OFFSET
0,2
COMMENTS
A subset of Pascal's triangle A007318 with only even elements.
REFERENCES
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 224.
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)
EXAMPLE
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)
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Ralf Stephan, Feb 06 2005
STATUS
approved