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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.
6
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).
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
KEYWORD
easy,nonn
AUTHOR
Benoit Cloitre, Apr 28 2002
EXTENSIONS
Edited by Dean Hickerson, May 08 2002
STATUS
approved