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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.

%I A069921 #73 Feb 16 2025 08:32:46
%S A069921 1,5,10,29,73,194,505,1325,3466,9077,23761,62210,162865,426389,
%T A069921 1116298,2922509,7651225,20031170,52442281,137295677,359444746,
%U A069921 941038565,2463670945,6449974274,16886251873,44208781349,115740092170,303011495165,793294393321
%N 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.
%C A069921 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).
%C A069921 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
%C A069921 Re(C(n)) = A014742(n)/a(n), n >= 0. For Im(C(n)) see the name. - _Wolfdieter Lang_, Jul 16 2014
%H A069921 Kival Ngaokrajang, <a href="/A069921/a069921_1.pdf">Illustration of initial terms</a>
%H A069921 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/DissectionFallacy.html">Dissection Fallacy</a>
%H A069921 Wikipedia, <a href="http://en.wikipedia.org/wiki/Missing_square_puzzle">Missing square puzzle</a>
%H A069921 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-1).
%F A069921 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.
%F A069921 a(n) = ceiling(3/5*(g/2)^(n+1))-(1+(-1)^n)/2, with g = 3 + sqrt(5).
%F A069921 a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3). - _Vladeta Jovovic_, May 06 2002
%F A069921 G.f.: (1+3*x-2*x^2)/((1+x)*(1-3*x+x^2)). - _Vladeta Jovovic_, May 06 2002
%F A069921 a(n) = F(n)^2 + F(n+2)^2. - _Ron Knott_, Aug 02 2004
%F A069921 a(n-1) = (A000045(n)^2 + A000032(n)^2)/2. - _Roger L. Bagula_, Nov 17 2008
%F A069921 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
%F A069921 Equals the logarithmic derivative of A224415. - _Paul D. Hanna_, Apr 05 2013
%F A069921 2*a(n) = Fibonacci(n+1)^2 + Lucas(n+1)^2. - _Bruno Berselli_, Sep 26 2017
%t A069921 a[n_] := 3Fibonacci[n]Fibonacci[n+2]+(-1)^n
%t A069921 (*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 *)
%t A069921 LinearRecurrence[{2, 2, -1}, {1, 5, 10}, 70] (* _Vladimir Joseph Stephan Orlovsky_, Feb 08 2012 *)
%o A069921 (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
%o A069921 (Magma) [3*Fibonacci(n)*Fibonacci(n+2)+(-1)^n: n in [0..40]]; // _Vincenzo Librandi_, Sep 24 2015
%Y A069921 Cf. A000032, A000045, A014742, A059929, A069959-A069963, A224415.
%K A069921 nonn,easy
%O A069921 0,2
%A A069921 _Benoit Cloitre_, May 05 2002
%E A069921 Edited by _Dean Hickerson_, May 08 2002