A069960 Define C(n) by the recursion C(0) = 3*i where i^2 = -1, C(n+1) = 1/(1 + C(n)), then a(n) = 3*(-1)^n/Im(C(n)) where Im(z) denotes the imaginary part of the complex number z.

1, 10, 13, 45, 106, 289, 745, 1962, 5125, 13429, 35146, 92025, 240913, 630730, 1651261, 4323069, 11317930, 29630737, 77574265, 203092074, 531701941, 1392013765, 3644339338, 9541004265, 24978673441, 65395016074, 171206374765, 448224108237, 1173465949930

Offset: 0

Author

Benoit Cloitre, Apr 28 2002

Keywords

• easy
• nonn

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) + 9*F(n-1)) + 3*i*(-1)^n)/(F(n+1)^2 + 9*F(n)^2), where F(n) = Fibonacci(n), for which Im(C(n)) = 3*(-1)^n/(F(n+1)^2 + 9*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).

Crossrefs

Cf. A000045, A069921, A069959, A069961, A069962, A069963.

Programs

Magma

F:=Fibonacci; [F(n+1)^2 +9*F(n)^2: n in [0..40]]; // G. C. Greubel, Aug 17 2022

Mathematica

a[n_] := 9Fibonacci[n]^2+Fibonacci[n+1]^2
9*First[#]+Last[#]&/@(Partition[Fibonacci[Range[0,30]],2,1]^2) (* Harvey P. Dale, Mar 06 2012 *)

PARI

a(n) = round((2^(-1-n)*(-(-1)^n*2^(5+n)-(3-sqrt(5))^n*(-21+sqrt(5))+(3+sqrt(5))^n*(21+sqrt(5))))/5) \\ Colin Barker, Sep 30 2016

PARI

Vec(-(x-1)*(9*x+1)/((x+1)*(x^2-3*x+1)) + O(x^30)) \\ Colin Barker, Sep 30 2016

SageMath

f=fibonacci; [f(n+1)^2 +9*f(n)^2 for n in (0..40)] # G. C. Greubel, Aug 17 2022

Formula

a(n) = 9*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+9*x) / ((1+x)*(1-3*x+x^2)). (End)

a(n) = (2^(-1-n)*(-(-1)^n*2^(5+n) - (3-sqrt(5))^n*(-21+sqrt(5)) + (3+sqrt(5))^n*(21+sqrt(5))))/5. - Colin Barker, Sep 30 2016

Extensions

Edited by Dean Hickerson, May 08 2002