A296769 -id:A296769 - cp's OEIS frontend

A296662 Table read by rows, the odd rows of A296664, T(n, k) for n >= 0 and 0 <= k <= 2n.

1, 2, 3, 2, 5, 9, 10, 9, 5, 14, 28, 34, 35, 34, 28, 14, 42, 90, 117, 125, 126, 125, 117, 90, 42, 132, 297, 407, 451, 461, 462, 461, 451, 407, 297, 132, 429, 1001, 1430, 1638, 1703, 1715, 1716, 1715, 1703, 1638, 1430, 1001, 429

Offset: 0

Peter Luschny, Dec 20 2017

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 diagonal next to the main diagonal of M(2n+1)^(2n+1).

			The triangle starts:
0: [  1]
1: [  2,   3,   2]
2: [  5,   9,  10,   9,   5]
3: [ 14,  28,  34,  35,  34,  28,  14]
4: [ 42,  90, 117, 125, 126, 125, 117,  90,  42]
5: [132, 297, 407, 451, 461, 462, 461, 451, 407, 297, 132]

Cf. A000108, A001700, A050158, A296664, A296666, A296769 (row sums).

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+1)^(2*n+1), 1),list), n=0..6);

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

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

T(n, n) = A001700(n).
T(n, 0) = T(n, 2*n) = A000108(n+1).
T(n, k) = binomial(2*n+1, n+1) - binomial(2*n+1, n-k-1) for k=0..n.
T(n, k) = binomial(2*n+1, n+1) - binomial(2*n+1, k-n-1) for k=n+1..2*n and n>0.

cp's OEIS Frontend

A296662 Table read by rows, the odd rows of A296664, T(n, k) for n >= 0 and 0 <= k <= 2n.

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