bibkey: hanna2023a363184 authors: Paul D. Hanna year: 2023 title: “OEIS A363184, g.f. satisfying 4 = Sum_{n in Z} (-1)^n x^n (4 A(x) + x^(2n-1))^(n+1)” doi: null url: https://oeis.org/A363184 claim: “Expansion of g.f. A(x) satisfying 4 = Sum_{n=-oo..+oo} (-1)^n * x^n * (4A(x) + x^(2n-1))^(n+1). Conjecture: g.f. A(x) == theta_3(x^2) (mod 4); a(n) == 2 (mod 4) if n = 2*k^2 for integer k > 0, and a(n) == 0 (mod 4) if floor(n/2) is nonsquare.” strata_touched:

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

OEIS A363184

Paul D. Hanna’s entry, dated May 20, 2023, specifies the bilateral generating equation and the conjecture quoted above. The sequence has offset zero, with initial coefficients 1, 4, 18, 88, 452.

The terms at indices zero and minus one contain opposite Laurent monomials x^(-1) and -x^(-1). Cancelling them leaves 4*A. For n=-m-2, factoring out the Laurent monomial gives the power-series term (-1)^(m+2) * X^(2*m^2+6*m+3) * invOfUnit(1+4*A*X^(2*m+5),1)^(m+1). The positive term at n=m+1 is (-1)^(m+1) * X^(m+1) * (4*A+X^(2*m+1))^(m+2). Both terms have order at least m+1, so the exact coefficientwise meaning of the equation uses the finite integer window -N-2 <= n <= N at degree N. It requires no bilateral infinite-sum operation on formal power series.

The module constructs the unique integral solution by degree contraction. Linearization modulo sixteen and a weighted telescoping sum yield an integral identity 1=A*expand(2,thetaSeries)+4*Q. Cancelling four over the integers before reduction, and then using the self-inverse property of 1+2*S modulo four, proves the theta congruence. Here S indicates positive doubled squares. Consequently the positive coefficient is two modulo four exactly at n=2*k^2 with k>0, and zero modulo four otherwise. The nonsquare-floor clause follows from this classification, and the constant coefficient is one.

Verified locator

• URL: https://oeis.org/A363184