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A069962
Define C(n) by the recursion C(0) = 5*i where i^2 = -1, C(n+1) = 1/(1 + C(n)), then a(n) = 5*(-1)^n/Im(C(n)) where Im(z) denotes the imaginary part of z.
5
1, 26, 29, 109, 250, 689, 1769, 4666, 12181, 31925, 83546, 218761, 572689, 1499354, 3925325, 10276669, 26904634, 70437281, 184407161, 482784250, 1263945541, 3309052421, 8663211674, 22680582649, 59378536225, 155455026074, 406986541949, 1065504599821
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) is defined with C(0) = 5*i in C(n) = (F(n) + C(0)*F(n-1))/(F(n+1) + C(0)*F(n)) = (F(n)*(F(n+1) + 25*F(n-1)) + (-1)^n*2*i)/(F(n+1)^2 + 25*F(n)^2), where F(n) = Fibonacci(n), for which Im(C(n)) = 2*(-1)^n/(F(n+1)^2 + 25*F(n)^2).
FORMULA
a(n) = 25*F(n)^2 + F(n+1)^2, where F(n) = A000045(n).
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+25*x) / ((1+x)*(1-3*x+x^2)). (End)
a(n) = (2^(-1-n)*(-3*(-1)^n*2^(5+n) - (3-sqrt(5))^n*(-53+sqrt(5)) + (3+sqrt(5))^n*(53+sqrt(5))))/5. - Colin Barker, Oct 01 2016
MATHEMATICA
a[n_]:= 25*Fibonacci[n]^2+Fibonacci[n+1]^2; Table[a[n], {n, 0, 30}]
LinearRecurrence[{2, 2, -1}, {1, 26, 29}, 30] (* Harvey P. Dale, Mar 18 2018 *)
PROG
(PARI) a(n) = round((2^(-1-n)*(-3*(-1)^n*2^(5+n)-(3-sqrt(5))^n*(-53+sqrt(5))+(3+sqrt(5))^n*(53+sqrt(5))))/5) \\ Colin Barker, Oct 01 2016
(PARI) Vec(-(x-1)*(25*x+1)/((x+1)*(x^2-3*x+1)) + O(x^30)) \\ Colin Barker, Oct 01 2016
(Magma) F:=Fibonacci; [F(n+1)^2 + 25*F(n)^2: n in [0..40]]; // G. C. Greubel, Aug 17 2022
(SageMath) f=fibonacci; [f(n+1)^2 +25*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