cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A069959 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.

Original entry on oeis.org

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, 255323954617, 668446791325, 1750016419364
Offset: 0

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Author

Benoit Cloitre, Apr 28 2002

Keywords

Comments

If we define C(n) with C(0) = i in the recurrence C(n+1) = 1/(1 + C(n)) then Im(C(n)) = 1/Fibonacci(2*n+1).
Here, C(n) is defined with C(0) = 2*i in C(n) = (F(n) + C(0)*F(n-1))/(F(n+1) + C(0)*F(n)) = (F(n)*(F(n+1) + 4*F(n-1)) + (-1)^n*2*i)/(F(n+1)^2 + 4*F(n)^2), where F(n) = Fibonacci(n), for which Im(C(n)) = 2*(-1)^n/(F(n+1)^2 + 4*F(n)^2).

Crossrefs

Programs

  • Magma
    F:=Fibonacci; [F(n+1)^2 +4*F(n)^2: n in [0..40]]; // G. C. Greubel, Aug 17 2022
    
  • Mathematica
    a[n_]:= 4Fibonacci[n]^2+Fibonacci[n+1]^2;
    4#[[1]]^2+#[[2]]^2&/@Partition[Fibonacci[Range[0,30]],2,1] (* or *) LinearRecurrence[{2,2,-1},{1,5,8},30] (* Harvey P. Dale, Aug 25 2017 *)
  • PARI
    a(n) = round((2^(-1-n)*(-3*(-1)^n*2^(2+n)-(3-sqrt(5))^n*(-11+sqrt(5))+(3+sqrt(5))^n*(11+sqrt(5))))/5) \\ Colin Barker, Sep 28 2016
    
  • PARI
    Vec(-(x-1)*(4*x+1)/((x+1)*(x^2-3*x+1)) + O(x^40)) \\ Colin Barker, Sep 28 2016
    
  • SageMath
    f=fibonacci; [f(n+1)^2+4*f(n)^2 for n in (0..40)] # G. C. Greubel, Aug 17 2022

Formula

a(n) = 4*F(n)^2 + F(n+1)^2, where F(n) = A000045(n) is the n-th Fibonacci number.
From Colin Barker, Jun 14 2013: (Start)
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
G.f.: (1-x)*(1+4*x) / ((1+x)*(1-3*x+x^2)). (End)
a(n) = (2^(-1-n)*(-3*(-1)^n*2^(2+n) - (3-sqrt(5))^n*(-11+sqrt(5)) + (3+sqrt(5))^n*(11+sqrt(5))))/5. - Colin Barker, Sep 28 2016

Extensions

Edited by Dean Hickerson, May 08 2002