bibkey: hanna2026a395833 authors: Paul D. Hanna year: 2026 title: “OEIS A395833, g.f. A(x) satisfies [x^n] A(x/A(x)^(2n-1)) = 0 for n > 1” doi: null url: https://oeis.org/A395833 claim: “G.f. A(x) satisfies [x^n] A( x/A(x)^(2n-1) ) = 0 for n > 1. Conjecture: a(n) == 0 (mod 3) for n >= 2.” strata_touched:

• D5/S1/Recurrence/Residue/NegativePowerDiagonalModPrime license: citation-only triage: anchor

OEIS A395833

Paul D. Hanna’s entry, dated May 7, 2026, gives the generating equation and the congruence conjecture quoted above. The normalization is a(0) = a(1) = 1. Its first coefficients are 1, 1, 3, 30, 567.

The formal module constructs the unique normalized integer series for the family with exponent (p - 1)*(n - 1) + 1. Its specialization at p = 3 has exponent 2*n - 1 for n > 1 and proves the quoted mod-three clause. Unit inverses give the exact meaning of the quotient in the substitution. The substitution coefficient has leading term a(n) with multiplier one, giving a triangular construction and uniqueness. Modulo p, the residual of 1 + x in degree r + 1 is (-1)^r * choose(p*r,r). The identity choose(p*r,r) = p * choose(p*r-1,r-1) for r > 0 makes it zero.

The same general theorem proves (C2) of A266489, a different sequence. No equivalence of the two sequences is asserted.

Verified locator

• URL: https://oeis.org/A395833