A329913 - OEIS

%I #20 Jun 24 2022 09:30:09

%S 0,1,36,540,6080,56250,455112,3342192,22809600,146988270,904475000,

%T 5358254616,30750385536,171773279860,937514244240,5014575000000,

%U 26351064760320,136319273714070,695429503781400,3503580441563400,17452918098000000,86055711108818220

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

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

%H Robert Israel, <a href="/A329913/b329913.txt">Table of n, a(n) for n = 0..1638</a>

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

%F 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

%F (-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

%p 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

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

%o (PARI) a(n) = sum(k=0, n, k^5*binomial(n, k)^2); \\ _Michel Marcus_, Nov 24 2019

%o (Magma) [(&+[Binomial(n,k)^2*k^5: k in [0..n]]): n in [0..30]]; // _G. C. Greubel_, Jun 23 2022

%o (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

%Y Cf. A074334, A294486.

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

%K nonn

%O 0,3

%A _Nikita D. Gogin_, Nov 24 2019