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v[1] = 1; v[2] = 1 - x; for(n = 3, N, my(i = n); v[i] = ((2*n - 4 - (n - 2)*x)*v[i - 1])/(n - 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":"A118800","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:   1,    -1;\n  2:   2,    -3,    1;\n  3:   4,    -8,    5,    -1;\n  4:   8,   -20,   18,    -7,    1;\n  5:  16,   -48,   56,   -32,    9,    -1;\n  6:  32,  -112,  160,  -120,   50,   -11,   1;\n  7:  64,  -256,  432,  -400,  220,   -72,  13,   -1;\n  8: 128,  -576, 1120, -1232,  840,  -364,  98,  -15,  1;\n  9: 256, -1280, 2816, -3584, 2912, -1568, 560, -128, 17, -1;\n","recurrence_sentence":"The row polynomials satisfy (n-1)*P(n,x) = (2*n-2-(n-1)*x)*P(n-1,x) for n >= 2, with P(0,x) = 1, P(1,x) = 1 - x.","mathematica":"P[0, x_] := 1;\nP[1, x_] := 1 - x;\nP[n_, x_] := P[n, x] = Expand[Together[((2*n - 2 - (n - 1)*x)*P[n - 1, x])/(n - 1)]];\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] = 1 - x; for(n = 2, N, my(i = n + 1); v[i] = ((2*n - 2 - (n - 1)*x)*v[i - 1])/(n - 1)); 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-7c1541c019618244","steps":[],"edges":[],"text":"A118800"},"headline":"A118800: Triangle read by rows: T satisfies the matrix products: C*T*C = T^-1 and T*C*T = C^-1, where C is Pascal's triangle.","chain":null}