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

Author

Paul D. Hanna, Feb 06 2013

Keywords

• nonn

Examples

			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.
		

Crossrefs

Cf. A222050, A222051.

Programs

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),", "))

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).