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begins:\n   1: 1;\n   2: 1,  2;\n   3: 1,  4,   2;\n   4: 1,  6,   8,   2;\n   5: 1,  8,  18,  12,   2;\n   6: 1, 10,  32,  38,  16,    2;\n   7: 1, 12,  50,  88,  66,   20,   2;\n   8: 1, 14,  72, 170, 192,  102,  24,   2;\n   9: 1, 16,  98, 292, 450,  360, 146,  28,  2;\n  10: 1, 18, 128, 462, 912, 1002, 608, 198, 32, 2;\n","recurrence_sentence":"The row polynomials satisfy P(n,x) = (1+x)*P(n-1,x) + x*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 + x)*P[n - 1, x] + x*P[n - 2, x]];\nTable[CoefficientList[P[n, x], x], {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 + x)*v[i - 1] + 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":"A035607","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,  2;\n  2: 1,  4,   2;\n  3: 1,  6,   8,   2;\n  4: 1,  8,  18,  12,   2;\n  5: 1, 10,  32,  38,  16,    2;\n  6: 1, 12,  50,  88,  66,   20,   2;\n  7: 1, 14,  72, 170, 192,  102,  24,   2;\n  8: 1, 16,  98, 292, 450,  360, 146,  28,  2;\n  9: 1, 18, 128, 462, 912, 1002, 608, 198, 32, 2;\n","recurrence_sentence":"The row polynomials satisfy P(n,x) = (1+x)*P(n-1,x) + x*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 + x)*P[n - 1, x] + 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] = 1 + 2*x; for(n = 2, N, my(i = n + 1); v[i] = (1 + x)*v[i - 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}},{"oeis_id":"A113413","target":"reversed_triangle","first_row":1,"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: 2,  4,   1;\n  3: 2,  8,   6,   1;\n  4: 2, 12,  18,   8,    1;\n  5: 2, 16,  38,  32,   10,   1;\n  6: 2, 20,  66,  88,   50,  12,   1;\n  7: 2, 24, 102, 192,  170,  72,  14,   1;\n  8: 2, 28, 146, 360,  450, 292,  98,  16,  1;\n  9: 2, 32, 198, 608, 1002, 912, 462, 128, 18, 1;\n","recurrence_sentence":"Row n lists the coefficients of P(n,x) in descending powers of x, where P(n,x) = (1+x)*P(n-1,x) + x*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 + x)*P[n - 1, x] + x*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 + x)*v[i - 1] + x*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":1,"offset":0,"reversed":true,"trim_zeros":"none","regular_triangle":true}}],"derivation":{"root":"n-e073e5e0b021f36a","steps":[],"edges":[],"text":"A035607"},"headline":"A035607: Table a(d,m) of number of points of L1 norm m in cubic lattice Z^d, read by antidiagonals (d >= 1, m >= 0).","chain":null}