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Search: id:A105774
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| A105774 |
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A "fractal" transform of the Fibonacci's numbers : a(1)=1 then if F(n)<k<=F(n+1) a(k)=F(n+1)-a(k-F(n)) where F(n)=A000045(n). |
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4
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| 1, 1, 2, 4, 4, 7, 7, 6, 12, 12, 11, 9, 9, 20, 20, 19, 17, 17, 14, 14, 15, 33, 33, 32, 30, 30, 27, 27, 28, 22, 22, 23, 25, 25, 54, 54, 53, 51, 51, 48, 48, 49, 43, 43, 44, 46, 46, 35, 35, 36, 38, 38, 41, 41, 40, 88, 88, 87, 85, 85, 82, 82, 83, 77, 77, 78, 80, 80, 69, 69, 70, 72, 72
(list; graph; listen)
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OFFSET
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1,3
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COMMENT
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Let tau=(1+sqrt(5))/2 then the missing numbers 3,5,8,10,13,16,18,21,... are given by (round(tau^2*k))_{k>0} (A004937 ). Indices n such that a(n)=a(n+1) are given by (floor(tau^2*k)-1)_{k>0} (A003622). n such that a(n) differs from a(n+1) are given by (floor(tau*k+1/tau))_{k>0} (A022342). Indices n giving isolated terms (a(n) differs from a(n-1) and a(n+1)) are given by (floor(tau*floor(tau^2*k))_{k>0} (A003623). Remove 0's form the first differences of sorted values then you get a version of the infinite Fibonacci's word (A001468). I.e. sorted values are 1,1,2,4,4,6,7,7,9,9,11,12,12,..., first differences are 0,1,2,0,2,1,0,2,0,2,1,0,2,0,1,..., remove 0's gives 1,2,2,1,2,2,1,2,1,2,2,1,2,1,2,2,1,2,...#{ k : a(k)=k}=infty
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FORMULA
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a(A000045(n))=A006498(n); limsup a(n)/n=tau and liminf a(n)/n=1/tau where tau=(1+sqrt(5))/2
a(n) mod 2 = A085002(n) - Benoit Cloitre (benoit7848c(AT)orange.fr), May 10 2005
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EXAMPLE
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For 1=F(2)<k<=F(3)=2 the rule gives a(2)=2-a(1)=1 ... if 5=F(5)<k<=F(6)=8 the rule forces a(6)=8-a(6-5)=8-a(1)=7; a(7)=8-a(2)=7; a(8)=8-a(3)=6
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CROSSREFS
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Cf. A105669, A105670, A105672, A093347, A093348.
Sequence in context: A082515 A062855 A103622 this_sequence A130805 A023831 A023844
Adjacent sequences: A105771 A105772 A105773 this_sequence A105775 A105776 A105777
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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 04 2005
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