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Array T(n,k) = k^2 + (2n-4)*floor(k/2)^2, n >= 0, k >= 0, read by descending antidiagonals.
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%I #39 Sep 26 2022 01:31:12

%S 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,

%T 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,

%U 49,64,67,72,49,48,19,16,1,0,21,50,81,96,85,90,57,56,21,18,1,0

%N Array T(n,k) = k^2 + (2n-4)*floor(k/2)^2, n >= 0, k >= 0, read by descending antidiagonals.

%C Column k is an arithmetic progression with difference 2*A008794(k).

%C Odd rows of A133728 triangle are contained in row 0.

%C For i = 0 through 4, row i is 0 and the diagonal of A319929, A322630 = A213037, A003991, A322744, and A327259, respectively. In general, row i is 0 and the diagonal of array U(i;n,k) described in A327263.

%H David Lovler, <a href="/A354596/b354596.txt">Table of n, a(n) for n = 0..5150</a>

%F T(n,k) = U(n;k,k) (see A327263).

%F 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.

%F 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.

%F 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).

%e T(n,k) begins:

%e 0, 1, 0, 5, 0, 9, 0, 13, ...

%e 0, 1, 2, 7, 8, 17, 18, 31, ...

%e 0, 1, 4, 9, 16, 25, 36, 49, ...

%e 0, 1, 6, 11, 24, 33, 54, 67, ...

%e 0, 1, 8, 13, 32, 41, 72, 85, ...

%e 0, 1, 10, 15, 40, 49, 90, 103, ...

%e 0, 1, 12, 17, 48, 57, 108, 121, ...

%e ...

%t 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 *)

%o (PARI) T(n,k) = k^2 + (2*n-4)*(k\2)^2;

%Y Cf. A000290, A008794, A133728, A213037, A247375.

%Y Cf. A266222, A266439.

%Y Cf. A319929, A322630, A322744, A327259, A327263, A354594, A354595.

%K nonn,tabl,easy

%O 0,7

%A _David Lovler_, Jun 01 2022