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

Author

Clark Kimberling, Mar 11 2015

Keywords

• nonn
• easy
• tabl
• frac

Comments

(See Comments at A255811.)

Examples

			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,...
		

Links

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

Crossrefs

Cf. A244811, A000012.

Programs

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*)

Formula

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