%I #25 Jun 13 2015 00:53:37

%S 1,0,-1,-2,-5,-12,-29,-70,-169,-408,-985,-2378,-5741,-13860,-33461,

%T -80782,-195025,-470832,-1136689,-2744210,-6625109,-15994428,

%U -38613965,-93222358,-225058681,-543339720,-1311738121,-3166815962,-7645370045

%N Expansion of 1+(1-2*x)/(-1+2*x+x^2).

%C It is essentially A000129, A077985 and A069306 except for signs and offsets.

%C a(n)=a(n - 1) - sqrt(2*a(n - 1)^2 + (-1)^n) = a(n-1)*(1-sqrt(2-(-1)^n/a(n-1)^2)) for n>0.

%C So in the limit of n->infinity, a(n+1)/a(n)= 1-Sqrt(2).

%C This matches the following formula in A000045:

%C Fibonacci(n) = Fibonacci(n - 1)*(1 + sqrt( 5+4*(-1)^(n-1)/Fibonacci(n-1)^2) )/2.

%H Vincenzo Librandi, <a href="/A176981/b176981.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec">Index entries for linear recurrences with constant coefficients</a>, signature (2,1).

%F a(0)=1, a(1)=0, a(2)=-1, a(n)=2*a(n-1)+a(n-2) [From Harvey P. Dale, Dec 24 2011]

%F G.f.: 1 / (1 + x^2 / (1 - 2*x / (1 - x / (1 + x)))). - _Michael Somos_, Jan 03 2013

%F G.f.: 1 - Q(0)*x^2/2 , where Q(k) = 1 + 1/(1 - x*(4*k+2 + x)/( x*(4*k+4 + x) + 1/Q(k+1) )); (continued fraction). - _Sergei N. Gladkovskii_, Sep 08 2013

%e 1 - x^2 - 2*x^3 - 5*x^4 - 12*x^5 - 29*x^6 - 70*x^7 - 169*x^8 - 408*x^9 - 985*x^10 + ...

%t a[0] = 1; a[n_] := a[n] = a[n - 1] - Sqrt[2*a[n - 1]^2 + (-1)^n]; Table[a[n], {n, 0, 30}]

%t Join[{1}, LinearRecurrence[{2, 1}, {0, -1}, 30]] (* or *) Join[{1}, Rest[ CoefficientList[Series[1 + (1 - 2 x)/(-1 + 2 x + x^2), {x, 0, 30}] ,x]]] (* _Harvey P. Dale_, Dec 24 2011 *)

%Y Cf. A000129.

%K sign,easy

%O 0,4

%A _Roger L. Bagula_, Apr 30 2010