A354596 Array T(n,k) = k^2 + (2n-4)*floor(k/2)^2, n >= 0, k >= 0, read by descending antidiagonals.
0, 1, 0, 0, 1, 0, 5, 2, 1, 0, 0, 7, 4, 1, 0, 9, 8, 9, 6, 1, 0, 0, 17, 16, 11, 8, 1, 0, 13, 18, 25, 24, 13, 10, 1, 0, 0, 31, 36, 33, 32, 15, 12, 1, 0, 17, 32, 49, 54, 41, 40, 17, 14, 1, 0, 0, 49, 64, 67, 72, 49, 48, 19, 16, 1, 0, 21, 50, 81, 96, 85, 90, 57, 56, 21, 18, 1, 0
Offset: 0
Examples
T(n,k) begins: 0, 1, 0, 5, 0, 9, 0, 13, ... 0, 1, 2, 7, 8, 17, 18, 31, ... 0, 1, 4, 9, 16, 25, 36, 49, ... 0, 1, 6, 11, 24, 33, 54, 67, ... 0, 1, 8, 13, 32, 41, 72, 85, ... 0, 1, 10, 15, 40, 49, 90, 103, ... 0, 1, 12, 17, 48, 57, 108, 121, ... ...
Links
- David Lovler, Table of n, a(n) for n = 0..5150
Crossrefs
Programs
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Mathematica
T[n_, k_] := k^2 + (2*n - 4)*Floor[k/2]^2; Table[T[n - k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Amiram Eldar, Jun 20 2022 *) -
PARI
T(n,k) = k^2 + (2*n-4)*(k\2)^2;
Formula
T(n,k) = U(n;k,k) (see A327263).
For each row, T(n,k) = T(n,k-1) + 2*T(n,k-2) - 2*T(n,k-3) - T(n,k-4) + T(n,k-5), k >= 5.
G.f. for row n: x*(1 + (2*n-1)*x + 3*x^2 + (2*n-3)*x^3)/((1 - x)^3*(1 + x)^2). When n = 2, this reduces to x*(1 + x)/(1 - x)^3.
E.g.f. for row n: (((4-n)*x + n*x^2)*cosh(x) + (n-2 + n*x + n*x^2)*sinh(x))/2. When n = 2, this reduces to (x + x^2)*cosh(x) + (x + x^2)*sinh(x) = (x + x^2)*exp(x).
Comments