A255812 Rectangular array: row n gives the denominators in the positive convolutory n-th root of (1,1,1,...).

1, 1, 1, 1, 2, 1, 1, 8, 3, 1, 1, 16, 9, 4, 1, 1, 128, 81, 32, 5, 1, 1, 256, 243, 128, 25, 6, 1, 1, 1024, 729, 2048, 125, 72, 7, 1, 1, 2048, 6561, 8192, 625, 1296, 49, 8, 1, 1, 32768, 19683, 65536, 15625, 31104, 343, 128, 9, 1, 1, 65536, 59049, 262144, 78125

Offset

1,5

Comments

(See Comments at A255811.)

Links

Clark Kimberling, Antidiagonals n = 1..60, flattened

Formula

G.f. of s: (1 - t)^(-1/n).

Example

First, regarding the numbers numerator/denominator, we have
row 1:  1,1,1,1,1,1,1,1,1,1,1,1,..., the 0th self-convolution of (1,1,1,...);
row 2:  1,1/2,3/8,5/16,35/128,63/256, ..., convolutory sqrt of (1,1,1,...);
row 3:  1,1/3,2/9,14/81,35/243,91/729,..., convolutory 3rd root
row 4:  1,1/4,5/32,15/128,195/2048,663/8192,..., convolutory 4th root.
Taking only denominators:
row 1:  1,1,1,1,1,1,1,...
row 2:  1,2,8,16,128,...
row 3:  1,3,9,81,243,729,...
row 4:  1,4,32,128,2048,8192,...

Mathematica

z = 15; t[n_] := CoefficientList[Normal[Series[(1 - t)^(-1/n), {t, 0, z}]], t];
u = Table[Numerator[t[n]], {n, 1, z}]
TableForm[Table[u[[n, k]], {n, 1, z}, {k, 1, z}]]     (*A255811 array*)
Table[u[[n - k + 1, k]], {n, z}, {k, n, 1, -1}] // Flatten (*A255811 sequence*)
v = Table[Denominator[t[n]], {n, 1, z}]
TableForm[Table[v[[n, k]], {n, 1, z}, {k, 1, z}]]     (*A255812 array*)
Table[v[[n - k + 1, k]], {n, z}, {k, n, 1, -1}] // Flatten  (*A255812 sequence*)

Crossrefs

Cf. A244811, A000012.

Sequence in context: A077058 A053373 A297733 * A249141 A353953 A102875

Adjacent sequences: A255809 A255810 A255811 * A255813 A255814 A255815

Keyword

nonn,easy,tabl,frac

Author

Clark Kimberling, Mar 11 2015

Status

approved