A222052 a(n) = A222051(n)/binomial(2*n,n), the central terms in rows of triangle A220178 divided by the central binomial coefficients.

1, 3, 25, 210, 1881, 17303, 162214, 1540710, 14776281, 142774455, 1387743525, 13553773500, 132906406950, 1307654814222, 12902933709922, 127632756058610, 1265251299930585, 12566655467547195, 125025126985317013, 1245750306517239978, 12429515281592007781

Offset

0,2

Links

Table of n, a(n) for n=0..20.

Formula

a(n) = [x^n] 1/sqrt(1-2*x-3*x^2))^(2*n+1).

a(n) = (2*n+1)*A222050(n), where g.f. G(x) of A222050 satisfies: G(x) = sqrt(1 + 2*x*G(x)^4 + 3*x^2*G(x)^6).

Example

G.f.: A(x) = 1 + 3*x + 25*x^2 + 210*x^3 + 1881*x^4 + 17303*x^5 +...
Illustrate a(n) = [x^n] 1/sqrt(1-2*x-3*x^2))^(2*n+1):
Let G(x) = 1/sqrt(1-2*x-3*x^2) be the g.f. of A002426, then
the array of coefficients of x^k in G(x)^(2*n+1) begins:
G(x)^1 : [1,  1,   3,    7,    19,    51,    141,     393,...];
G(x)^3 : [1,  3,  12,   40,   135,   441,   1428,    4572,...];
G(x)^5 : [1,  5,  25,  105,   420,  1596,   5880,   21120,...];
G(x)^7 : [1,  7,  42,  210,   966,  4158,  17094,   67782,...];
G(x)^9 : [1,  9,  63,  363,  1881,  9009,  40755,  176319,...];
G(x)^11: [1, 11,  88,  572,  3289, 17303,  85228,  398684,...];
G(x)^13: [1, 13, 117,  845,  5330, 30498, 162214,  814606,...];
G(x)^15: [1, 15, 150, 1190,  8160, 50388, 287470, 1540710,...]; ...
in which the main diagonal forms this sequence.

Prog

(PARI) {a(n)=polcoeff(1/sqrt(1-2*x-3*x^2+x*O(x^n))^(2*n+1), n)}
for(n=0, 25, print1(a(n), ", "))

Crossrefs

Cf. A222050, A222051.

Sequence in context: A037783 A037587 A280970 * A230718 A112240 A155640

Adjacent sequences: A222049 A222050 A222051 * A222053 A222054 A222055

Keyword

nonn

Author

Paul D. Hanna, Feb 06 2013

Status

approved