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
Keywords
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.
Programs
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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),", "))