A294653 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of Product_{j>=1} (1-j^(k*j)*x^j) in powers of x.

1, 1, -1, 1, -1, -1, 1, -1, -4, 0, 1, -1, -16, -23, 0, 1, -1, -64, -713, -229, 1, 1, -1, -256, -19619, -64807, -2761, 0, 1, -1, -1024, -531185, -16757533, -9688425, -42615, 1, 1, -1, -4096, -14347883, -4294435855, -30499541197, -2165979799, -758499, 0

Offset: 0

Author

Seiichi Manyama, Nov 06 2017

Keywords

• sign
• tabl

Examples

			Square array begins:
    1,    1,      1,         1,           1, ...
   -1,   -1,     -1,        -1,          -1, ...
   -1,   -4,    -16,       -64,        -256, ...
    0,  -23,   -713,    -19619,     -531185, ...
    0, -229, -64807, -16757533, -4294435855, ...
		

Links

• Seiichi Manyama, Antidiagonals n = 0..52, flattened

Crossrefs

Columns k=0..1 give A010815, A292312.

Rows n=0..2 give A000012, (-1)*A000012, (-1)*A000302.

Cf. A283675, A294758.

Programs

Mathematica

rows = 10;
col[k_] := col[k] = CoefficientList[Product[(1 - j^(k*j)*x^j), {j, 1, rows + 3}] + O[x]^(rows + 3), x];
A[n_, k_] := col[k][[n + 1]];
(* or: *)
A[0, ] = 1; A[n, k_] := A[n, k] = -(1/n)*Sum[DivisorSum[j, #^(1 + k*j) &]*A[n - j, k], {j, 1, n}];
Table[A[n - k, k], {n, 0, rows - 1}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Nov 10 2017 *)

Formula

A(0,k) = 1 and A(n,k) = -(1/n) * Sum_{j=1..n} (Sum_{d|j} d^(1+k*j)) * A(n-j,k) for n > 0.