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%I #16 Dec 11 2019 23:41:27
%S 1,1,0,2,2,1,4,6,3,0,8,16,12,4,1,16,40,40,20,5,0,32,96,120,80,30,6,1,
%T 64,224,336,280,140,42,7,0,128,512,896,896,560,224,56,8,1,256,1152,
%U 2304,2688,2016,1008,336,72,9,0,512,2560,5760,7680,6720,4032,1680,480,90,10
%N Triangle T(n,k) read by rows: T(n,k) = Sum_{i=0..n} C(n,2i)*C(2i,k).
%C Gives the general solution to a(n) = 2*a(n-1) + k(k+2)*a(n-2), a(0) = a(1) = 1. The value k=1 gives the row sums of the triangle, or 1,1,5,13,... This is A046717, the solution to a(n) = 2*a(n-1) + 3*a(n-2), a(0)=a(1)=1.
%C Product of A007318 and A007318 with every odd-indexed row set to zero. - _Paul Barry_, Nov 08 2005
%D A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, identity 156.
%D J-L. Kim, Relation between weight distribution and combinatorial identities, Bulletin of the Institute of Combinatorics and its Applications, Canada, 31, 2001, pp. 69-79.
%H Michael De Vlieger, <a href="/A080928/b080928.txt">Table of n, a(n) for n = 0..11475</a> (rows 0 <= n <= 150).
%H Saul Schleimer, Bert Wiest, <a href="https://arxiv.org/abs/1807.01500">On the conjugacy problem in braid groups: Garside theory and subsurfaces</a>, arXiv:1807.01500 [math.GT], 2018.
%F T(n, n) = (n+1) mod 2, T(n, k) = C(n, k)*2^(n-k-1).
%F T(n, 0) = A011782(n), T(n, k)=0, k>n, T(2n, 2n)=1, T(2n-1, 2n-1)=0, T(n+1, n)=n+1. Otherwise T(n, k) = T(n-1, k-1) + 2T(n-1, k). Rows are the coefficients of the polynomials in the expansion of (1-x)/((1+kx)(1-(k+2)x). The main diagonal is 1, 0, 1, 0, 1, 0, ... with g.f. 1/(1-x^2). Subsequent subdiagonals are given by A011782(k)*C(n+k, k) with g.f. A011782(k)/(1-x)^k.
%F T(n, k) = Sum_{j=0..n} C(n, j)*C(j, k)*(1+(-1)^j)/2; T(n, k) = 2^(n-k-1)*(C(n, k) + (-1)^n*C(0, n-k)). - _Paul Barry_, Nov 08 2005
%e Triangle begins:
%e 1;
%e 1, 0;
%e 2, 2, 1;
%e 4, 6, 3, 0;
%e 8, 16, 12, 4, 1;
%e 16, 40, 40, 20, 5, 0;
%e 32, 96, 120, 80, 30, 6, 1;
%e 64, 224, 336, 280, 140, 42, 7, 0;
%e 128, 512, 896, 896, 560, 224, 56, 8, 1;
%e 256, 1152, 2304, 2688, 2016, 1008, 336, 72, 9, 0; etc.
%t Table[Sum[Binomial[n, 2 i] Binomial[2 i, k], {i, 0, n}], {n, 0, 10}, {k, 0, n}] // Flatten (* _Michael De Vlieger_, Oct 11 2018 *)
%Y Apart from k=n, T(n, k) equals (1/2)*A038207(n, k).
%Y Columns include A011782, 2*A001792, A080929, 4*A080930. Row sums are in A046717.
%K nonn,tabl,easy
%O 0,4
%A _Paul Barry_, Feb 26 2003
%E Edited by _Ralf Stephan_, Feb 04 2005
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