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
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
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).
Comments