%I #16 Jan 09 2023 05:35:22
%S 1,2,9,22,114,308,1717,4902,28526,84284,504410,1525500,9311140,
%T 28638504,177367949,552532742,3460680278,10888711788,68810154286,
%U 218247846932,1389230689596,4435391806552,28403131135554,91183193910620,586877613154252,1892886293567192
%N a(n) = Jacobi P-Polynomial P_n(alpha=1, beta=1, x=sqrt(2)) multiplied by 2^(n/2+floor(n/2)) and divided by n+1.
%H Alois P. Heinz, <a href="/A025176/b025176.txt">Table of n, a(n) for n = 0..1000</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Jacobi_polynomials">Jacobi polynomials</a>
%F (n+2) * a(n) = 2^(2+floor(n/2)-floor((n+1)/2)) * (2n+1) * a(n-1) - 4 * (n-1) * a(n-2), with a(0)=1 and a(1)=2. - _Sean A. Irvine_, Aug 14 2019
%F From _Vaclav Kotesovec_, Jan 09 2023: (Start)
%F Recurrence: (n+1)*(n+2)*(2*n-3)*a(n) = 12*(2*n-1)*(2*n^2 - 2*n - 1)*a(n-2) - 16*(n-3)*(n-2)*(2*n+1)*a(n-4).
%F a(n) ~ 2^(n-1) * (1 + sqrt(2))^(n + 1/2) * (3 + 2*sqrt(2) + (-1)^n) / (sqrt(Pi) * n^(3/2)). (End)
%p A025176 := proc(n)
%p JacobiP(n,1,1,sqrt(2)) ;
%p %*2^(n/2+floor(n/2))/(n+1) ;
%p simplify(%) ;
%p end proc: # _R. J. Mathar_, Feb 05 2013
%t Table[ 2^(n/2+Floor[ n/2 ])/(n+1)JacobiP[ n, 1, 1, Sqrt[ 2 ] ], {n, 0, 24} ]
%K nonn
%O 0,2
%A _Wouter Meeussen_