bibkey: hanna2016a266489 authors: Paul D. Hanna year: 2016 title: “OEIS A266489, g.f. A(x) satisfies: [x^n] A(x/A(x)^n) = 0 for n>1” doi: null url: https://oeis.org/A266489 claim: “G.f. A(x) satisfies: [x^n] A( x/A(x)^n ) = 0 for n>1. (C2) a(n) == 0 (mod 2) for n>=2.” strata_touched:

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

OEIS A266489

Paul D. Hanna’s entry, dated February 7, 2016, gives the generating equation and the congruence conjecture (C2) quoted above. The normalization is a(0) = a(1) = 1. Its first coefficients are 1, 1, 2, 12, 132.

The formal module constructs the unique normalized integer series for the family with exponent (p - 1)*(n - 1) + 1. Its specialization at p = 2 has exponent n for n > 1 and proves (C2). 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 one congruence clause of A395833, a different sequence. No equivalence of the two sequences is asserted.

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• URL: https://oeis.org/A266489