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begins:\n   1:  1;\n   2:  3,   1;\n   3:  5,   5,    1;\n   4:  7,  14,    7,    1;\n   5:  9,  30,   27,    9,    1;\n   6: 11,  55,   77,   44,   11,    1;\n   7: 13,  91,  182,  156,   65,   13,   1;\n   8: 15, 140,  378,  450,  275,   90,  15,   1;\n   9: 17, 204,  714, 1122,  935,  442, 119,  17,  1;\n  10: 19, 285, 1254, 2508, 2717, 1729, 665, 152, 19, 1;\n","recurrence_sentence":"The row polynomials satisfy P(n,x) = (2+x)*P(n-1,x) - P(n-2,x) for n >= 3, with P(1,x) = 1, P(2,x) = 3 + x.","mathematica":"P[1, x_] := 1;\nP[2, x_] := 3 + x;\nP[n_, x_] := P[n, x] = Expand[(2 + x)*P[n - 1, 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] = 3 + x; for(n = 3, N, my(i = n); v[i] = (2 + x)*v[i - 1] - 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":"A082985","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: 1,  3;\n  2: 1,  5,   5;\n  3: 1,  7,  14,   7;\n  4: 1,  9,  27,  30,    9;\n  5: 1, 11,  44,  77,   55,   11;\n  6: 1, 13,  65, 156,  182,   91,   13;\n  7: 1, 15,  90, 275,  450,  378,  140,   15;\n  8: 1, 17, 119, 442,  935, 1122,  714,  204,  17;\n  9: 1, 19, 152, 665, 1729, 2717, 2508, 1254, 285, 19;\n","recurrence_sentence":"Row n lists the coefficients of P(n,x) in descending powers of x, where P(n,x) = (2+x)*P(n-1,x) - P(n-2,x) for n >= 2, with P(0,x) = 1, P(1,x) = 3 + x.","mathematica":"P[0, x_] := 1;\nP[1, x_] := 3 + x;\nP[n_, x_] := P[n, x] = Expand[(2 + x)*P[n - 1, 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] = 3 + x; for(n = 2, N, my(i = n + 1); v[i] = (2 + x)*v[i - 1] - 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}},{"oeis_id":"A111125","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:  3,   1;\n  2:  5,   5,    1;\n  3:  7,  14,    7,    1;\n  4:  9,  30,   27,    9,    1;\n  5: 11,  55,   77,   44,   11,    1;\n  6: 13,  91,  182,  156,   65,   13,   1;\n  7: 15, 140,  378,  450,  275,   90,  15,   1;\n  8: 17, 204,  714, 1122,  935,  442, 119,  17,  1;\n  9: 19, 285, 1254, 2508, 2717, 1729, 665, 152, 19, 1;\n","recurrence_sentence":"The row polynomials satisfy P(n,x) = (2+x)*P(n-1,x) - P(n-2,x) for n >= 2, with P(0,x) = 1, P(1,x) = 3 + x.","mathematica":"P[0, x_] := 1;\nP[1, x_] := 3 + x;\nP[n_, x_] := P[n, x] = Expand[(2 + x)*P[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] = 3 + x; for(n = 2, N, my(i = n + 1); v[i] = (2 + x)*v[i - 1] - 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-318700a428d7b97b","steps":[],"edges":[],"text":"A111125"},"headline":"A111125: Triangle read by rows: T(k,s) = ((2*k+1)/(2*s+1))*binomial(k+s,2*s), 0 <= s <= k.","chain":null}