|
%I #26 Jun 14 2022 03:12:16
%S 0,0,0,0,0,288,3504,26936,168000,930240,4775232,23279256,109368864,
%T 499928000,2237835600,9854764920,42836127360,184246957440,
%U 785668464000,3326326610400,13998420079488,58611422003904,244341952079456,1014823115578800,4201232634318720,17343550105777280,71420954783418240,293472577948946760,1203572398002156000
%N a(n) = 288*binomial(2*n,n-5) + 8*Sum_{i=1..n-5} binomial(2*n,n-5-i)*(5+i).
%H Olivia Beckwith, Steven J. Miller and Karen Shen, <a href="http://arxiv.org/abs/1112.3719">Distribution of Eigenvalues of Weighted, Structured Matrix Ensembles</a>, arXiv preprint arXiv:1112.3719 [math.PR], 2011-2012.
%H Olivia Beckwith, Victor Luo, Stephen J. Miller, Karen Shen and Nicholas Triantafillou, <a href="https://www.emis.de/journals/INTEGERS/papers/p21/p21.Abstract.html">Distribution of Eigenvalues of Weighted, Structured Matrix Ensembles</a>, Electronic Journal of Combinatorial Number Theory, Volume 15 (2015) #A21.
%p f:=n->288*binomial(2*n,n-5)+8*add(binomial(2*n,n-5-i)*(5+i),i=1..n-5);
%p [seq(f(n),n=0..40)];
%t Table[288*Binomial[2n,n-5]+8*Sum[Binomial[2n,n-5-i](5+i),{i,n-5}],{n,0,30}] (* _Harvey P. Dale_, Apr 10 2019 *)
%o (PARI) a(n) = 288*binomial(2*n,n-5) + 8*sum(i=1, n-5, binomial(2*n,n-5-i)*(5+i)); \\ _Michel Marcus_, Apr 05 2019; corrected Jun 14 2022
%Y Cf. A182025.
%K nonn
%O 0,6
%A _N. J. A. Sloane_, Apr 07 2012
|
