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Search: id:A069959
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| A069959 |
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Define C(n) by the recursion C(0)=2*I where I^2=-1, C(n+1)=1/(1+C(n)); then a(n)=2*(-1)^n/Im(C(n)) where Im(z) denotes the imaginary part of the complex number z. |
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+0
6
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| 1, 5, 8, 25, 61, 164, 425, 1117, 2920, 7649, 20021, 52420, 137233, 359285, 940616, 2462569, 6447085, 16878692, 44188985, 115688269, 302875816, 792939185, 2075941733, 5434886020, 14228716321, 37251262949, 97525072520
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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)+4*F(n-1))+(-1)^n*2*I)/(F(n+1)^2+4*F(n)^2).
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FORMULA
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a(n) = 4 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_] := 4Fibonacci[n]^2+Fibonacci[n+1]^2
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CROSSREFS
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Cf. A069921, A069960-A069963.
Sequence in context: A105963 A140113 A025623 this_sequence A038250 A067973 A101584
Adjacent sequences: A069956 A069957 A069958 this_sequence A069960 A069961 A069962
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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(AT)math.ucdavis.edu), May 08 2002
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