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x])/(4*n^2 + 14*n + 12)]];\nTable[CoefficientList[P[n, x], x], {n, 0, 9}] // Flatten","pari":"P(N) = {my(v = vector(N + 1)); v[1] = 6 + 8*x + x^2; for(n = 1, N, my(i = n + 1); v[i] = ((15*n^2 + 46*n + 35 + (12*n^2 + 35*n + 25)*x)*v[i - 1] + (n + 2 + (7*n + 11)*x - (8*n + 13)*x^2)*deriv(v[i - 1], x))/(4*n^2 + 14*n + 12)); v};\nrow(n, p) = {my(c = Vecrev(p)); c};\nT(N) = {my(v = P(N)); concat(vector(#v, i, row(i - 1, v[i])))};\nT(9)","reading":{"first_row":0,"offset":0,"reversed":false,"trim_zeros":"none","regular_triangle":false}}],"derivation":{"root":"n-f46f66d0d087b029","steps":[],"edges":[],"text":"A110608"},"headline":"Number triangle T(n,k) = binomial(n,k)*binomial(2n,n-k).","chain":null}