0,4

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.

Product of A007318 and A007318 with every odd indexed row set to zero. - Paul Barry (pbarry(AT)wit.ie), Nov 08 2005

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, identity 156.

J-L. Kim, Relation between weight distribution and combinatorial identities, Bulletin of the Institute of Combinatorics and its Applications, Canada, 31, Jan. 2001, pp. 69-79.

T(n, n) = (n+1) mod 2, T(n, k) = C(n, k)*2^(n-k-1).

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.

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 (pbarry(AT)wit.ie), Nov 08 2005

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

64,224,336,280,140,42,7,0

128,512,896,896,560,224,56,8,1

256,1152,2304,2688,2016,1008,336,72,9,0

Apart from k=n, T(n, k) equals (1/2)*A038207(n, k).

Columns include A011782, 2*A001792, A080929, 4*A080930. Row sums are in A046717.

Sequence in context: A115313 A048942 A121484 this_sequence A068957 A119468 A091869

Adjacent sequences: A080925 A080926 A080927 this_sequence A080929 A080930 A080931

easy,nonn,tabl

Paul Barry (pbarry(AT)wit.ie), Feb 26 2003

Edited by Ralf Stephan, Feb 04 2005