A255811 Rectangular array: row n gives the numerators in the positive convolutory n-th root of (1,1,1,...).

1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 5, 2, 1, 1, 1, 35, 14, 5, 1, 1, 1, 63, 35, 15, 3, 1, 1, 1, 231, 91, 195, 11, 7, 1, 1, 1, 429, 728, 663, 44, 91, 4, 1, 1, 1, 6435, 1976, 4641, 924, 1729, 20, 9, 1, 1, 1, 12155, 5434, 16575, 4004, 8645, 110, 51, 5, 1, 1, 1

Offset: 1

Author

Clark Kimberling, Mar 11 2015

Keywords

• nonn
• easy
• tabl
• frac

Comments

The convolution n times of the sequence comprising row n is the constant sequence (1,1,1,...) = A000012.

It appears that if n+1 is a prime (A000040), then most of the terms in row n are divisible by n+1. Taking n = 4 for an example, 968 of the first 1000 terms are divisible by 5.

Is (column 4) = A175485, the numerators of averages of squares of 1,...,n?

Examples

			First, regarding the numbers numerator/denominator, we have
row 1:  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,..., convolutoary 4th root.
Taking only numerators:
row 1:  1,1,1,1,1,1,1,...
row 2:  1,1,3,5,35,63,...
row 3:  1,1,2,14,35,91,...
row 4:  1,1,5,15,195,663,...
		

Links

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

Crossrefs

Cf. A244812, A000040, 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).