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A107921
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Smallest positive integer m such that the length of continued fraction for 1/n + Sum_{k=1..m} 1/k is n, or 0 if no such m exists.
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0
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1, 1, 2, 5, 4, 7, 5, 0, 9, 11, 9, 14, 0, 18, 24, 0, 27, 18, 13, 26, 0, 26, 22, 0, 32, 31, 0, 40, 28, 40, 0, 40, 0, 0, 0, 0, 48, 48, 0, 45, 45, 51, 58, 54, 0, 54, 74, 0, 55, 54, 0, 53, 74, 0, 0, 65, 59, 72, 60, 61, 70, 65, 0, 76, 79, 0, 81, 96, 96, 77, 87, 0, 95, 0, 0, 83, 88, 0, 89, 98
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OFFSET
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1,3
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LINKS
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EXAMPLE
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a(4) = 5 because sum 1/4 + (1/1 +...+1/5)= 38/15 and continued fraction for 38/15 has terms {2,1,1,7} and length 4.
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MATHEMATICA
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f[n_] := Block[{k = 1}, While[ Length[ ContinuedFraction[1/n + HarmonicNumber[k]]] != n && k < 1000, k++ ]; If[k == 1000, 0, k]]; Table[ f[n], {n, 80}] (* Robert G. Wilson v, Jun 03 2005 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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