A329521 The sixth moments of the alternated squared binomial coefficients; a(n) = Sum_{m=0..n} (-1)^m*m^6*binomial(n, m)^2.

0, -1, 60, -162, -5280, 20250, 128520, -569380, -1854720, 9338490, 20097000, -113704668, -181621440, 1142905764, 1447926480, -10042461000, -10529925120, 79859881530, 71384175720, -587933314540, -457825368000, 4070529226764

Offset: 0

Author

Nikita D. Gogin, Nov 15 2019

Keywords

• sign

References

• H. W. Gould, Combinatorial Identities, 1972.

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A074334, A294486, A329444.

Programs

Magma

[&+[(-1)^m*m^6*(Binomial(n,m))^2:m in [0..n]]:n in [0..21]]; // Marius A. Burtea, Nov 15 2019

Mathematica

Table[Sum[(-1)^m*m^6*(Binomial[n, m])^2, {m, 0, n}], {n, 21}]

PARI

a(n) = sum(m=0, n, (-1)^m*m^6*binomial(n , m)^2); \\ Michel Marcus, Nov 15 2019

Formula

a(n) = (-1)^((n+2)/2)*binomial(n, n/2)*(n^3*(n+1)*(3n-1)/4), if n is even,

a(n) = (-1)^((n-1)/2)*binomial(n,((n+1)/2))*(n^2*( n+1)*(n^3+n^2-9n+3)/8), if n is odd.

G.f.: x*(-1 + 60*x - 188*x^2 - 3720*x^3 + 15752*x^4 + 8400*x^5 - 90928*x^6 + 79680*x^7 + 42112*x^8 - 69120*x^9 + 17408*x^10)/(1+4*x^2)^(13/2). - Stefano Spezia, Nov 15 2019