bibkey: hanna2024a222013 authors: Paul D. Hanna year: 2024 title: “OEIS A222013, g.f. A(x) = Sum n! x^n A(x)^(n(n+1)/2) / Product (1 + k x A(x)^k), parity conjecture (Dec 2024)” doi: null url: https://oeis.org/A222013 claim: “G.f. satisfies: A(x) = Sum_{n>=0} n! * x^n * A(x)^(n*(n+1)/2) / Product_{k=1..n} (1 + kxA(x)^k). Conjecture: a(n) is odd iff n = 2^k - 1 for some k >= 0. - Paul D. Hanna, Dec 06 2024” strata_touched:

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

OEIS A222013

Paul D. Hanna created this entry on February 4, 2013. The parity conjecture quoted above is dated December 6, 2024. The coefficient sequence has offset zero and begins 1, 1, 2, 7, 32, 172, 1038, 6865, 49098, 376942, 3094812, 27129690, 253821716.

The formal equation uses the finite window 0 <= r <= N at coefficient degree N. Every denominator factor has constant coefficient one, and division denotes its formal unit inverse. The summand contains X^r, so every omitted index has zero coefficient at degree N. This gives the exact meaning of the infinite sum without an infinite-sum operation on formal series.

Degree contraction constructs the unique integer solution. Modulo two, all summands with r >= 2 vanish because 2 divides r!. The remaining terms give C = 1 + X*C^2. Thus X*C equals the zero-constant Catalan series modulo two, and binary Catalan support gives Odd(a(n)) exactly when n+1 = 2^k. Since powers of two are positive, this is equivalent to n = 2^k - 1.

The formal dependency is D5/S1/Recurrence/Invariants/CatalanCompositionSquareParity, specifically catalan_equation and binary_catalan. Its freeze prerequisite has statement ID sha256:0033f4a50a501546a5f332ce1203c8cf1a936428e7a155a06a0443bb0f0f6232.

Verified locator

• URL: https://oeis.org/A222013