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v[1] = 1; for(n = 2, N, my(i = n); v[i] = (((15/2)*n^2 - (11/2)*n + 1 + (6*n^2 - 8*n + 2)*x)*v[i - 1] + ((1/2)*n - 1/2 + ((7/2)*n - 1/2)*x - (4*n - 1)*x^2)*deriv(v[i - 1], x))/(2*n^2 + n)); 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":"A108767","target":"reversed_triangle","first_row":1,"oeis_offset":1,"reversed":true,"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,  6,    5;\n   4: 1, 12,   28,    14;\n   5: 1, 20,   90,   120,     42;\n   6: 1, 30,  220,   550,    495,    132;\n   7: 1, 42,  455,  1820,   3003,   2002,    429;\n   8: 1, 56,  840,  4900,  12740,  15288,   8008,   1430;\n   9: 1, 72, 1428, 11424,  42840,  79968,  74256,  31824,   4862;\n  10: 1, 90, 2280, 23940, 122094, 325584, 465120, 348840, 125970, 16796;\n","recurrence_sentence":"Row n lists the coefficients of P(n,x) in descending powers of x, where (2*n^2+n)*P(n,x) = ((15/2)*n^2-(11/2)*n+1+(6*n^2-8*n+2)*x)*P(n-1,x) + ((1/2)*n-1/2+((7/2)*n-1/2)*x-(4*n-1)*x^2)*d/dx P(n-1,x) for n >= 2, with P(1,x) = 1.","mathematica":"P[1, x_] := 1;\nP[n_, x_] := P[n, x] = Expand[Together[(((15/2)*n^2 - (11/2)*n + 1 + (6*n^2 - 8*n + 2)*x)*P[n - 1, x] + ((1/2)*n - 1/2 + ((7/2)*n - 1/2)*x - (4*n - 1)*x^2)*D[P[n - 1, x], x])/(2*n^2 + n)]];\nTable[Reverse[PadRight[CoefficientList[P[n, x], x], n]], {n, 1, 10}] // Flatten","pari":"P(N) = {my(v = vector(N)); v[1] = 1; for(n = 2, N, my(i = n); v[i] = (((15/2)*n^2 - (11/2)*n + 1 + (6*n^2 - 8*n + 2)*x)*v[i - 1] + ((1/2)*n - 1/2 + ((7/2)*n - 1/2)*x - (4*n - 1)*x^2)*deriv(v[i - 1], x))/(2*n^2 + n)); v};\nrow(n, p) = {my(c = Vecrev(p)); c = concat(c, vector(n - #c)); c = Vecrev(c); c};\nT(N) = {my(v = P(N)); concat(vector(#v, i, row(i, v[i])))};\nT(10)","reading":{"first_row":1,"offset":1,"reversed":true,"trim_zeros":"none","regular_triangle":true}},{"oeis_id":"A120986","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:     5,      6,      1;\n  3:    14,     28,     12,      1;\n  4:    42,    120,     90,     20,      1;\n  5:   132,    495,    550,    220,     30,      1;\n  6:   429,   2002,   3003,   1820,    455,     42,     1;\n  7:  1430,   8008,  15288,  12740,   4900,    840,    56,    1;\n  8:  4862,  31824,  74256,  79968,  42840,  11424,  1428,   72,  1;\n  9: 16796, 125970, 348840, 465120, 325584, 122094, 23940, 2280, 90, 1;\n","recurrence_sentence":"The row polynomials satisfy (2*n^2+5*n+3)*P(n,x) = ((15/2)*n^2+(19/2)*n+3+(6*n^2+4*n)*x)*P(n-1,x) + ((1/2)*n+((7/2)*n+3)*x-(4*n+3)*x^2)*d/dx P(n-1,x) for n >= 1, with P(0,x) = 1.","mathematica":"P[0, x_] := 1;\nP[n_, x_] := P[n, x] = Expand[Together[(((15/2)*n^2 + (19/2)*n + 3 + (6*n^2 + 4*n)*x)*P[n - 1, x] + ((1/2)*n + ((7/2)*n + 3)*x - (4*n + 3)*x^2)*D[P[n - 1, x], x])/(2*n^2 + 5*n + 3)]];\nTable[PadRight[CoefficientList[P[n, x], x], n + 1], {n, 0, 9}] // Flatten","pari":"P(N) = {my(v = vector(N + 1)); v[1] = 1; for(n = 1, N, my(i = n + 1); v[i] = (((15/2)*n^2 + (19/2)*n + 3 + (6*n^2 + 4*n)*x)*v[i - 1] + ((1/2)*n + ((7/2)*n + 3)*x - (4*n + 3)*x^2)*deriv(v[i - 1], x))/(2*n^2 + 5*n + 3)); 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-4479b0484a11b442","steps":[],"edges":[],"text":"A120986"},"headline":"A120986: Triangle read by rows: T(n,k) is the number of ternary trees with n edges and having k middle edges (n >= 0, k >= 0).","chain":null}
