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A119558 a(1)=0,a(2)=0,a(3)=1 then a(n)=abs(a(n-1)-a(n-2))+(-1)^n*a(n-3). 1

%I #3 Mar 30 2012 18:39:24

%S 0,0,1,1,0,2,1,1,-2,4,5,-1,2,8,7,3,-4,14,15,-3,4,22,21,5,-6,32,33,-5,

%T 6,44,43,7,-8,58,59,-7,8,74,73,9,-10,92,93,-9,10,112,111,11,-12,134,

%U 135,-11,12,158,157,13,-14,184,185,-13,14,212,211,15,-16,242,243,-15,16,274,273,17,-18,308,309,-17,18,344

%N a(1)=0,a(2)=0,a(3)=1 then a(n)=abs(a(n-1)-a(n-2))+(-1)^n*a(n-3).

%C See A104156 for an order 2 example.

%D B. Cloitre, On strange predictible recursions, preprint 2006

%F for n>=1 a(4n)=(-1)^n*(2*floor(n/2)-1), a(4*n+1)=(-1)^(n+1)*2*floor(n/2), a(4*n+2)=n^2-n+2, a(4*n+3)=n^2-n+2+(-1)^n

%K sign

%O 0,6

%A _Benoit Cloitre_, May 30 2006

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