Keyboard shortcuts

Press ← or → to navigate between chapters

Press ? to show this help

Press Esc to hide this help


bibkey: hanna2023a363104 authors: Paul D. Hanna year: 2023 title: “OEIS A363104, g.f. satisfying 4 = Sum_{n in Z} (-x)^n (4 A(x) + x^(n-1))^(n+1)” doi: null url: https://oeis.org/A363104 claim: “Expansion of g.f. A(x) satisfying 4 = Sum_{n=-oo..+oo} (-x)^n * (4*A(x) + x^(n-1))^(n+1). Conjecture: g.f. A(x) == theta_3(x) (mod 4); a(n) == 2 (mod 4) iff n is a nonzero square and a(n) == 0 (mod 4) iff n is nonsquare.” strata_touched:

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

OEIS A363104

Paul D. Hanna’s entry, dated May 21, 2023, specifies the generating equation and conjecture quoted above. The sequence has offset zero and constant term one.

The formal reading first cancels the Laurent monomials at indices zero and minus one, leaving 4*A at index zero and zero at index minus one. The remaining terms are power series. Pair index m+1 with -m-2, for natural m: their terms independent of A cancel, and the pair is divisible by four over the integers. At degree N, the exact window is -N-2 <= n <= N; every index outside this window has zero coefficient. No bilateral infinite-sum operation on formal series is asserted.

Dividing the paired remainder by four gives an integral degree-contracting iteration. Its stabilized coefficients define the unique solution of the finite-window equation. Modulo sixteen, the terms linear in A telescope, giving 4 = 4*A*theta_3 + 16*Q for an integer power series Q. Cancellation over the integers gives 1 = A*theta_3 + 4*Q. The reduction of theta_3 is self-inverse modulo four, proving the series congruence and both coefficient biconditionals at every positive index.

Verified locator

  • URL: https://oeis.org/A363104