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Square array T(n,k), n >= 1, k >= 1, read by antidiagonals downwards, where T(n,k) = Sum_{j=1..n} floor(n/(2*j-1))^k.
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%I #45 Dec 17 2021 11:07:20

%S 1,1,2,1,4,4,1,8,10,5,1,16,28,17,7,1,32,82,65,27,9,1,64,244,257,127,

%T 41,11,1,128,730,1025,627,225,55,12,1,256,2188,4097,3127,1313,353,70,

%U 15,1,512,6562,16385,15627,7809,2419,522,93,17,1,1024,19684,65537,78127,46721,16841,4114,759,115,19

%N Square array T(n,k), n >= 1, k >= 1, read by antidiagonals downwards, where T(n,k) = Sum_{j=1..n} floor(n/(2*j-1))^k.

%F G.f. of column k: (1/(1 - x)) * Sum_{j>=1} (j^k - (j - 1)^k) * x^j/(1 - x^(2*j)).

%F T(n,k) = Sum_{j=1..n} Sum_{d|j, j/d odd} d^k - (d - 1)^k.

%e Square array begins:

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

%e 2, 4, 8, 16, 32, 64, 128, ...

%e 4, 10, 28, 82, 244, 730, 2188, ...

%e 5, 17, 65, 257, 1025, 4097, 16385, ...

%e 7, 27, 127, 627, 3127, 15627, 78127, ...

%e 9, 41, 225, 1313, 7809, 46721, 280065, ...

%e 11, 55, 353, 2419, 16841, 117715, 823673, ...

%t T[n_, k_] := Sum[Floor[n/(2*j - 1)]^k, {j, 1, n}]; Table[T[k, n - k + 1], {n, 1, 11}, {k, 1, n}] // Flatten (* _Amiram Eldar_, Dec 17 2021 *)

%o (PARI) T(n, k) = sum(j=1, n, (n\(2*j-1))^k);

%o (PARI) T(n, k) = sum(j=1, n, sumdiv(j, d, j/d%2*(d^k-(d-1)^k)));

%Y Columns k=1..3 give A060831, A350143, A350144.

%Y T(n,n) gives A350145.

%Y Cf. A344725.

%K nonn,tabl

%O 1,3

%A _Seiichi Manyama_, Dec 16 2021