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7983360,  31933440,  74844000, 107775360,  94303440,  46189440,   9743085;\n  9: 112640, 3041280, 27371520, 127733760, 359251200, 646652160, 754427520, 554273280, 233834040, 43302600;\n","recurrence_sentence":"The row polynomials satisfy (2*n+3*n^2*x)*P(n,x) = (4*n+12+(6*n^2+30*n+36)*x+(9*n^2+36*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[((4*n + 12 + (6*n^2 + 30*n + 36)*x + (9*n^2 + 36*n + 27)*x^2)*P[n - 1, x])/(2*n + 3*n^2*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] = ((4*n + 12 + (6*n^2 + 30*n + 36)*x + (9*n^2 + 36*n + 27)*x^2)*v[i - 1])/(2*n + 3*n^2*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-a820c1859f68e71c","steps":["basis A013610","\u03b8+1"],"edges":["e-24fdbe8219e7135e","e-fd21c1c3934f9ea2"],"text":"A178820 \u2192 basis A013610 \u2192 \u03b8+1"},"headline":"A178820 \u2192 basis A013610 \u2192 \u03b8+1","chain":null}