%I #9 Jul 05 2015 14:21:51
%S 1,1,2,1,8,3,1,18,25,4,1,32,98,56,5,1,50,270,336,105,6,1,72,605,1320,
%T 891,176,7,1,98,1183,4004,4719,2002,273,8,1,128,2100,10192,18590,
%U 13728,4004,400,9,1,162,3468,22848,59670,68068,34476,7344,561,10,1,200,5415
%N Triangle T(n,k), 0 <= k <= n, read by rows defined by: T(n,k) = (binomial(2*n,2*k) + binomial(2*n+1,2*k))/2.
%C Sum_{k=0..n}T(n,k)*x^k = A164111(n), A000012(n), A002001(n), A001653(n+1), A001019(n), A166965(n) for x =-1, 0, 1, 2, 4, 9 respectively.
%F G.f.: (1-x)/(1-2*x*(1+y)+x^2*(1-y)^2).
%F T(n,k) = 2*T(n-1,k)+2*T(n-1,k-1)+2*T(n-2,k-1)-T(n-2,k)-T(n-2,k-2), T(0,0)=T(1,0)=1, T(1,1)=2, T(n,k)=0 if k<0 or if k>n.
%F T(n,k) = (A086645(n,k) + A091042(n,k))/2.
%F T(n,k) = binomial(2*n,2*k)*(2*n+1-k)/(2*n+1-2*k). - _Peter Luschny_, Nov 26 2013
%e Triangle begins:
%e 1
%e 1, 2
%e 1, 8, 3
%e 1, 18, 25, 4
%e 1, 32, 98, 56, 5
%e 1, 50, 270, 336, 105, 6
%e 1, 72, 605, 1320, 891, 176, 7
%e 1, 98, 1183, 4004, 4719, 2002, 273, 8
%e 1, 128, 2100, 10192, 18590, 13728, 4004, 400, 9
%p T := (n,k) -> binomial(2*n, 2*k)*(2*n+1-k)/(2*n+1-2*k);
%p seq(seq(T(n,k), k=0..n), n=0..9); # _Peter Luschny_, Nov 26 2013
%t Flatten[Table[(Binomial[2n,2k]+Binomial[2n+1,2k])/2,{n,0,10},{k,0,n}]] (* _Harvey P. Dale_, Jul 05 2015 *)
%Y Cf. A086645, A091042
%Y Cf. Columns : A000012, A001105, A180324 ; Diagonals: A000027, A131423
%Y Cf. T(2*n,n): A228329, Row sums : A002001
%K nonn,tabl
%O 0,3
%A _Philippe Deléham_, Nov 25 2013