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%I #5 Mar 30 2012 18:39:22
%S 1,3,2,2,9,5,8,3,3,19,10,7,649,15,4,4,33,17,170,9,55,197,24,5,5,51,26,
%T 127,9801,11,1520,17,23,35,6,6,73,37,25,19,2049,13,3482,199,161,24335,
%U 48,7,7,99,50,649,66249,485,89,15,151,19603,530,31
%N Least integer x>0 such that x^2=ceiling(x*r*floor(x/r)) where r=sqrt(n).
%C For r=sqrt(n), the solutions to x^2=ceiling(x*r*floor(x/r)) appear to be given by the Chebyshev polynomial T(k,x) k>0 evaluated at x=a(n). These solutions also seem to be given by a sequence (b(k))_(k>=0) satisfying the recurrence b(k)= 2*A002350(n)*b(k-1)-b(k-2).
%o (PARI) a(n)=if(n<2,1,x=1;while(abs(x^2-ceil(sqrt(n)*x*floor(x/ sqrt(n))))>0,x++);x)
%K nonn
%O 1,2
%A _Benoit Cloitre_, Feb 14 2004
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