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.

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

0,2

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).

Links

Colin Barker, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

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

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 *)

Prog

(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
(Magma) F:=Fibonacci; [F(n+1)^2 +4*F(n)^2: n in [0..40]]; // G. C. Greubel, Aug 17 2022
(SageMath) f=fibonacci; [f(n+1)^2+4*f(n)^2 for n in (0..40)] # G. C. Greubel, Aug 17 2022

Crossrefs

Cf. A000045, A069921, A069960, A069961, A069962, A069963.

Sequence in context: A286571 A234334 A025623 * A038250 A067973 A101584

Adjacent sequences: A069956 A069957 A069958 * A069960 A069961 A069962

Keyword

easy,nonn

Author

Benoit Cloitre, Apr 28 2002

Extensions

Edited by Dean Hickerson, May 08 2002

Status

approved