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

%I A222052 #8 Nov 02 2024 03:22:20
%S A222052 1,3,25,210,1881,17303,162214,1540710,14776281,142774455,1387743525,
%T A222052 13553773500,132906406950,1307654814222,12902933709922,
%U A222052 127632756058610,1265251299930585,12566655467547195,125025126985317013,1245750306517239978,12429515281592007781
%N A222052 a(n) = A222051(n)/binomial(2*n,n), the central terms in rows of triangle A220178 divided by the central binomial coefficients.
%F A222052 a(n) = [x^n] 1/(sqrt(1-2*x-3*x^2))^(2*n+1).
%F A222052 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 A222052 G.f.: A(x) = 1 + 3*x + 25*x^2 + 210*x^3 + 1881*x^4 + 17303*x^5 +...
%e A222052 Illustrate a(n) = [x^n] 1/(sqrt(1-2*x-3*x^2))^(2*n+1):
%e A222052 Let G(x) = 1/sqrt(1-2*x-3*x^2) be the g.f. of A002426, then
%e A222052 the array of coefficients of x^k in G(x)^(2*n+1) begins:
%e A222052 G(x)^1 : [1,  1,   3,    7,    19,    51,    141,     393,...];
%e A222052 G(x)^3 : [1,  3,  12,   40,   135,   441,   1428,    4572,...];
%e A222052 G(x)^5 : [1,  5,  25,  105,   420,  1596,   5880,   21120,...];
%e A222052 G(x)^7 : [1,  7,  42,  210,   966,  4158,  17094,   67782,...];
%e A222052 G(x)^9 : [1,  9,  63,  363,  1881,  9009,  40755,  176319,...];
%e A222052 G(x)^11: [1, 11,  88,  572,  3289, 17303,  85228,  398684,...];
%e A222052 G(x)^13: [1, 13, 117,  845,  5330, 30498, 162214,  814606,...];
%e A222052 G(x)^15: [1, 15, 150, 1190,  8160, 50388, 287470, 1540710,...]; ...
%e A222052 in which the main diagonal forms this sequence.
%o A222052 (PARI) {a(n)=polcoeff(1/sqrt(1-2*x-3*x^2+x*O(x^n))^(2*n+1),n)}
%o A222052 for(n=0,25,print1(a(n),", "))
%Y A222052 Cf. A222050, A222051.
%K A222052 nonn
%O A222052 0,2
%A A222052 _Paul D. Hanna_, Feb 06 2013