A078069 - OEIS

%I #37 Apr 20 2019 03:34:10

%S 1,-3,4,-2,-4,12,-16,8,16,-48,64,-32,-64,192,-256,128,256,-768,1024,

%T -512,-1024,3072,-4096,2048,4096,-12288,16384,-8192,-16384,49152,

%U -65536,32768,65536,-196608,262144,-131072,-262144,786432,-1048576,524288,1048576,-3145728,4194304,-2097152,-4194304

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

%C Pisano period lengths: 1, 1, 8, 1, 4, 8, 24, 1, 24, 4, 40, 8, 12, 24, 8, 1, 16, 24, 72, 4,... - _R. J. Mathar_, Aug 10 2012

%H Vincenzo Librandi, <a href="/A078069/b078069.txt">Table of n, a(n) for n = 0..1000</a>

%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>

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

%F a(n) = (-2)*(a(n-1)+a(n-2)), n>1 ; a(0)=1, a(1)=-3. - _Philippe Deléham_, Nov 19 2008

%F a(n) = A108520(n)-A108520(n-1). - _R. J. Mathar_, Aug 11 2012

%F G.f.: G(0)*(1 - x)/(2*(1 + x)), where G(k)= 1 + 1/(1 - x*(k+1)/(x*(k+2) - 1/G(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, Jun 20 2013

%F a(n) = (1/2 - I)*(-1 - I)^n + (1/2 + I)*(-1 + I)^n, n>=0. _Taras Goy_, Apr 20 2019

%t CoefficientList[Series[(1-x)/(1+2x+2x^2),{x,0,50}],x] (* or *) LinearRecurrence[{-2,-2},{1,-3},50] (* _Harvey P. Dale_, Jan 19 2012 *)

%o (PARI) Vec((1-x)/(1+2*x+2*x^2)+O(x^99)) \\ _Charles R Greathouse IV_, Sep 25 2012

%Y Cf. A090131.

%K sign,easy

%O 0,2

%A _N. J. A. Sloane_, Nov 17 2002