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Search: id:A073545
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| A073545 |
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Least k such that 1/tau(k) + 1/tau(k+1) + 1/tau(k+2) + ... + 1/tau(k+n) is equal to 1 (where tau(k)=A000005(k) is the number of divisors of k). |
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+0
1
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| 1, 2, 6, 25, 54, 243, 1204, 3549, 19544, 81829, 104663, 663490, 743764, 7925355
(list; graph; listen)
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OFFSET
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0,2
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EXAMPLE
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a(2)=6 because 1/tau(6)+1/tau(7)+1/tau(8) = 1/4+1/2+1/4 = 1.
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MATHEMATICA
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a[n_] := For[k=1, True, k++, If[Sum[1/DivisorSigma[0, k+i], {i, 0, n}]==1, Return[k]]]
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CROSSREFS
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Sequence in context: A089718 A123150 A086591 this_sequence A103063 A030228 A066317
Adjacent sequences: A073542 A073543 A073544 this_sequence A073546 A073547 A073548
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KEYWORD
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nonn
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AUTHOR
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Benoit Cloitre (benoit7848c(AT)orange.fr), Aug 27 2002
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EXTENSIONS
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Edited by Dean Hickerson (dean(AT)math.ucdavis.edu), Sep 03 2002
2 more terms from Ryan Propper (rpropper(AT)stanford.edu), Sep 04 2005
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