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Search: id:A069961
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| A069961 |
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Define C(n) by the recursion C(0)=4*I where I^2=-1, C(n+1)=1/(1+C(n)); then a(n)=4*(-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, 17, 20, 73, 169, 464, 1193, 3145, 8212, 21521, 56321, 147472, 386065, 1010753, 2646164, 6927769, 18137113, 47483600, 124313657, 325457401, 852058516, 2230718177, 5840095985, 15289569808, 40028613409, 104796270449
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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)+16*F(n-1))+(-1)^n*4*I)/(F(n+1)^2+16*F(n)^2).
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FORMULA
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a(n) = 16 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_] := 16Fibonacci[n]^2+Fibonacci[n+1]^2
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CROSSREFS
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Cf. A069921, A069959, A069960, A069962, A069963.
Adjacent sequences: A069958 A069959 A069960 this_sequence A069962 A069963 A069964
Sequence in context: A116037 A081643 A045020 this_sequence A140146 A039502 A039505
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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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