A286509 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of k-th power of continued fraction 1/(1 + x/(1 + x^2/(1 + x^3/(1 + x^4/(1 + x^5/(1 + ...)))))).

1, 1, 0, 1, -1, 0, 1, -2, 1, 0, 1, -3, 3, 0, 0, 1, -4, 6, -2, -1, 0, 1, -5, 10, -7, -1, 1, 0, 1, -6, 15, -16, 3, 4, -1, 0, 1, -7, 21, -30, 15, 6, -6, 1, 0, 1, -8, 28, -50, 40, 0, -17, 6, 0, 0, 1, -9, 36, -77, 84, -26, -30, 24, -3, -1, 0, 1, -10, 45, -112, 154, -90, -30, 64, -21, -2, 2, 0, 1, -11, 55, -156, 258, -217, 15, 125, -81, 6, 9, -3, 0

Offset: 0

Author

Ilya Gutkovskiy, May 10 2017

Keywords

• sign
• tabl

Examples

			Square array begins:
1,  1,  1,  1,   1,   1,  ...
0, -1, -2, -3,  -4,  -5,  ...
0,  1,  3,  6,  10,  15,  ...
0,  0, -2, -7, -16, -30,  ...
0, -1, -1,  3,  15,  40,  ...
0,  1,  4,  6,   0, -26,  ...
		

Links

• Seiichi Manyama, Antidiagonals n = 0..139, flattened
• Eric Weisstein's World of Mathematics, Rogers-Ramanujan Continued Fraction

Crossrefs

Columns k=0-5 give: A000007, A007325, A055101, A055102, A055103, A078905 (with offset 0).

Rows n=0-2 give: A000012, A001489, A000217.

Main diagonal gives A291651.

Antidiagonal sums give A302015.

Programs

Mathematica

Table[Function[k, SeriesCoefficient[1/(1 + ContinuedFractionK[x^i, 1, {i, 1, n}])^k, {x, 0, n}]][j - n], {j, 0, 12}, {n, 0, j}] // Flatten
Table[Function[k, SeriesCoefficient[Product[(1 - x^(5 i - 1)) (1 - x^(5 i - 4))/((1 - x^(5 i - 2)) (1 - x^(5 i - 3))), {i, n}]^k, {x, 0, n}]][j - n], {j, 0, 12},{n, 0, j}] // Flatten

Formula

G.f. of column k: Product_{j>=1} ((1 - x^(5*j-1))*(1 - x^(5*j-4)) / ((1 - x^(5*j-2))*(1 - x^(5*j-3))))^k.