043a38afc3ac9d0a739e594942fb5328c725b - The imported public declarations are
thetaSeriesandcoeff_thetaSeries.