A080928 Triangle T(n,k) read by rows: T(n,k) = Sum_{i=0..n} C(n,2i)*C(2i,k).

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, 512, 2560, 5760, 7680, 6720, 4032, 1680, 480, 90, 10

Offset: 0

Author

Paul Barry, Feb 26 2003

Keywords

• nonn
• tabl
• easy

Comments

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, Nov 08 2005

Examples

			Triangle begins:
    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; etc.
		

References

• 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, 2001, pp. 69-79.

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150).
• Saul Schleimer, Bert Wiest, On the conjugacy problem in braid groups: Garside theory and subsurfaces, arXiv:1807.01500 [math.GT], 2018.

Crossrefs

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.

Programs

Mathematica

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 *)

Formula

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+k*x)*(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, Nov 08 2005

Extensions

Edited by Ralf Stephan, Feb 04 2005