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Search: id:A079053
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| A079053 |
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Recaman Fibonacci variation: a(1)=1; a(2)=2; for n > 2, a(n) = a(n-1)+a(n-2)-F(n) if that number is positive and not already in the sequence, otherwise a(n) = a(n-1)+a(n-2)+F(n) where F(n) denote the n-th Fibonacci number. |
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+0
7
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| 1, 2, 5, 4, 14, 10, 11, 42, 19, 6, 114, 264, 145, 32, 787, 1806, 996, 218, 5395, 12378, 6827, 1494, 36978, 84840, 46793, 10240, 253451, 581502, 320724, 70186, 1737179, 3985674, 2198275, 481062, 11906802, 27318216, 15067201, 3297248, 81610435
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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Starting with other initial values a(1)=x a(2)=y gives the same kind of recurrence relations.
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FORMULA
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For n>2, if n==0 or 2 (mod 4) a(n)=2*a(n-1)-a(n-2)-a(n-4); if n==1 or 3 (mod 4) a(n)=a(n-2)+2*a(n-3)+a(n-4) lim n ->infinity a(4n)/a(4n-1)=2.29433696806047607330083539....; lim n ->infinity a(4n-1)/a(4n-2)=24.7510757456062014116731647..; lim n ->infinity a(4n-2)/a(4n-3)=0.218836132868832627648170038...; lim n ->infinity a(4n-3)/a(4n-4)=0.551544105222898180785441647...
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EXAMPLE
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a(10)=6 because a(9)+a(8)-F(10)=19+42-55=6 and 6 is not already in the sequence. a(11)=42 because a(10)+a(9)-F(11)=6+19-89 < 0 then a(11)=6+19+89=114.
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PROGRAM
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(PARI) an=vector(100); an[1]=1; an[2]=2; a(n)=if(n<0, 0, an[n]); for(n=3, 100, an[n]=if(abs(sign(a(n-1)+a(n-2)-fibonacci(n))-1)+setsearch(Set(vector(n-1, i, a(i))), a(n-1)+a(n-2)-fibonacci(n)), a(n-1)+a(n-2)+fibonacci(n), a(n-1)+a(n-2)-fibonacci(n)))
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CROSSREFS
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Cf. A005132.
Sequence in context: A117824 A122212 A102468 this_sequence A002518 A093727 A065160
Adjacent sequences: A079050 A079051 A079052 this_sequence A079054 A079055 A079056
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KEYWORD
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nice,nonn
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AUTHOR
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Benoit Cloitre (benoit7848c(AT)orange.fr), Feb 02 2003
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