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A341196
a(n) = Sum_{k=0..n} k^4 * (n-k)! * binomial(n,k)^2.
1
0, 1, 20, 243, 2800, 33425, 424116, 5762155, 83891648, 1306561185, 21709011700, 383654149571, 7189789929840, 142465285362673, 2976697773182420, 65417312929686075, 1508567496105346816, 36425941390897897025, 919100609186531702868
OFFSET
0,3
COMMENTS
In general, for m >= 0, Sum_{k=0..n} k^m * (n-k)! * binomial(n,k)^2 ~ exp(2*sqrt(n) - n - 1/2) * n^(n + (2*m+1)/4) / sqrt(2). - Vaclav Kotesovec, Feb 20 2021
FORMULA
E.g.f.: x*(1+4*x-5*x^2+x^3)*exp(x/(1-x))/(1-x)^5.
a(n) = n*n!*F([2, 2, 1 - n], [1, 1, 1], -1), where F is the generalized hypergeometric function. - Stefano Spezia, Feb 06 2021
a(n) ~ exp(2*sqrt(n) - n - 1/2) * n^(n + 9/4) / sqrt(2). - Vaclav Kotesovec, Feb 20 2021
MATHEMATICA
a[n_] := Sum[k^4 * (n-k)! * Binomial[n, k]^2, {k, 0, n}]; Array[a, 19, 0] (* Amiram Eldar, Feb 06 2021 *)
PROG
(PARI) a(n) = sum(k=0, n, k^4*(n-k)!*binomial(n, k)^2);
(PARI) my(N=20, x='x+O('x^N)); concat(0, Vec(serlaplace(x*(1+4*x-5*x^2+x^3)*exp(x/(1-x))/(1-x)^5)))
CROSSREFS
Column 4 of A341200.
Sequence in context: A073398 A040075 A138442 * A140124 A123954 A125432
KEYWORD
nonn
AUTHOR
Seiichi Manyama, Feb 06 2021
STATUS
approved