cp's OEIS Frontend

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.

%I A069962 #29 Aug 18 2022 01:43:30
%S A069962 1,26,29,109,250,689,1769,4666,12181,31925,83546,218761,572689,
%T A069962 1499354,3925325,10276669,26904634,70437281,184407161,482784250,
%U A069962 1263945541,3309052421,8663211674,22680582649,59378536225,155455026074,406986541949,1065504599821
%N 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.
%C A069962 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.
%C A069962 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).
%H A069962 Colin Barker, <a href="/A069962/b069962.txt">Table of n, a(n) for n = 0..1000</a>
%H A069962 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-1).
%F A069962 a(n) = 25*F(n)^2 + F(n+1)^2, where F(n) = A000045(n).
%F A069962 From _Colin Barker_, Jun 14 2013: (Start)
%F A069962 a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
%F A069962 G.f.: (1-x)*(1+25*x) / ((1+x)*(1-3*x+x^2)). (End)
%F A069962 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
%t A069962 a[n_]:= 25*Fibonacci[n]^2+Fibonacci[n+1]^2; Table[a[n], {n,0,30}]
%t A069962 LinearRecurrence[{2,2,-1},{1,26,29},30] (* _Harvey P. Dale_, Mar 18 2018 *)
%o A069962 (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
%o A069962 (PARI) Vec(-(x-1)*(25*x+1)/((x+1)*(x^2-3*x+1)) + O(x^30)) \\ _Colin Barker_, Oct 01 2016
%o A069962 (Magma) F:=Fibonacci; [F(n+1)^2 + 25*F(n)^2: n in [0..40]]; // _G. C. Greubel_, Aug 17 2022
%o A069962 (SageMath) f=fibonacci; [f(n+1)^2 +25*f(n)^2 for n in (0..40)] # _G. C. Greubel_, Aug 17 2022
%Y A069962 Cf. A000045, A069921, A069959, A069960, A069961, A069963.
%K A069962 easy,nonn
%O A069962 0,2
%A A069962 _Benoit Cloitre_, Apr 28 2002
%E A069962 Edited by _Dean Hickerson_, May 08 2002