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A368343 Square array T(n,k), n >= 3, k >= 0, read by antidiagonals downwards, where T(n,k) = Sum_{j=0..n} k^(n-j) * floor(j/3). 4

%I #25 Dec 22 2023 16:06:47

%S 1,1,1,1,2,1,1,3,3,2,1,4,7,5,2,1,5,13,16,7,2,1,6,21,41,34,9,3,1,7,31,

%T 86,125,70,12,3,1,8,43,157,346,377,143,15,3,1,9,57,260,787,1386,1134,

%U 289,18,4,1,10,73,401,1562,3937,5547,3405,581,22,4

%N Square array T(n,k), n >= 3, k >= 0, read by antidiagonals downwards, where T(n,k) = Sum_{j=0..n} k^(n-j) * floor(j/3).

%H Seiichi Manyama, <a href="/A368343/b368343.txt">Antidiagonals n = 3..142, flattened</a>

%F T(n,k) = T(n-3,k) + Sum_{j=0..n-3} k^j.

%F T(n,k) = 1/(k-1) * Sum_{j=0..n} floor(k^j/(k^2+k+1)) = Sum_{j=0..n} floor(k^j/(k^3-1)) for k > 1.

%F T(n,k) = (k+1)*T(n-1,k) - k*T(n-2,k) + T(n-3,k) - (k+1)*T(n-4,k) + k*T(n-5,k).

%F G.f. of column k: x^3/((1-x) * (1-k*x) * (1-x^3)).

%F T(n,k) = 1/(k-1) * (floor(k^(n+1)/(k^3-1)) - floor((n+1)/3)) for k > 1.

%e Square array begins:

%e 1, 1, 1, 1, 1, 1, 1, ...

%e 1, 2, 3, 4, 5, 6, 7, ...

%e 1, 3, 7, 13, 21, 31, 43, ...

%e 2, 5, 16, 41, 86, 157, 260, ...

%e 2, 7, 34, 125, 346, 787, 1562, ...

%e 2, 9, 70, 377, 1386, 3937, 9374, ...

%e 3, 12, 143, 1134, 5547, 19688, 56247, ...

%o (PARI) T(n, k) = sum(j=0, n, k^(n-j)*(j\3));

%Y Columns k=0..4 give A002264, A130518, A178455, A368344, A368345.

%Y Cf. A055129, A368296.

%K nonn,tabl

%O 3,5

%A _Seiichi Manyama_, Dec 22 2023

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