%I #35 Nov 25 2015 04:43:07
%S 1,2,3,4,7,8,15,16,31,32,63,64,127,128,255,256,511,512,1023,1024,2047,
%T 2048,4095,4096,8191,8192,16383,16384,32767,32768,65535,65536,131071,
%U 131072,262143,262144,524287,524288,1048575,1048576,2097151,2097152
%N Expansion of (1 + 2x - 2x^3) / (1 - 3x^2 + 2x^4).
%C Union of A000079 and A000225 without 0 = 2^0 - 1. - _Reinhard Zumkeller_, Jan 18 2005
%C Let f(0)=1, f(1)=1, and f(n) = f(n - 1 - (1 + (-1)^n)/2) + f(n-2); then a(n-1) = f(n). - _John M. Campbell_, May 22 2011
%C The same sequence is obtained iteratively by setting a(0)=1, a(1)=2, and a(n) = a(n-1) + a(n-2) - (GCD(1+a(n-1), a(n-2))-1), showing a kind of affinity to Fibonacci numbers. - _Stanislav Sykora_, Oct 16 2015
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,3,0,-2).
%F G.f.: 2*(1+x)/(1-2*x^2)-1/(1-x^2);
%F a(n) = 3*a(n-2) - 2*a(n-4);
%F a(n) = (1+sqrt(2)/2)*(sqrt(2))^n + (1/2-sqrt(2)/2)*(-sqrt(2))^n - (1+(-1)^n)/2;
%F a(n) = Sum_{k=0..n} binomial(floor(n/2), floor(k/2)).
%F a(n) = 2^floor((n+2)/2) - 1 + (n mod 2). - _Reinhard Zumkeller_, Jan 18 2005
%p seq(op([2^n-1,2^n]),n=1..100); # _Robert Israel_, Oct 16 2015
%t t={1}; Do[AppendTo[t,t[[-1]]+1]; AppendTo[t,t[[-1]]+t[[-2]]],{n,10}]; t (* _Vladimir Joseph Stephan Orlovsky_, Jan 27 2012 *)
%t CoefficientList[Series[(1 + 2*x - 2*x^3)/(1 - 3*x^2 + 2*x^4), {x, 0, 40}], x] (* _T. D. Noe_, Jan 27 2012 *)
%o (PARI) a=vector(1000);a[0]=1;a[1]=2;for(n=3,#a,a[n]=a[n-1]+a[n-2]+1-gcd(1+a[n-1],a[n-2])) \\ _Stanislav Sykora_, Oct 16 2015
%o (PARI) Vec(2*(1+x)/(1-2*x^2)-1/(1-x^2) + O(x^100)) \\ _Altug Alkan_, Oct 16 2015
%Y Cf. A000045, A000079, A000225.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Jul 25 2004, corrected Sep 05 2006
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