%I #22 Nov 12 2023 12:07:52
%S 1,7,55,465,4047,35673,316521,2819295,25173855,225157881,2016242265,
%T 18070920255,162071863425,1454320387575,13055422263255,
%U 117237213829953,1053070838993151,9461217421304505,85019389336077225,764113545253570191,6868417199986308129
%N a(n) = Sum_{k=0..n} C(n,k)^2*C(2k,k)*(2k+1), where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).
%C Zhi-Wei Sun proved that for any n > 0 we have Sum_{k=0..n-1} a(k) = n^2*A086618(n-1), and (Sum_{k=0..n-1}a(k,x))/n is a polynomial with integer coefficients, where a(k,x) = sum_{j=0..k}C(k,j)^2*C(2j,j)*(2j+1)*x^j.
%H Zhi-Wei Sun, <a href="/A246459/b246459.txt">Table of n, a(n) for n = 0..160</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 (obtained via the Zeilberger algorithm): 9*(n+1)^2*a(n) - (19*n^2+74*n+87)*a(n+1) + (n+3)*(11*n+29)*a(n+2) - (n+3)^2*a(n+3) = 0.
%F a(n) ~ 3^(2*n+1/2) / Pi. - _Vaclav Kotesovec_, Aug 27 2014
%F a(n) = (4*n+3)*A002893(n)/3. - _Mark van Hoeij_, Nov 12 2023
%e a(2) = 55 since Sum_{k=0,1,2} C(2,k)^2*C(2k,k)(2k+1) = 1 + 8*3 + 6*5 = 55.
%p A246459:=n->add(binomial(n,k)^2*binomial(2*k,k)*(2*k+1), k=0..n): seq(A246459(n), n=0..20); # _Wesley Ivan Hurt_, Aug 26 2014
%t a[n_]:=Sum[Binomial[n,k]^2*Binomial[2k,k](2k+1),{k,0,n}]
%t Table[a[n],{n,0,20}]
%Y Cf. A086618, A245769, A246065, A246138.
%K nonn
%O 0,2
%A _Zhi-Wei Sun_, Aug 26 2014