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.

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

Offset: 0

Author

Benoit Cloitre, May 05 2002

Keywords

• nonn
• easy

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 A000045(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

• 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).

Crossrefs

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

Programs

Magma

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

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 *)

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

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

Extensions

Edited by Dean Hickerson, May 08 2002