%I #17 Mar 18 2023 18:35:23
%S 1,3,2,5,43,522,1104509,60248974744,2075863890266492169136,
%T 10942918579397694712648387271683911959312808,
%U 30436613005235318097155473477154291219175029919236526500330140104415890363628017565032
%N a(1) = 1 and, for n >= 2, a(n) is the least integer such that the numerator of the continued fraction [a(1),a(2),...,a(n)] is a perfect square.
%F a(n) = (A086541(n) - A086541(n-2)) / A086541(n-1) (n >= 3). - _Hiroaki Yamanouchi_, Oct 03 2014
%o (PARI) v=[1];for(k=1,6,m=1;while(issquare(contfracpnqn(concat(v,[m]))[1,1])==0,m++);v=concat(v,[m]));a(n)=if(n<2,1,v[n]);
%Y Cf. A086541
%K nonn
%O 1,2
%A _Benoit Cloitre_, Jan 15 2013
%E a(8)-a(11) from _Max Alekseyev_, Mar 18 2023