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 A069962 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. 2
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS 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). LINKS FORMULA a(n) = 25 F(n)^2 + F(n+1)^2, where F(n) = A000045(n) is the n-th Fibonacci number. MATHEMATICA a[n_] := 25Fibonacci[n]^2+Fibonacci[n+1]^2 CROSSREFS Cf. A069921, A069959-A069961, A069963. Sequence in context: A055109 A106552 A106550 * A178098 A045163 A046292 Adjacent sequences:  A069959 A069960 A069961 * A069963 A069964 A069965 KEYWORD easy,nonn AUTHOR Benoit Cloitre, Apr 28 2002 EXTENSIONS Edited by Dean Hickerson (dean.hickerson(AT)yahoo.com), May 08 2002 STATUS approved

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