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3, x] + (1 + 4*x)*P[n - 1, x]];\nTable[CoefficientList[P[n, x], x], {n, 0, 9}] // Flatten","pari":"P(N) = {my(v = vector(N + 1)); v[1] = 1 + 4*x; v[2] = 1 + 8*x + 16*x^2; v[3] = 1 + 12*x + 48*x^2 + 64*x^3; for(n = 3, N, my(i = n + 1); v[i] = (1 + 8*x + 16*x^2)*v[i - 3] + (1 + 4*x)*v[i - 1]); 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-0d55e0bb3c384646","steps":["reverse","basis A013611"],"edges":["e-0e9e334768ea0a35","e-dee121250e432a7f"],"text":"A102547 \u2192 reverse \u2192 basis A013611"},"headline":"A102547 \u2192 reverse \u2192 basis A013611","chain":null}