A329913 The fifth moments of the squared binomial coefficients; a(n) = Sum_{m=0..n} m^5*binomial(n, m)^2.

0, 1, 36, 540, 6080, 56250, 455112, 3342192, 22809600, 146988270, 904475000, 5358254616, 30750385536, 171773279860, 937514244240, 5014575000000, 26351064760320, 136319273714070, 695429503781400, 3503580441563400, 17452918098000000, 86055711108818220

Offset: 0

Author

Nikita D. Gogin, Nov 24 2019

Keywords

• nonn

References

• H. W. Gould, Combinatorial Identities, 1972. (See formulas 3.77, 3.78, and 3.79 on page 31.)

Links

• Robert Israel, Table of n, a(n) for n = 0..1638

Crossrefs

Cf. A074334, A294486.

Cf. A000984, A002457, A037966, A037972, A074334, A329444.

Programs

Magma

[(&+[Binomial(n,k)^2*k^5: k in [0..n]]): n in [0..30]]; // G. C. Greubel, Jun 23 2022

Maple

seq( binomial(2*n,n)*n^4*(n^3 + 3*n^2 - 3*n - 5)/((16*n-8)*(2*n-3)),n=0..30); # Robert Israel, Jan 26 2020

Mathematica

Table[Sum[m^5*(Binomial[n, m])^2, {m, 0, n}], {n, 21}]

PARI

a(n) = sum(k=0, n, k^5*binomial(n, k)^2); \\ Michel Marcus, Nov 24 2019

SageMath

[n^4*(n+1)*(n^3+3*n^2-3*n-5)/(8*(2*n-1)*(2*n-3))*catalan_number(n) for n in (0..30)] # G. C. Greubel, Jun 23 2022

Formula

a(n) = binomial(2*n,n)*n^4*(n^3 + 3*n^2 - 3*n - 5)/(8*(2*n-1)*(2*n-3)).

G.f.: x*(1 + 14*x - 54*x^2 + 404*x^3 - 1544*x^4 + 2880*x^5 - 2160*x^6)/(1-4*x)^(11/2). - Stefano Spezia, Jan 03 2020

(-12960 + 8640*n)*a(n) + (7200 - 13680*n)*a(n + 1) + (3920 + 9056*n)*a(n + 2) + (-4184 - 3160*n)*a(n + 3) + (1404 + 620*n)*a(n + 4) + (-584 - 110*n)*a(n + 5) + (14 + 10*n)*a(n + 6) + (n + 6)*a(n + 7) = 0. - Robert Israel, Jan 26 2020