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A087793 Least k such that S^k(n)=n^2 where S(x)=n*ceiling(sqrt(x)). 0

%I #5 Mar 30 2012 18:39:21

%S 0,2,3,4,4,4,5,5,5,5,6,6,6,6,7,7,6,7,7,7,7,7,7,7,7,7,8,8,8,8,8,8,8,8,

%T 8,8,8,8,8,8,8,8,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,

%U 9,9,9,9,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10

%N Least k such that S^k(n)=n^2 where S(x)=n*ceiling(sqrt(x)).

%C For all m>k, S^m(n)=n^2

%F a(n)=sqrt(n)+o(sqrt(n))

%o (PARI) a(n)=if(n<0,0,z=1; c=0; while(abs(z-n^2)>0,z=n*ceil(sqrt(z)); c++); c)

%K nonn

%O 0,2

%A _Benoit Cloitre_, Oct 07 2003

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