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A069961
Define C(n) by the recursion C(0) = 4*i where i^2 = -1, C(n+1) = 1/(1 + C(n)), then a(n) = 4*(-1)^n/Im(C(n)) where Im(z) denotes the imaginary part of z.
5
1, 17, 20, 73, 169, 464, 1193, 3145, 8212, 21521, 56321, 147472, 386065, 1010753, 2646164, 6927769, 18137113, 47483600, 124313657, 325457401, 852058516, 2230718177, 5840095985, 15289569808, 40028613409, 104796270449, 274360197908, 718284323305, 1880492771977
OFFSET
0,2
COMMENTS
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) + 16*F(n-1)) + (-1)^n*4*i)/(F(n+1)^2 + 16*F(n)^2), where F(n) = Fibonacci(n), for which Im(C(n)) = 4*(-1)^n/(F(n+1)^2 + 16*F(n)^2).
FORMULA
a(n) = 16*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+16*x) / ((1+x)*(1-3*x+x^2)). (End)
a(n) = (2^(-1-n)*(-15*(-1)^n*2^(2+n) - (3-sqrt(5))^n*(-35+sqrt(5)) + (3+sqrt(5))^n*(35+sqrt(5))))/5. - Colin Barker, Oct 01 2016
MATHEMATICA
a[n_]:= 16Fibonacci[n]^2+Fibonacci[n+1]^2; Array[a, 30, 0]
16First[#]^2+Last[#]^2&/@Partition[Fibonacci[Range[0, 30]], 2, 1] (* Harvey P. Dale, Nov 08 2011 *)
PROG
(PARI) a(n) = round((2^(-1-n)*(-15*(-1)^n*2^(2+n)-(3-sqrt(5))^n*(-35+sqrt(5))+(3+sqrt(5))^n*(35+sqrt(5))))/5) \\ Colin Barker, Oct 01 2016
(PARI) Vec(-(x-1)*(16*x+1)/((x+1)*(x^2-3*x+1)) + O(x^30)) \\ Colin Barker, Oct 01 2016
(Magma) F:=Fibonacci; [F(n+1)^2 + 16*F(n)^2: n in [0..40]]; // G. C. Greubel, Aug 17 2022
(SageMath) f=fibonacci; [f(n+1)^2 +16*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