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A069963 Define C(n) by the recursion C(0)=6*I where I^2=-1, C(n+1)=1/(1+C(n)); then a(n)=6*(-1)^n/Im(C(n)) where Im(z) denotes the imaginary part of the complex number z. 6
1, 37, 40, 153, 349, 964, 2473, 6525, 17032, 44641, 116821, 305892, 800785, 2096533, 5488744, 14369769, 37620493, 98491780, 257854777, 675072621, 1767363016, 4627016497, 12113686405, 31714042788, 83028441889, 217371282949, 569085406888, 1489884937785 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

C(n) = (F(n)+F(n-1)*C(0))/(F(n+1)+F(n)*C(0)) = (F(n)*(F(n+1)+36*F(n-1))+(-1)^n*6*I)/(F(n+1)^2+36*F(n)^2).

If we define C(n) with C(0)=I then Im(C(n))=1/F(2n+1) where F(k) are the Fibonacci numbers.

LINKS

Colin Barker, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

FORMULA

a(n) = 36 F(n)^2 + F(n+1)^2, where F(n) = A000045(n) is the n-th Fibonacci number.

a(n) = 2*a(n-1)+2*a(n-2)-a(n-3). G.f.: -(x-1) *(36*x+1) / ((x+1)*(x^2-3*x+1)). - Colin Barker, Jun 14 2013

a(n) = (2^(-1-n)*(-35*(-1)^n*2^(2+n) - (3-sqrt(5))^n*(-75+sqrt(5)) + (3+sqrt(5))^n*(75+sqrt(5))))/5. - Colin Barker, Sep 28 2016

MATHEMATICA

a[n_] := 36Fibonacci[n]^2+Fibonacci[n+1]^2

PROG

(PARI) a(n)=36*fibonacci(n)^2+fibonacci(n+1)^2 \\ Charles R Greathouse IV, Jun 14 2013

(PARI) a(n) = round((2^(-1-n)*(-35*(-1)^n*2^(2+n)-(3-sqrt(5))^n*(-75+sqrt(5))+(3+sqrt(5))^n*(75+sqrt(5))))/5) \\ Colin Barker, Sep 28 2016

(PARI) Vec(-(x-1)*(36*x+1)/((x+1)*(x^2-3*x+1)) + O(x^30)) \\ Colin Barker, Sep 28 2016

CROSSREFS

Cf. A069921, A069959, A069960, A069961, A069962.

Sequence in context: A295801 A111198 A081645 * A071855 A137675 A161725

Adjacent sequences:  A069960 A069961 A069962 * A069964 A069965 A069966

KEYWORD

easy,nonn

AUTHOR

Benoit Cloitre, Apr 28 2002

EXTENSIONS

Edited by Dean Hickerson, May 08 2002

STATUS

approved

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Last modified September 24 07:01 EDT 2020. Contains 337317 sequences. (Running on oeis4.)