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%I #5 Feb 06 2013 20:14:46
%S 1,3,25,210,1881,17303,162214,1540710,14776281,142774455,1387743525,
%T 13553773500,132906406950,1307654814222,12902933709922,
%U 127632756058610,1265251299930585,12566655467547195,125025126985317013,1245750306517239978,12429515281592007781
%N a(n) = A222051(n)/binomial(2*n,n), the central terms in rows of triangle A220178 divided by the central binomial coefficients.
%F a(n) = [x^n] 1/sqrt(1-2*x-3*x^2))^(2*n+1).
%F 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).
%e G.f.: A(x) = 1 + 3*x + 25*x^2 + 210*x^3 + 1881*x^4 + 17303*x^5 +...
%e Illustrate a(n) = [x^n] 1/sqrt(1-2*x-3*x^2))^(2*n+1):
%e Let G(x) = 1/sqrt(1-2*x-3*x^2) be the g.f. of A002426, then
%e the array of coefficients of x^k in G(x)^(2*n+1) begins:
%e G(x)^1 : [1, 1, 3, 7, 19, 51, 141, 393,...];
%e G(x)^3 : [1, 3, 12, 40, 135, 441, 1428, 4572,...];
%e G(x)^5 : [1, 5, 25, 105, 420, 1596, 5880, 21120,...];
%e G(x)^7 : [1, 7, 42, 210, 966, 4158, 17094, 67782,...];
%e G(x)^9 : [1, 9, 63, 363, 1881, 9009, 40755, 176319,...];
%e G(x)^11: [1, 11, 88, 572, 3289, 17303, 85228, 398684,...];
%e G(x)^13: [1, 13, 117, 845, 5330, 30498, 162214, 814606,...];
%e G(x)^15: [1, 15, 150, 1190, 8160, 50388, 287470, 1540710,...]; ...
%e in which the main diagonal forms this sequence.
%o (PARI) {a(n)=polcoeff(1/sqrt(1-2*x-3*x^2+x*O(x^n))^(2*n+1),n)}
%o for(n=0,25,print1(a(n),", "))
%Y Cf. A222050, A222051.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Feb 06 2013
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