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%I #15 Nov 11 2019 00:43:49
%S 0,0,1,1,0,0,-1,1,2,2,-1,1,0,2,1,1,-2,2,3,3,-2,2,1,3,0,2,-1,3,2,2,-3,
%T 3,4,4,-3,3,2,4,-1,3,0,4,1,3,-2,4,3,3,-4,4,5,5,-4,4,3,5,-2,4,1,5,0,4,
%U -1,5,2,4,-3,5,4,4,-5,5,6,6,-5,5,4,6,-3,5,2,6,-1,5,0,6,1,5,-2,6,3,5,-4,6,5,5,-6,6,7,7
%N a(1)=0,a(2)=0,a(3)=1 then a(n)=abs(a(n-1)-a(n-2))-a(n-3).
%C See A104156 for an order 2 example.
%C b(n) = a(2n)-floor(sqrt(n))+1 is an infinite binary word consisting of a sequence of block (0,1) and single 0's where 0's occur when n is of form k^2-1, k>=2 i.e. b(n) begins for n>=1 : (0,1),0,(0,1),(0,1),0,(0,1),(0,1),(0,1),0,(0,1),... and single 0's occur at n=3,8,15,...
%D B. Cloitre, On strange predictible recursions, preprint 2006
%F abs(a(2n-1)) = A004738(n)-1 where sign(a(2*n-1)) alternates between 2 consecutive zeros.
%o (PARI) an=vector(10000); an[1]=0; an[2]=0; an[3]=1; a(n)=if(n<0, 0, an[n]);
%o for(n=4, 10000, an[n]=abs(a(n-1)-a(n-2))-a(n-3))
%o an
%Y Cf. A104156.
%K sign
%O 0,9
%A _Benoit Cloitre_, May 30 2006
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