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Search: id:A069962
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| A069962 |
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Define C(n) by the recursion C(0)=5*I where I^2=-1, C(n+1)=1/(1+C(n)); then a(n)=5*(-1)^n/Im(C(n)) where Im(z) denotes the imaginary part of the complex number z. |
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3
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| 1, 26, 29, 109, 250, 689, 1769, 4666, 12181, 31925, 83546, 218761, 572689, 1499354, 3925325, 10276669, 26904634, 70437281, 184407161, 482784250, 1263945541, 3309052421, 8663211674, 22680582649, 59378536225, 155455026074
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OFFSET
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0,2
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COMMENT
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If we define C(n) with C(0)=I then Im(C(n))=1/F(2n+1) where F(k) are the Fibonacci numbers.
C(n) = (F(n)+F(n-1)*C(0))/(F(n+1)+F(n)*C(0)) = (F(n)*(F(n+1)+25*F(n-1))+(-1)^n*5*I)/(F(n+1)^2+25*F(n)^2).
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FORMULA
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a(n) = 25 F(n)^2 + F(n+1)^2, where F(n) = A000045(n) is the n-th Fibonacci number.
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MATHEMATICA
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a[n_] := 25Fibonacci[n]^2+Fibonacci[n+1]^2
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CROSSREFS
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Cf. A069921, A069959-A069961, A069963.
Sequence in context: A055109 A106552 A106550 this_sequence A045163 A046292 A134252
Adjacent sequences: A069959 A069960 A069961 this_sequence A069963 A069964 A069965
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KEYWORD
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easy,nonn
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AUTHOR
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Benoit Cloitre (benoit7848c(AT)orange.fr), Apr 28 2002
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EXTENSIONS
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Edited by Dean Hickerson (dean.hickerson(AT)yahoo.com), May 08 2002
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