A054651
Triangle T(n,k) read by rows giving coefficients in expansion of n! * Sum_{i=0..n} C(x,i) in descending powers of x.
Original entry on oeis.org
1, 1, 1, 1, 1, 2, 1, 0, 5, 6, 1, -2, 11, 14, 24, 1, -5, 25, 5, 94, 120, 1, -9, 55, -75, 304, 444, 720, 1, -14, 112, -350, 1099, 364, 3828, 5040, 1, -20, 210, -1064, 3969, -4340, 15980, 25584, 40320, 1, -27, 366, -2646, 12873, -31563, 79064, 34236, 270576, 362880
Offset: 0
The first few polynomials are:
1, 1+x, 2+x+x^2, 6+5*x+x^3, 24+14*x+11*x^2-2*x^3+x^4, ...
So the triangle begins:
1;
1, 1;
1, 1, 2;
1, 0, 5, 6;
1, -2, 11, 14, 24;
1, -5, 25, 5, 94, 120;
1, -9, 55, -75, 304, 444, 720;
1, -14, 112, -350, 1099, 364, 3828, 5040;
1, -20, 210, -1064, 3969, -4340, 15980, 25584, 40320;
...
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c[n_, k_] := Product[n-i, {i, 0, k-1}]/k!; row[n_] := CoefficientList[ n!*Sum[c[x, k], {k, 0, n}], x] // Reverse; Table[ row[n], {n, 0, 9}] // Flatten (* Jean-François Alcover, Oct 04 2012 *)
A347987
a(n) = [x^n] (2*n)! * Sum_{k=0..2*n} binomial(x,k).
Original entry on oeis.org
1, 1, 11, -75, 3969, -140595, 7374191, -435638203, 30421321073, -2409092861175, 214562251828275, -21195275581114635, 2301157855016159905, -272330254968023391035, 34894294917147760652775, -4812715265513253499593675, 710922905477027337578759265, -111981455662673544130741177455
Offset: 0
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Table[(2*n)!/n! * SeriesCoefficient[Log[1+x]^n/(1-x), {x, 0, 2*n}], {n, 0, 20}] (* Vaclav Kotesovec, May 25 2025 *)
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a(n) = (2*n)!*polcoef(sum(k=n, 2*n, binomial(x, k)), n);
A348063
Coefficient of x^2 in expansion of n!* Sum_{k=0..n} binomial(x,k).
Original entry on oeis.org
1, 0, 11, 5, 304, 364, 15980, 34236, 1368936, 4429656, 173699712, 771653376, 30605906304, 175622947200, 7149130156800, 50800930272000, 2137822335475200, 18241636315507200, 796397873127782400, 7971407298921830400, 361615771356450508800, 4168685961862906982400, 196587429737202833817600
Offset: 2
-
a(n) = n!*polcoef(sum(k=2, n, binomial(x, k)), 2);
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a(n) = if(n<2, 0, a(n-1)+(n-1)^2*a(n-2)+(-1)^n*(n-2)!);
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N=40; x='x+O('x^N); Vec(serlaplace(log(1+x)^2/(2*(1-x))))
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from sympy.abc import x
from sympy import ff, expand
def A348063(n): return sum(ff(n,n-k)*expand(ff(x,k)).coeff(x**2) for k in range(2,n+1)) # Chai Wah Wu, Sep 27 2021
A348064
Coefficient of x^3 in expansion of n!* Sum_{k=0..n} binomial(x,k).
Original entry on oeis.org
1, -2, 25, -75, 1099, -4340, 79064, -382060, 8550916, -48306984, 1303568760, -8346754416, 266955481584, -1894529909376, 70785236377728, -547468189825536, 23610353987137536, -196402650598402560, 9679304091074250240, -85687212859582878720, 4785340778000524477440
Offset: 3
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a(n) = n!*polcoef(sum(k=3, n, binomial(x, k)), 3);
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N=40; x='x+O('x^N); Vec(serlaplace(log(1+x)^3/(6*(1-x))))
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from sympy.abc import x
from sympy import ff, expand
def A348064(n): return sum(ff(n,n-k)*expand(ff(x,k)).coeff(x**3) for k in range(3,n+1)) # Chai Wah Wu, Sep 27 2021
A348065
Coefficient of x^4 in expansion of n!* Sum_{k=0..n} binomial(x,k).
Original entry on oeis.org
1, -5, 55, -350, 3969, -31563, 408050, -3920950, 58206676, -657328100, 11111159696, -144321864960, 2747845864464, -40364369180016, 856755330487200, -14042902728462624, 329258021171239296, -5956512800554963584, 153050034289602269952, -3028534064042216488704, 84691080748928315003904
Offset: 4
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a(n) = n!*polcoef(sum(k=4, n, binomial(x, k)), 4);
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N=40; x='x+O('x^N); Vec(serlaplace(log(1+x)^4/(24*(1-x))))
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from sympy.abc import x
from sympy import ff, expand
def A348065(n): return sum(ff(n,n-k)*expand(ff(x,k)).coeff(x**4) for k in range(4,n+1)) # Chai Wah Wu, Sep 27 2021
A348068
Coefficient of x^5 in expansion of n!* Sum_{k=0..n} binomial(x,k).
Original entry on oeis.org
1, -9, 112, -1064, 12873, -140595, 1870385, -23551110, 351042406, -5043110072, 84074954600, -1361614072000, 25218570009424, -455365645674480, 9298765013106384, -185409487083100320, 4144212593899945056, -90492302454898284864, 2199399908894486591040
Offset: 5
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a(n) = n!*polcoef(sum(k=5, n, binomial(x, k)), 5);
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N=40; x='x+O('x^N); Vec(serlaplace(log(1+x)^5/(120*(1-x))))
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from sympy.abc import x
from sympy import ff, expand
def A348068(n): return sum(ff(n,n-k)*expand(ff(x,k)).coeff(x**5) for k in range(5,n+1)) # Chai Wah Wu, Sep 27 2021
Showing 1-6 of 6 results.
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