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%I #8 Feb 04 2019 07:27:54
%S 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,
%T 1,1024,729,2048,125,72,7,1,1,2048,6561,8192,625,1296,49,8,1,1,32768,
%U 19683,65536,15625,31104,343,128,9,1,1,65536,59049,262144,78125
%N Rectangular array: row n gives the denominators in the positive convolutory n-th root of (1,1,1,...).
%C (See Comments at A255811.)
%H Clark Kimberling, <a href="/A255812/b255812.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,..., 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,..., convolutory 4th root.
%e Taking only denominators:
%e row 1: 1,1,1,1,1,1,1,...
%e row 2: 1,2,8,16,128,...
%e row 3: 1,3,9,81,243,729,...
%e row 4: 1,4,32,128,2048,8192,...
%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. A244811, A000012.
%K nonn,easy,tabl,frac
%O 1,5
%A _Clark Kimberling_, Mar 11 2015
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