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A069921 Define C(n) by the recursion C(0) = 1 + I where I^2 = -1, C(n+1) = 1/(1+C(n)); then a(n) = (-1)^n/Im(C(n)) where Im(z) is the imaginary part of the complex number z. 9
1, 5, 10, 29, 73, 194, 505, 1325, 3466, 9077, 23761, 62210, 162865, 426389, 1116298, 2922509, 7651225, 20031170, 52442281, 137295677, 359444746, 941038565, 2463670945, 6449974274, 16886251873, 44208781349, 115740092170, 303011495165, 793294393321 (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)) = (3*F(n)*F(n+1) + (-1)^n*(1+I))/(3*F(n)*F(n+2) + (-1)^n).

a(n) = F(n+2)^2 + F(n)^2 is the square of the short sides L(n) of the parallelogram appearing in the dissection fallacy of the square F(n+3) X F(n+3), where F(n) is the Fibonacci number A000040(n). For n >= 0, floor(L(n)/h(n)) = A014742(n) (see the proof there), where h(n) is the perpendicular distance between the long sides LL(n) = L(n+1). a(n) is also the first difference of A014742(n). See the link with an illustration. - Kival Ngaokrajang, Jun 27 2014, edited by Wolfdieter Lang, Jul 16 2014

Re(C(n)) = A014742(n)/a(n), n >= 0. For Im(C(n)) see the name. - Wolfdieter Lang, Jul 16 2014

LINKS

Table of n, a(n) for n=0..28.

Kival Ngaokrajang, Illustration of initial terms

Eric Weisstein's World of Mathematics, Dissection Fallacy

Wikipedia, Missing square puzzle

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

FORMULA

a(n) = 3*F(n)*F(n+2) + (-1)^n = 3*A059929(n) +(-1)^n, where F(n) = A000045(n) is the n-th Fibonacci number.

a(n) = ceiling(3/5*(g/2)^(n+1))-(1+(-1)^n)/2, with g = 3 + sqrt(5).

a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3). - Vladeta Jovovic, May 06 2002

G.f.: (1+3*x-2*x^2)/((1+x)*(1-3*x+x^2)). - Vladeta Jovovic, May 06 2002

a(n) = F(n)^2 + F(n+2)^2. - Ron Knott, Aug 02 2004

a(n-1) = (A000045(n)^2 + A000032(n)^2)/2. - Roger L. Bagula, Nov 17 2008

a(n) = 2*F(n)*F(n+2) + F(n+1)^2 = F(n+1)*F(n+3) + F(n)^2 +(-1)^(n-1). - J. M. Bergot, Sep 15 2012

Equals the logarithmic derivative of A224415. - Paul D. Hanna, Apr 05 2013

2*a(n) = Fibonacci(n+1)^2 + Lucas(n+1)^2. - Bruno Berselli, Sep 26 2017

MATHEMATICA

a[n_] := 3Fibonacci[n]Fibonacci[n+2]+(-1)^n

(*A000045*) F[n_] := (((1 + Sqrt[5])/2)^n - ((1 - Sqrt[5])/2)^n)/Sqrt[5]; (*A000032*) L[n_] := ((1 + Sqrt[5])/2)^n + ((1 - Sqrt[5])/2)^n; Table[FullSimplify[ExpandAll[(F[n]^2 + L[n]^2)/2]], {n, 0, 50}] (* Roger L. Bagula, Nov 17 2008 *)

LinearRecurrence[{2, 2, -1}, {1, 5, 10}, 70] (* Vladimir Joseph Stephan Orlovsky, Feb 08 2012 *)

PROG

(PARI) a(n)=([0, 1, 0; 0, 0, 1; -1, 2, 2]^n*[1; 5; 10])[1, 1] \\ Charles R Greathouse IV, Sep 23 2015

(MAGMA) [3*Fibonacci(n)*Fibonacci(n+2)+(-1)^n: n in [0..40]]; // Vincenzo Librandi, Sep 24 2015

CROSSREFS

Cf. A000032, A000045, A014742, A059929, A069959-A069963, A224415.

Sequence in context: A093029 A105505 A005514 * A053818 A294286 A133629

Adjacent sequences:  A069918 A069919 A069920 * A069922 A069923 A069924

KEYWORD

nonn,easy

AUTHOR

Benoit Cloitre, May 05 2002

EXTENSIONS

Edited by Dean Hickerson, May 08 2002

STATUS

approved

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Last modified November 13 15:41 EST 2019. Contains 329106 sequences. (Running on oeis4.)