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v[i])))};\nT(9)","reading":{"first_row":0,"offset":0,"reversed":false,"trim_zeros":"none","regular_triangle":false}},{"oeis_id":"A078812","target":"reversed_triangle","first_row":0,"oeis_offset":0,"reversed":true,"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:  3,   4,   1;\n  3:  4,  10,   6,    1;\n  4:  5,  20,  21,    8,    1;\n  5:  6,  35,  56,   36,   10,    1;\n  6:  7,  56, 126,  120,   55,   12,   1;\n  7:  8,  84, 252,  330,  220,   78,  14,   1;\n  8:  9, 120, 462,  792,  715,  364, 105,  16,  1;\n  9: 10, 165, 792, 1716, 2002, 1365, 560, 136, 18, 1;\n","recurrence_sentence":"Row n lists the coefficients of P(n,x) in descending powers of x, where P(n,x) = (1+2*x)*P(n-1,x) - x^2*P(n-2,x) for n >= 2, with P(0,x) = 1, P(1,x) = 1 + 2*x.","mathematica":"P[0, x_] := 1;\nP[1, x_] := 1 + 2*x;\nP[n_, x_] := P[n, x] = Expand[(1 + 2*x)*P[n - 1, x] - x^2*P[n - 2, x]];\nTable[Reverse[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 + 2*x; for(n = 2, N, my(i = n + 1); v[i] = (1 + 2*x)*v[i - 1] - x^2*v[i - 2]); v};\nrow(n, p) = {my(c = Vecrev(p)); c = concat(c, vector(n + 1 - #c)); c = Vecrev(c); 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":true,"trim_zeros":"none","regular_triangle":true}},{"oeis_id":"A172431","target":"triangle","first_row":0,"oeis_offset":1,"reversed":false,"trim_zeros":"none","regular_triangle":true,"rows_shown":10,"polytool_commit":"c48629374597092cc41ef4d6273f37991516f984","triangle_example":"Triangle begins:\n   1: 1;\n   2: 1,  2;\n   3: 1,  4,   3;\n   4: 1,  6,  10,   4;\n   5: 1,  8,  21,  20,    5;\n   6: 1, 10,  36,  56,   35,    6;\n   7: 1, 12,  55, 120,  126,   56,    7;\n   8: 1, 14,  78, 220,  330,  252,   84,   8;\n   9: 1, 16, 105, 364,  715,  792,  462, 120,   9;\n  10: 1, 18, 136, 560, 1365, 2002, 1716, 792, 165, 10;\n","recurrence_sentence":"The row polynomials satisfy P(n,x) = (1+2*x)*P(n-1,x) - x^2*P(n-2,x) for n >= 3, with P(1,x) = 1, P(2,x) = 1 + 2*x.","mathematica":"P[1, x_] := 1;\nP[2, x_] := 1 + 2*x;\nP[n_, x_] := P[n, x] = Expand[(1 + 2*x)*P[n - 1, x] - x^2*P[n - 2, x]];\nTable[PadRight[CoefficientList[P[n, x], x], n], {n, 1, 10}] // Flatten","pari":"P(N) = {my(v = vector(N)); v[1] = 1; v[2] = 1 + 2*x; for(n = 3, N, my(i = n); v[i] = (1 + 2*x)*v[i - 1] - x^2*v[i - 2]); v};\nrow(n, p) = {my(c = Vecrev(p)); c = concat(c, vector(n - #c)); c};\nT(N) = {my(v = P(N)); concat(vector(#v, i, row(i, v[i])))};\nT(10)","reading":{"first_row":0,"offset":1,"reversed":false,"trim_zeros":"none","regular_triangle":true}}],"derivation":{"root":"n-79ebafb266da3b41","steps":[],"edges":[],"text":"A172431"},"headline":"A172431: Even row Pascal-square read by antidiagonals.","chain":null}