OFFSET
0,2
COMMENTS
For each positive integer k, the sequence obtained by reducing a(n) modulo k is a periodic sequence with period dividing k. For example, modulo 5 the sequence becomes [1, 2, 2, 1, 0, 1, 2, 2, 1, 0, ...] with period 5. In particular, a(5*n+4) == 0 (mod 5). Cf. A047974. - Peter Bala, Mar 13 2025
FORMULA
a(n) ~ 2^(2*n + 1/2) * n^(n-1) / exp(n - 1/4).
From Peter Luschny, Nov 23 2021: (Start)
a(n) = n! * [x^n](exp(x)*(1 - sqrt(1 - 4*x))/(2*x)).
a(n) = (4*(n-1)*(n-2)*a(n - 3) - (n-1)*(8*n-5)*a(n - 2) + 4*n^2*a(n - 1))/(n + 1) for n >= 4.
a(n-1) = A224500(n) / n for n >= 1. (End)
a(n) = hypergeom([-n, 1/2, 1], [2], -4). - Peter Bala, Mar 13 2025
MAPLE
gf := exp(x)*(1 - sqrt(1 - 4*x))/(2*x): ser := series(gf, x, 24):
seq(n!*coeff(ser, x, n), n = 0..19);
# Alternative:
a := n -> `if`(n < 4, [1, 2, 7, 46][n + 1], ((4*n^2 - 12*n + 8)*a(n - 3) - (8*n^2 - 13*n + 5)*a(n - 2) + 4*n^2*a(n - 1))/(n + 1)):
seq(a(n), n = 0..19); # Peter Luschny, Nov 23 2021
# Alternative:
seq(simplify(hypergeom([-n, 1/2, 1], [2], -4)), n = 0..19); # Peter Bala, Mar 13 2025
MATHEMATICA
Table[Sum[Binomial[n, j]*CatalanNumber[j]*j!, {j, 0, n}], {n, 0, 20}]
PROG
(PARI) a(n) = sum(k=0, n, binomial(n, k) * (binomial(2*k, k)/(k+1)) * k!); \\ Michel Marcus, Nov 23 2021
CROSSREFS
KEYWORD
nonn
AUTHOR
Vaclav Kotesovec, Nov 23 2021
STATUS
approved