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Search: id:A026355
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(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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Let f(1)=1, f(2)=q, and f(k+2) = f(k+1)+f(k)-n; a(n) is the smallest positive integer q such that f(k) -> infinity as k -> infinity. - Benoit Cloitre (benoit7848c(AT)orange.fr), Aug 04 2002
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FORMULA
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For n>0, a(n) = floor((n-1)*phi) + 2, where phi=(1+sqrt(5))/2.
Recurrences: a(n+1) = a(n)+(3 + sign(phi*n-a(n)))/2 for n>=0. Also a(n+1) = a(n) + 1 + A005614(n-2) for n>=2. - Benoit Cloitre (benoit7848c(AT)orange.fr), Aug 04 2002
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CROSSREFS
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Cf. A000201, A005614, A026351. Different from A007067.
Sequence in context: A138390 A047448 A029921 this_sequence A099267 A007067 A092979
Adjacent sequences: A026352 A026353 A026354 this_sequence A026356 A026357 A026358
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KEYWORD
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nonn,easy
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AUTHOR
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Clark Kimberling (ck6(AT)evansville.edu)
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