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v[1] = 1; v[2] = -2 + x; for(n = 3, N, my(i = n); v[i] = (n - 4 + x)*v[i - 1] - (n - 2)*x*v[i - 2]); v};\nrow(n, p) = {my(c = Vecrev(p)); c};\nT(N) = {my(v = P(N)); concat(vector(#v, i, row(i, v[i])))};\nT(10)","reading":{"first_row":1,"offset":1,"reversed":false,"trim_zeros":"none","regular_triangle":false}},{"oeis_id":"A132014","target":"triangle","first_row":1,"oeis_offset":0,"reversed":false,"trim_zeros":"none","regular_triangle":true,"rows_shown":10,"polytool_commit":"c48629374597092cc41ef4d6273f37991516f984","triangle_example":"Triangle begins:\n  0:  1;\n  1: -2,  1;\n  2:  2, -4,  1;\n  3:  0,  6, -6,  1;\n  4:  0,  0, 12, -8,   1;\n  5:  0,  0,  0, 20, -10,   1;\n  6:  0,  0,  0,  0,  30, -12,   1;\n  7:  0,  0,  0,  0,   0,  42, -14,   1;\n  8:  0,  0,  0,  0,   0,   0,  56, -16,   1;\n  9:  0,  0,  0,  0,   0,   0,   0,  72, -18, 1;\n","recurrence_sentence":"The row polynomials satisfy P(n,x) = (n-3+x)*P(n-1,x) - (n-1)*x*P(n-2,x) for n >= 2, with P(0,x) = 1, P(1,x) = -2 + x.","mathematica":"P[0, x_] := 1;\nP[1, x_] := -2 + x;\nP[n_, x_] := P[n, x] = Expand[(n - 3 + x)*P[n - 1, x] - (n - 1)*x*P[n - 2, x]];\nTable[PadRight[CoefficientList[P[n, x], x], n + 1], {n, 0, 9}] // Flatten","pari":"P(N) = {my(v = vector(N + 1)); v[1] = 1; v[2] = -2 + x; for(n = 2, N, my(i = n + 1); v[i] = (n - 3 + x)*v[i - 1] - (n - 1)*x*v[i - 2]); v};\nrow(n, p) = {my(c = Vecrev(p)); c = concat(c, vector(n + 1 - #c)); c};\nT(N) = {my(v = P(N)); concat(vector(#v, i, row(i - 1, v[i])))};\nT(9)","reading":{"first_row":1,"offset":0,"reversed":false,"trim_zeros":"none","regular_triangle":true}}],"derivation":{"root":"n-6fea5aff4f44fa54","steps":["D\u00b2\u22122D+1 (shift -1)"],"edges":["e-082a4f27a6bb712f"],"text":"A023531 \u2192 D\u00b2\u22122D+1 (shift -1)"},"headline":"A132014: T(n,j) for double application of an iterated mixed order Laguerre transform: Coefficients of Laguerre polynomial (-1)^n*n!*L(n,2-n,x).","chain":"A023531 \u2192 D\u00b2\u22122D+1 (shift -1)"}