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Search: id:A138467
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| A138467 |
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a(1)=1, then for n>=2 a(n)=n-floor((1/3)*a(a(n-1))). |
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1
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| 1, 2, 3, 3, 4, 5, 6, 7, 7, 8, 9, 10, 11, 11, 12, 13, 14, 15, 15, 16, 17, 18, 18, 19, 20, 21, 22, 22, 23, 24, 25, 26, 26, 27, 28, 29, 30, 30, 31, 32, 33, 34, 34, 35, 36, 37, 37, 38, 39, 40, 41, 41, 42, 43, 44, 45, 45, 46, 47, 48, 49, 49, 50, 51, 52, 53, 53, 54, 55, 56, 56, 57, 58
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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for p in IN*, a(1)=1 ; a(n)=n-floor((1/p)*a(a(n-1))) - Yalcin Aktar (aktaryalcin(AT)msn.com), Jul 13 2008
I conjecture that a(n)=floor(r(p)*(n+1)) with r(p)=(1/2)*(sqrt(p*(p+4))-p) - Yalcin Aktar (aktaryalcin(AT)msn.com), Jul 13 2008
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REFERENCES
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B. Cloitre, On some generalisations of A005206, in preparation 2008
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FORMULA
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For n>=1, a(n)=floor(r*(n+1)) where r=(3/2)*(sqrt(7/3)-1)
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PROGRAM
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(PARI) a(n)=floor((3/2)*(sqrt(7/3)-1)*(n+1))
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CROSSREFS
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Cf. A005206, A135414.
Sequence in context: A083920 A066508 A053207 this_sequence A127036 A108789 A091960
Adjacent sequences: A138464 A138465 A138466 this_sequence A138468 A138469 A138470
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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), May 09 2008
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EXTENSIONS
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More terms from Yalcin Aktar (aktaryalcin(AT)msn.com), Jul 13 2008
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