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
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
Programs
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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.