A296664 Table read by rows, diagonals of powers of Toeplitz matrices generated by the characteristic function of 1, T(n, k) for n >= 0 and 0 <= k <= 2*floor(n/2).

1, 1, 1, 2, 1, 2, 3, 2, 2, 5, 6, 5, 2, 5, 9, 10, 9, 5, 5, 14, 19, 20, 19, 14, 5, 14, 28, 34, 35, 34, 28, 14, 14, 42, 62, 69, 70, 69, 62, 42, 14, 42, 90, 117, 125, 126, 125, 117, 90, 42, 42, 132, 207, 242, 251, 252, 251, 242, 207, 132, 42

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(n)^n if n is even or the diagonal next to the main diagonal if n is odd. Note that the antidiagonals of M(n)^n are the rows of Pascal's triangle A007318.

Examples

			The first few matrices M(n)^n are:
n=0   n=1     n=2       n=3         n=4
|1|  |0 1|  |1 0 1|  |0 2 0 1|  |2 0 3 0 1|
     |1 0|  |0 2 0|  |2 0 3 0|  |0 5 0 4 0|
            |1 0 1|  |0 3 0 2|  |3 0 6 0 3|
                     |1 0 2 0|  |0 4 0 5 0|
                                |1 0 3 0 2|
The triangle starts:
0: [ 1]
1: [ 1]
2: [ 1,  2,   1]
3: [ 2,  3,   2]
4: [ 2,  5,   6,   5,   2]
5: [ 5,  9,  10,   9,   5]
6: [ 5, 14,  19,  20,  19,  14,  5]
7: [14, 28,  34,  35,  34,  28,  14]
8: [14, 42,  62,  69,  70,  69,  62, 42, 14]
9: [42, 90, 117, 125, 126, 125, 117, 90, 42]
		

Crossrefs

Cf. A000108, A001405, A208355, A296663 (row sums), A296662 (odd rows), A296666 (even rows).

Programs

Maple

v := n -> `if`(n=1, 1, 0):
M := n -> LinearAlgebra:-ToeplitzMatrix([seq(v(j), j=0..n)], symmetric):
seq(convert(ArrayTools:-Diagonal(M(n)^n, n mod 2), 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[n], Mod[n, 2]]];
Table[d[n], {n, 0, 10}] // Flatten

Sage

def T(n, k):
    h, e = n//2, n%2 == 0
    a = binomial(n, h) if e else binomial(2*h+1, h+1)
    if k > h:
        b = binomial(n, k-h-1) if e else binomial(2*h+1, k-h-1)
    else:
        b = binomial(n, h+k+1) if e else binomial(2*h+1, h-k-1)
    return a - b
for n in (0..9): print([T(n, k) for k in (0..2*(n//2))])

Formula

T(n, 0) = T(n, 2*floor(n/2)) = A208355(n) = A000108(floor((n+1)/2)).

T(n, floor(n/2)) = A001405(n).

Further formulas can be found in A296662 and A296666 for the cases n odd and n even.