A103328 Triangle T(n, k) read by rows: binomial(2n, 2k+1).

0, 2, 0, 4, 4, 0, 6, 20, 6, 0, 8, 56, 56, 8, 0, 10, 120, 252, 120, 10, 0, 12, 220, 792, 792, 220, 12, 0, 14, 364, 2002, 3432, 2002, 364, 14, 0, 16, 560, 4368, 11440, 11440, 4368, 560, 16, 0, 18, 816, 8568, 31824, 48620, 31824, 8568, 816, 18, 0, 20, 1140, 15504

Offset: 0

Author

Ralf Stephan, Feb 06 2005

Keywords

• nonn
• tabl

Comments

A subset of Pascal's triangle A007318 with only even elements.

Examples

			Triangle begins
   0;
   2,   0;
   4,   4,    0;
   6,  20,    6,     0;
   8,  56,   56,     8,     0;
  10, 120,  252,   120,    10,    0;
  12, 220,  792,   792,   220,   12,   0;
  14, 364, 2002,  3432,  2002,  364,  14,  0;
  16, 560, 4368, 11440, 11440, 4368, 560, 16, 0;
  ...
From _Peter Bala_, Jan 30 2022: (Start)
(1/2)*(N^2 + N)^2 = 2*Sum_{j = 1..N} j^3.
(1/2)*(N^2 + N)^4 = 4*Sum_{j = 1..N} j^5 + 4*Sum_{j = 1..N} j^7.
(1/2)*(N^2 + N)^6 = 6*Sum_{j = 1..N} j^7 + 20*Sum_{j = 1..N} j^9 + 6*Sum_{j = 1..N} j^11.
(1/2)*(N^2 + N)^8 = 8*Sum_{j = 1..N} j^9 + 56*Sum_{j = 1..N} j^11 + 56*Sum_{j = 1..N} j^13 + 8*Sum_{j = 1..N} j^15. (End)
		

References

• A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 224.

Crossrefs

Cf. A091042, A086645, A103327.

Formula

From Peter Bala, Jan 31 2022: (Start)

T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k) + 2*T(n-2,k-1) - T(n-2,k-2), with T(0,0) = 0, T(1,0) = 2 and T(n,k) = 0 if k < 0 or if k > n-1.

n-th row polynomial R(n,x) = (1/(2*sqrt(x)))*( (1 + sqrt(x))^(2*n) - (1 - sqrt(x))^(2*n) ).

O.g.f.: A(x,t) = 2*t/(1 - 2*(x + 1)*t + (x - 1)^2*t^2) = 2*t + (4 + 4*x)*t^2 + (6 + 20*x + 6*x^2)*t^3 + ....

G.f.: (1/sqrt(x))*sinh(t)*sinh(sqrt(x)*t) = 2*t^2/2! + (4 + 4*x)*t^4/4! + (6 + 20*x^2 + 6*x^3)*t^6/6! + ....

O.g.f. for n-th diagonal: ( Sum_{k = 0..n} binomial(2*n,2*k+1)*x^k )/(1 - x)^(2*n) = 1/(2*sqrt(x))*((1 - sqrt(x))^(-2*n) - (1 + sqrt(x))^(-2*n)).

With a different offset, 2/(x-4)*A(x/(x-4), t*(x-4)/4) = t/(1 + t*(2 - x) + t^2) is a g.f. of A053122.

Define S(r,N) = Sum_{j = 1..N} j^r. Then the following identity holds for n >= 1:

(1/2)*(N^2 + N)^(2*n) = T(n,0)*S(2*n+1,N) + T(n,1)*S(2*n+3,N) + ... + T(n,n-1)*S(4*n-1,N). Some examples are given below. (End)