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A349473
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Irregular triangle read by rows: the n-th row contains the elements in the continued fraction of the harmonic mean of the divisors of n.
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12
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1, 1, 3, 1, 2, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 3, 2, 7, 2, 2, 13, 2, 4, 2, 1, 1, 5, 2, 1, 1, 3, 1, 1, 6, 2, 3, 2, 2, 2, 1, 1, 2, 1, 1, 2, 1, 1, 8, 2, 1, 3, 3, 1, 1, 9, 2, 1, 6, 2, 1, 1, 1, 2, 2, 2, 4, 1, 1, 11, 3, 5, 2, 2, 2, 1, 1, 2, 2, 2, 10, 2, 1, 2, 3, 3, 1, 1, 14
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OFFSET
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1,3
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COMMENTS
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For an odd prime p > 3, the p-th row has a length 3 with a(p, 1) = a(p, 2) = 1 and a(p, 3) = (p-1)/2.
For a harmonic number m = A001599(k), the m-th row has a length 1 with a(k, 1) = A099377(m) = A001600(k).
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LINKS
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EXAMPLE
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The first ten rows of the triangle are:
1,
1, 3,
1, 2,
1, 1, 2, 2,
1, 1, 2,
2,
1, 1, 3,
2, 7, 2,
2, 13,
2, 4, 2
...
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MATHEMATICA
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row[n_] := ContinuedFraction[DivisorSigma[0, n] / DivisorSigma[-1, n]]; Table[row[k], {k, 1, 29}] // Flatten
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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