T(n,k) = Sum_{i = 0..n} C(2*n,2*i)*C(k+i,n).
O.g.f. for row n: 1/(1-x)^(n+1) * Sum_{k = 0..n} C(2*n,2*k)*x^k = 1/(1-x) * T(n,(1+x)/(1-x)), where T(n,x) denotes the Chebyshev polynomial of the first kind.
O.g.f. for the array: 1/(1-x) * {(1-t) - x*(1+t)}/{(1-t)^2 - x*(1+t)^2} = (1+x+x^2+x^3+...) + (1+3*x+5*x^2+7*x^3+...)*t + (1+9*x+25*x^2+49*x^3+...)*t^2 + ... .
Row n of the array has the form [p_n(0),p_n(1),p_n(2),...], where the polynomial function p_n(x) = Sum_{k = 0..n} C(2*n,2*k)*C(x+k,n). The first few are p_0(x) = 1, p_1(x) = 2*x+1, p_2(x) = (2*x+1)^2, p_3(x) = (2*x+1)*(8*x^2+8*x+3)/3 and p_4(x) = (2*x+1)^2*(4*x^2+4*x+3)/3.
Alternative expressions for p_n(x) include p_n(x) = Sum_{k = 0..n} 2^(2*k)*n/(n+k)*C(n+k,2*k)*C(x,k) and p_n(x) = Sum_{k = 1..n} 2^(k-1)*C(n-1,k-1)*C(2*x+1,k).
The polynomials p_n(x) satisfy the 3-term recurrence relation n*p_n(x) = 2*(2*x+1)*p_(n-1)(x)+(n-2)*p(n-2)(x) for n >= 2; their generating function is 1/2*((1+t)/(1-t))^(2*x+1) = 1/2 + (2*x+1)*t + (2*x+1)^2*t^2 + (2*x+1)*(8*x^2+8*x+3)/3*t^3 + ... . Thus p_n(x) is, apart from a constant factor, the Meixner polynomial of the first kind M_n(2*x+1;b,c) at b = 0, c = -1. Compare with
A142979.
The polynomial p_n(x) is the unique polynomial solution to the difference equation (2*x+1)*{f(x+1/2) - f(x-1/2)} = 2*n*f(x), normalized so that f(0) = 1. The function p_n(x) is also the unique polynomial solution to the difference equation (2*x+1)*{(x+1)*f(x+1) + x*f(x-1)} = ((2*x+1)^2 + 2*n^2)*f(x), normalized so that f(0) = 1.
The zeros of p_n(x) lie on the vertical line Re x = -1/2 in the complex plane, that is, the polynomials p_n(x-1), n = 1,2,3,..., satisfy a Riemann hypothesis (adapt the proof of the lemma on p.4 of [BUMP et al.]).
For n > 0, the entries in row n of the array occur in series acceleration formulas for log(2): 2*log(2) = 1 + (1/2 - 1/6 +...+(-1)^n/(n*(n-1))) + (-1)^(n+1)*Sum_{k >= 1} 1/(k*T(n,k-1)*T(n,k)). For example, the fourth row of the table (n = 3) gives 2*log(2) = 4/3 + 1/(1*1*19) + 1/(2*19*85) + 1/(3*85*231) + ... .
The corresponding result for column k is 2*log(2) = 1 + (1/(1*3) + 1/(2*3*5) +...+ 1/(k*(2*k-1)*(2k+1)) + (2*k+1)*Sum_{n >= 1} (-1)^(n+1)/(n*(n+1)*T(n,k)* T(n+1,k)).
For example, the third column of the table (k = 2) gives 2*log(2) = 41/30 + 5*(1/(1*2*5*25) - 1/(2*3*25*85) + 1/(3*4*85*225) - ... ).
For the main diagonal calculation suggests the result: 2*log(2) = 4/3 + Sum_{n >= 1} (-1)^(n+1)*(5*n+3)/(n*(n+1)*T(n,n)*T(n+1,n+1)).
Similar series acceleration formulas for log(2) come from the row, column and diagonal entries of the square array of Delannoy numbers,
A008288 (which may viewed as the array of crystal ball sequences for the product lattices A_1 x...x A_1). For corresponding results for the constants zeta(2) and zeta(3) see
A108625 and
A143007 respectively.
