A296666 Table read by rows, the even rows of A296664, T(n, k) for n >= 0 and 0 <= k <= 2n.

1, 1, 2, 1, 2, 5, 6, 5, 2, 5, 14, 19, 20, 19, 14, 5, 14, 42, 62, 69, 70, 69, 62, 42, 14, 42, 132, 207, 242, 251, 252, 251, 242, 207, 132, 42, 132, 429, 704, 858, 912, 923, 924, 923, 912, 858, 704, 429, 132

Offset: 0

Author

Peter Luschny, Dec 19 2017

Keywords

• nonn
• tabf

Comments

Let v be the characteristic function of 1 (A063524) and M(n) for n >= 0 the symmetric Toeplitz matrix generated by the initial segment of v, then row n is the main diagonal of M(2n)^(2n).

Seems to be A050157 + its reflection. - Andrey Zabolotskiy, Dec 19 2017

Examples

			0: [  1]
1: [  1,   2,   1]
2: [  2,   5,   6,   5,   2]
3: [  5,  14,  19,  20,  19,  14,   5]
4: [ 14,  42,  62,  69,  70,  69,  62,  42,  14]
5: [ 42, 132, 207, 242, 251, 252, 251, 242, 207, 132,  42]
6: [132, 429, 704, 858, 912, 923, 924, 923, 912, 858, 704, 429, 132]
		

Links

• Peter Luschny, Row n for n = 0..30

Crossrefs

Cf. A000108, A000984, A050157, A296662, A296664, A296665 (row sums).

Programs

Maple

v := n -> `if`(n=1, 1, 0);
B := n -> LinearAlgebra:-ToeplitzMatrix([seq(v(j), j=0..n)], symmetric):
seq(convert(ArrayTools:-Diagonal(B(2*n)^(2*n)), list), n=0..10);

Mathematica

v[n_] := If[n == 1, 1, 0];
m[n_] := MatrixPower[ToeplitzMatrix[Table[v[k], {k, 0, n}]], n];
d[n_] := If[n == 0, {1}, Diagonal[m[2 n]]];
Table[d[n], {n, 0, 6}] // Flatten

Sage

def T(n, k):
    if k > n:
        b = binomial(2*n, k - n - 1)
    else:
        b = binomial(2*n, n + k + 1)
    return binomial(2*n, n) - b
for n in (0..6):
    print([T(n, k) for k in (0..2*n)])

Formula

T(n, 0) = T(n, 2*n) = A000108(n).

T(n, n) are the central binomial coefficients A000984(n).

T(n, k) = binomial(2*n, n) - binomial(2*n, n+k+1) for k=0..n.

T(n, k) = binomial(2*n, n) - binomial(2*n, k-n-1) for k=n+1..2*n and n>0.