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248005800,   267846264,   156243654,    38263752;\n  8:  1679616,  40310784,  317447424,  1269789696,  2976069600,  4285540224,  3749847696,  1836660096,   387420489;\n  9: 10077696, 272097792, 2448880128, 11428107264, 32141551680, 57854793024, 67497258528, 49589822592, 20920706406, 3874204890;\n","recurrence_sentence":"The row polynomials satisfy (2+3*n*x)*P(n,x) = (12+(18*n+36)*x+(27*n+27)*x^2)*P(n-1,x) for n >= 1, with P(0,x) = 1.","mathematica":"P[0, x_] := 1;\nP[n_, x_] := P[n, x] = Expand[Together[((12 + (18*n + 36)*x + (27*n + 27)*x^2)*P[n - 1, x])/(2 + 3*n*x)]];\nTable[CoefficientList[P[n, x], x], {n, 0, 9}] // Flatten","pari":"P(N) = {my(v = vector(N + 1)); v[1] = 1; for(n = 1, N, my(i = n + 1); v[i] = ((12 + (18*n + 36)*x + (27*n + 27)*x^2)*v[i - 1])/(2 + 3*n*x)); 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-8d88ef9de74b305f","steps":["basis A013610","\u03b8+1"],"edges":["e-b88e4ed4246b1dea","e-8be16f3bdc419e18"],"text":"A038221 \u2192 basis A013610 \u2192 \u03b8+1"},"headline":"A038221 \u2192 basis A013610 \u2192 \u03b8+1","chain":null}