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A377217
Expansion of the o.g.f. A(x) defined by [x^n] A(x)^(6*n) = (3*n)!/n!^3 for n >= 0.
2
1, 1, 2, 14, 127, 1364, 16219, 206715, 2770342, 38567069, 553153830, 8126285739, 121758839828, 1854687918895, 28649693078544, 447912211497740, 7076246388778874, 112821090561117084, 1813395701702453669
OFFSET
0,3
COMMENTS
Compare with A000984(n) = [x^n] (1 + x)^(2*n) = (2*n)!/n!^2.
The central binomial coefficients A000984(n) satisfy the supercongruences u(n*p^k) == u(n*p^(k-1)) (mod p^(3*k)) for all primes p >= 5 and positive integers n and k.
More generally, for positive integers r and s, the sequence {u(r,s; n) : n >= 0} defined by u(r,s; n) = [x^(s*n)] (1 + x)^(r*n) = binomial(r*n, s*n) satisfies the same supercongruences (Meštrović, Section 6, equation 39).
Conjecture: for positive integers r and s, the sequence {v(r,s; n) : n >= 0} defined by v(r,s; n) = [x^(s*n)] A(x)^(r*n) also satisfies the same supercongruences.
FORMULA
O.g.f.: A(x) = ( x/(x * series_reversion(E(x)))^(1/6), where E(x) = exp(Sum_{n >= 1} (3*n)!/n!^3 *x^n/n) is the o.g.f. of A229451.
MAPLE
Order := 25:
E(x) := exp(add((3*n)!/n!^3 * x^n/n, n = 1..25)):
solve(series(x*E(x), x) = y, x):
convert(%, polynom):
g := taylor(y/%, y = 0, 25):
seq(coeftayl(g^(1/6), y = 0, n), n = 0..20);
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Oct 20 2024
STATUS
approved