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A255811
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Rectangular array: row n gives the numerators in the positive convolutory n-th root of (1,1,1,...).
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2
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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
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OFFSET
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1,8
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COMMENTS
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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?
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LINKS
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FORMULA
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G.f. of s: (1 - t)^(-1/n).
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EXAMPLE
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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,...
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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