|
%I #17 Aug 05 2019 05:33:12
%S 1,37,1225,43397,1563401,56309885,2020496185,72190600165,
%T 2569004841385,91095128385485,3220006254279233,113505318773615741,
%U 3991330807880182105,140050346341652428141,4904787249549605102233,171480516047539645266725
%N a(n) = Sum_{k=0..n} ((2k+1)*C(n,k)*C(n+k,k))^2, where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).
%C Zhi-Wei Sun noted that for any positive integer n we have Sum_{k=0..n-1} (2k+1)*a(k) = n^4*A246460(n).
%H Zhi-Wei Sun, <a href="/A246461/b246461.txt">Table of n, a(n) for n = 0..100</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1408.5381">Two new kinds of numbers and their arithmetic properties</a>, arXiv:1408.5381 [math.NT], 2014-2018.
%F Recurrence: n^3*(2*n-3)*(4*n^4 - 24*n^3 + 50*n^2 - 42*n + 11)*(6*n^4 - 36*n^3 + 67*n^2 - 39*n - 4)*a(n) = (2*n-1)*(840*n^11 - 10464*n^10 + 53192*n^9 - 137864*n^8 + 172296*n^7 - 19912*n^6 - 226019*n^5 + 271559*n^4 - 92324*n^3 - 42188*n^2 + 39128*n - 8466)*a(n-1) - (2*n-3)*(840*n^11 - 8016*n^10 + 28712*n^9 - 44872*n^8 + 15880*n^7 + 43992*n^6 - 64675*n^5 + 32567*n^4 + 1692*n^3 - 9364*n^2 + 4072*n - 606)*a(n-2) + (n-2)^3*(2*n-1)*(4*n^4 - 8*n^3 + 2*n^2 + 2*n - 1)*(6*n^4 - 12*n^3 - 5*n^2 + 11*n - 6)*a(n-3). - _Vaclav Kotesovec_, Aug 27 2014
%F a(n) ~ sqrt(24+17*sqrt(2)) * (17+12*sqrt(2))^n * sqrt(n) / (2*sqrt(2)*Pi^(3/2)). - _Vaclav Kotesovec_, Aug 27 2014
%e a(1) = 37 since Sum_{k=0..1} ((2k+1)*C(1,k)*C(1+k,k))^2 = 1^2 + (3*2)^2 = 37.
%p A246461:=n->add(((2*k+1)*binomial(n,k)*binomial(n+k,k))^2, k=0..n): seq(A246461(n), n=0..20); # _Wesley Ivan Hurt_, Aug 26 2014
%t a[n_]:=Sum[((2k+1)*Binomial[n,k]*Binomial[n+k,k])^2,{k,0,n}]
%t Table[a[n],{n,0,15}]
%Y Cf. A245769, A246065, A246138, A246459, A246460.
%K nonn
%O 0,2
%A _Zhi-Wei Sun_, Aug 26 2014
|
