A156890
Triangle formed by coefficients of the expansion of p(x, n), where p(x,n) = (1+x-x^2)^(n+1)*Sum_{j >= 0} (j+1)^n*(-x + x^2)^j.
Original entry on oeis.org
1, 1, 1, -1, 1, 1, -4, 5, -2, 1, 1, -11, 22, -23, 14, -3, 1, 1, -26, 92, -158, 145, -82, 32, -4, 1, 1, -57, 359, -906, 1265, -1135, 649, -238, 67, -5, 1, 1, -120, 1311, -4798, 9630, -12132, 10163, -5970, 2406, -620, 135, -6, 1, 1, -247, 4540, -24205, 66769, -113626, 131045, -106889, 62261, -26426, 8033, -1517, 268, -7, 1
Offset: 0
Irregular triangle begins as:
1;
1;
1, -1, 1;
1, -4, 5, -2, 1;
1, -11, 22, -23, 14, -3, 1;
1, -26, 92, -158, 145, -82, 32, -4, 1;
1, -57, 359, -906, 1265, -1135, 649, -238, 67, -5, 1;
1, -120, 1311, -4798, 9630, -12132, 10163, -5970, 2406, -620, 135, -6, 1;
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p[x_, n_]:= ((1+x-x^2)^(n+1))*Sum[(j+1)^n*(-x+x^2)^j, {j,0,Infinity}];
Table[CoefficientList[p[x, n], x], {n, 0, 10}]//Flatten
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def T(n,k): return ( (1+x-x^2)^(n+1)*sum((j+1)^n*(x^2-x)^j for j in (0..2*n+1)) ).series(x, 2*n+3).list()[k]
[1]+flatten([[T(n,k) for k in (0..2*n-2)] for n in (0..12)]) # G. C. Greubel, Jan 06 2022
A156901
Triangle formed by coefficients of the expansion of p(x, n), where p(x,n) = (1 + 2*x - x^2)^(n + 1)*Sum_{j >= 0} (j+1)^n*(-2*x + x^2)^j.
Original entry on oeis.org
1, 1, 1, -2, 1, 1, -8, 8, -4, 1, 1, -22, 55, -52, 23, -6, 1, 1, -52, 290, -472, 394, -188, 50, -8, 1, 1, -114, 1265, -3624, 4838, -3668, 1750, -536, 97, -10, 1, 1, -240, 4884, -24092, 49239, -56448, 40664, -19320, 6231, -1360, 180, -12, 1, 1, -494, 17419, -142124, 441625, -730898, 749723, -515944, 247067, -83122, 19673, -3244, 331, -14, 1
Offset: 0
Irregular triangle begins as:
1;
1;
1, -2, 1;
1, -8, 8, -4, 1;
1, -22, 55, -52, 23, -6, 1;
1, -52, 290, -472, 394, -188, 50, -8, 1;
1, -114, 1265, -3624, 4838, -3668, 1750, -536, 97, -10, 1;
1, -240, 4884, -24092, 49239, -56448, 40664, -19320, 6231, -1360, 180, -12, 1;
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p[x_, n_]= (1+2*x-x^2)^(n+1)*Sum[(k+1)^n*(-2*x+x^2)^k, {k,0,Infinity}];
Table[CoefficientList[p[x, n], x], {n,0,10}]//Flatten
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def T(n, k): return ( (1+2*x-x^2)^(n+1)*sum((j+1)^n*(x^2-2*x)^j for j in (0..2*n+1)) ).series(x, 2*n+2).list()[k]
flatten([1]+[[T(n, k) for k in (0..2*n-2)] for n in (1..12)]) # G. C. Greubel, Jan 07 2022
A156918
Triangle formed by coefficients of the expansion of p(x,n) = (1+x-x^2)^(n+1)*Sum_{j >= 0} (2*j+1)^n*(-x + x^2)^j.
Original entry on oeis.org
1, 1, -1, 1, 1, -6, 7, -2, 1, 1, -23, 46, -47, 26, -3, 1, 1, -76, 306, -536, 459, -232, 82, -4, 1, 1, -237, 1919, -5046, 6965, -5995, 3109, -958, 247, -5, 1, 1, -722, 11265, -44634, 91730, -113538, 90417, -49398, 17778, -3630, 737, -6, 1, 1, -2179, 62836, -381037, 1099549, -1878718, 2123525, -1658537, 898985, -346886, 93377, -13109, 2200, -7, 1
Offset: 0
Irregular triangle begins as:
1;
1, -1, 1;
1, -6, 7, -2, 1;
1, -23, 46, -47, 26, -3, 1;
1, -76, 306, -536, 459, -232, 82, -4, 1;
1, -237, 1919, -5046, 6965, -5995, 3109, -958, 247, -5, 1;
1, -722, 11265, -44634, 91730, -113538, 90417, -49398, 17778, -3630, 737, -6, 1;
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p[x_, n_] = (1+x-x^2)^(n+1)*Sum[(2*k+1)^n*(-x+x^2)^k, {k, 0, Infinity}];
Table[CoefficientList[p[x, n], x], {n,0,10}]//Flatten
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def T(n, k): return ( (1+x-x^2)^(n+1)*sum((2*j+1)^n*(x^2-x)^j for j in (0..2*n+1)) ).series(x, 2*n+2).list()[k]
flatten([1]+[[T(n, k) for k in (0..2*n)] for n in (1..12)]) # G. C. Greubel, Jan 07 2022
A156985
Triangle formed by coefficients of the expansion of p(x,n) = (1-x)^(2*n + 1)*Sum_{j >= 0} (1 +j +j^2)^n * x^j.
Original entry on oeis.org
1, 1, 0, 1, 1, 4, 14, 4, 1, 1, 20, 175, 328, 175, 20, 1, 1, 72, 1708, 9784, 17190, 9784, 1708, 72, 1, 1, 232, 14189, 199616, 884498, 1431728, 884498, 199616, 14189, 232, 1, 1, 716, 108250, 3353948, 31986447, 115907544, 176287788, 115907544, 31986447, 3353948, 108250, 716, 1
Offset: 0
Irregular triangle begins as:
1;
1, 0, 1;
1, 4, 14, 4, 1;
1, 20, 175, 328, 175, 20, 1;
1, 72, 1708, 9784, 17190, 9784, 1708, 72, 1;
1, 232, 14189, 199616, 884498, 1431728, 884498, 199616, 14189, 232, 1;
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p[x_, n_] = (1-x)^(2*n+1)*Sum[(1+k+k^2)^n*x^k, {k, 0, Infinity}];
Table[CoefficientList[p[x, n], x], {n, 0, 10}]//Flatten
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def T(n, k): return ( (1-x)^(2*n+1)*sum((j^2+j+1)^n*x^j for j in (0..2*n+1)) ).series(x, 2*n+2).list()[k]
flatten([1]+[[T(n, k) for k in (0..2*n)] for n in (1..12)]) # G. C. Greubel, Jan 07 2022
Showing 1-4 of 4 results.
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