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A069960
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Define C(n) by the recursion C(0) = 3*i where i^2 = -1, C(n+1) = 1/(1 + C(n)), then a(n) = 3*(-1)^n/Im(C(n)) where Im(z) denotes the imaginary part of the complex number z.
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5
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1, 10, 13, 45, 106, 289, 745, 1962, 5125, 13429, 35146, 92025, 240913, 630730, 1651261, 4323069, 11317930, 29630737, 77574265, 203092074, 531701941, 1392013765, 3644339338, 9541004265, 24978673441, 65395016074, 171206374765, 448224108237, 1173465949930
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OFFSET
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0,2
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COMMENTS
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If we define C(n) with C(0) = i then Im(C(n)) = 1/F(2*n+1) where F(k) are the Fibonacci numbers.
Here, C(n) = (F(n) + C(0)*F(n-1))/(F(n+1) + C(0)*F(n)) = (F(n)*(F(n+1) + 9*F(n-1)) + 3*i*(-1)^n)/(F(n+1)^2 + 9*F(n)^2), where F(n) = Fibonacci(n), for which Im(C(n)) = 3*(-1)^n/(F(n+1)^2 + 9*F(n)^2).
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LINKS
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FORMULA
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a(n) = 9*F(n)^2 + F(n+1)^2, where F(n) = A000045(n) is the n-th Fibonacci number.
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
G.f.: (1-x)*(1+9*x) / ((1+x)*(1-3*x+x^2)). (End)
a(n) = (2^(-1-n)*(-(-1)^n*2^(5+n) - (3-sqrt(5))^n*(-21+sqrt(5)) + (3+sqrt(5))^n*(21+sqrt(5))))/5. - Colin Barker, Sep 30 2016
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MATHEMATICA
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a[n_] := 9Fibonacci[n]^2+Fibonacci[n+1]^2
9*First[#]+Last[#]&/@(Partition[Fibonacci[Range[0, 30]], 2, 1]^2) (* Harvey P. Dale, Mar 06 2012 *)
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PROG
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(PARI) a(n) = round((2^(-1-n)*(-(-1)^n*2^(5+n)-(3-sqrt(5))^n*(-21+sqrt(5))+(3+sqrt(5))^n*(21+sqrt(5))))/5) \\ Colin Barker, Sep 30 2016
(PARI) Vec(-(x-1)*(9*x+1)/((x+1)*(x^2-3*x+1)) + O(x^30)) \\ Colin Barker, Sep 30 2016
(Magma) F:=Fibonacci; [F(n+1)^2 +9*F(n)^2: n in [0..40]]; // G. C. Greubel, Aug 17 2022
(SageMath) f=fibonacci; [f(n+1)^2 +9*f(n)^2 for n in (0..40)] # G. C. Greubel, Aug 17 2022
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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