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%I #7 Jan 27 2024 12:33:04
%S 1,12,97,696,4784,32230,214978,1426566,9441417,62405645,412278981,
%T 2723566163,17996243101,118957645301,786700165122,5205396517853,
%U 34461624895701,228274455988134,1512920531980961,10032446308837778
%N Column 2 of triangle A128592; a(n) = coefficient of q^(2n+6) in the q-analog of the odd double factorials (2n+5)!! for n>=0.
%F a(n) = [q^(2n+6)] Product_{j=1..n+3} (1-q^(2j-1))/(1-q) for n>=0.
%t a[n_] := SeriesCoefficient[Product[(1-q^(2j-1))/(1-q), {j, 1, n+3}], {q, 0, 2n+6}];
%t Table[a[n], {n, 0, 30}] (* _Jean-François Alcover_, Jan 27 2024 *)
%o (PARI) {a(n)=polcoeff(prod(j=1,n+3,(1-q^(2*j-1))/(1-q)),2*n+6,q)}
%Y Cf. A128592; A128080; A001147 ((2n-1)!!); A128593 (column 1), A128595 (row sums).
%K nonn
%O 0,2
%A _Paul D. Hanna_, Mar 12 2007