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%I #10 Feb 04 2019 07:28:24
%S 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,
%T 91,195,11,7,1,1,1,429,728,663,44,91,4,1,1,1,6435,1976,4641,924,1729,
%U 20,9,1,1,1,12155,5434,16575,4004,8645,110,51,5,1,1,1
%N Rectangular array: row n gives the numerators in the positive convolutory n-th root of (1,1,1,...).
%C The convolution n times of the sequence comprising row n is the constant sequence (1,1,1,...) = A000012.
%C 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.
%C Is (column 4) = A175485, the numerators of averages of squares of 1,...,n?
%H Clark Kimberling, <a href="/A255811/b255811.txt">Antidiagonals n = 1..60, flattened</a>
%F G.f. of s: (1 - t)^(-1/n).
%e First, regarding the numbers numerator/denominator, we have
%e row 1: 1,1,1,1,1,1,1,1,1,1,1,1,1,..., the 0th self-convolution of (1,1,1,...);
%e row 2: 1,1/2,3/8,5/16,35/128,63/256, ..., convolutory sqrt of (1,1,1,...);
%e row 3: 1,1/3,2/9,14/81,35/243,91/729,..., convolutory 3rd root;
%e row 4: 1,1/4,5/32,15/128,195/2048,663/8192,..., convolutoary 4th root.
%e Taking only numerators:
%e row 1: 1,1,1,1,1,1,1,...
%e row 2: 1,1,3,5,35,63,...
%e row 3: 1,1,2,14,35,91,...
%e row 4: 1,1,5,15,195,663,...
%t z = 15; t[n_] := CoefficientList[Normal[Series[(1 - t)^(-1/n), {t, 0, z}]], t];
%t u = Table[Numerator[t[n]], {n, 1, z}]
%t TableForm[Table[u[[n, k]], {n, 1, z}, {k, 1, z}]] (* A255811 array *)
%t Table[u[[n - k + 1, k]], {n, z}, {k, n, 1, -1}] // Flatten (* A255811 sequence *)
%t v = Table[Denominator[t[n]], {n, 1, z}]
%t TableForm[Table[v[[n, k]], {n, 1, z}, {k, 1, z}]] (* A255812 array *)
%t Table[v[[n - k + 1, k]], {n, z}, {k, n, 1, -1}] // Flatten (* A255812 sequence *)
%Y Cf. A244812, A000040, A000012.
%K nonn,easy,tabl,frac
%O 1,8
%A _Clark Kimberling_, Mar 11 2015
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