login
Square array A(n,k), n >= 1, k >= 0, read by antidiagonals downwards, where A(n,k) = Sum_{j=1..n} floor(n/j) * (-k)^(j-1).
7

%I #24 Jun 28 2024 05:05:38

%S 1,1,2,1,1,3,1,0,3,4,1,-1,5,2,5,1,-2,9,-4,4,6,1,-3,15,-20,13,4,7,1,-4,

%T 23,-52,62,-16,6,8,1,-5,33,-106,205,-174,49,4,9,1,-6,45,-188,520,-806,

%U 556,-88,7,10,1,-7,59,-304,1109,-2584,3291,-1660,173,7,11

%N Square array A(n,k), n >= 1, k >= 0, read by antidiagonals downwards, where A(n,k) = Sum_{j=1..n} floor(n/j) * (-k)^(j-1).

%H G. C. Greubel, <a href="/A344824/b344824.txt">Antidiagonals n = 1..50, flattened</a>

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

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

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

%F T(n, k) = Sum_{j=1..(k+1)} floor((k+1)/j) * (k-n+1)^(j-1), for n >= 1, 0 <= k <= n-1 (antidiagonal triangle). - _G. C. Greubel_, Jun 27 2024

%e Square array, A(n, k), begins:

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

%e 2, 1, 0, -1, -2, -3, -4, ...

%e 3, 3, 5, 9, 15, 23, 33, ...

%e 4, 2, -4, -20, -52, -106, -188, ...

%e 5, 4, 13, 62, 205, 520, 1109, ...

%e 6, 4, -16, -174, -806, -2584, -6636, ...

%e Antidiagonal triangle, T(n, k), begins:

%e 1;

%e 1, 2;

%e 1, 1, 3;

%e 1, 0, 3, 4;

%e 1, -1, 5, 2, 5;

%e 1, -2, 9, -4, 4, 6;

%e 1, -3, 15, -20, 13, 4, 7;

%e 1, -4, 23, -52, 62, -16, 6, 8;

%e 1, -5, 33, -106, 205, -174, 49, 4, 9;

%e 1, -6, 45, -188, 520, -806, 556, -88, 7, 10;

%t A[n_, k_] := Sum[If[k == 0 && j == 1, 1, (-k)^(j - 1)] * Quotient[n, j], {j, 1, n}]; Table[A[k, n - k], {n, 1, 10}, {k, 1, n}] // Flatten (* _Amiram Eldar_, May 29 2021 *)

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

%o (PARI) A(n, k) = sum(j=1, n, sumdiv(j, d, (-k)^(d-1)));

%o (Magma)

%o A:= func< n,k | k eq n select n else (&+[Floor(n/j)*(-k)^(j-1): j in [1..n]]) >;

%o A344824:= func< n,k | A(k+1, n-k-1) >;

%o [A344824(n,k): k in [0..n-1], n in [1..12]]; // _G. C. Greubel_, Jun 27 2024

%o (SageMath)

%o def A(n,k): return n if k==n else sum((n//j)*(-k)^(j-1) for j in range(1,n+1))

%o def A344824(n,k): return A(k+1, n-k-1)

%o flatten([[A344824(n,k) for k in range(n)] for n in range(1,13)]) # _G. C. Greubel_, Jun 27 2024

%Y Columns k=0..4 give A000027, A059851, A344817, A344818, A344819.

%Y A(n,n) gives A344820.

%Y Cf. A344821.

%K sign,tabl

%O 1,3

%A _Seiichi Manyama_, May 29 2021