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%I #9 Nov 16 2023 19:04:44
%S 3,6,8,11,12,13,15,18,20,24,27,29,30,35,38,39,40,41,42,48,51,53,56,58,
%T 61,63,66,68,72,73,74,80,83,84,85,87,89,90,97,99,102,104,105,110,120,
%U 123,125,130,132,143,146,147,148,150,152,156,157,168,171,173,182,185,193,195,198,200
%N Numbers k such that the period of the continued fraction for sqrt(k) is a prime.
%e 13 is a term because the continued fraction for sqrt(13) is (3;1,1,1,1,6,1,1,1,1,6,...), whose periodic part is (1,1,1,1,6); its length (the period) is 5 (a prime).
%t Select[Range@200,PrimeQ@Length@Last@ContinuedFraction[Sqrt[#]]&]
%o (PARI) isokf(n, p) = {localprec(p); my(cf = contfrac(sqrt(n))); setsearch(Set(cf), 2*cf[1]); }
%o f(n) = {if (issquare(n), 0, my(p=100); while (! isokf(n, p), p+=100); localprec(p); my(cf = contfrac(sqrt(n))); for (k=2, #cf, if (cf[k] == 2*cf[1], return (k-1))); ); } \\ A003285
%o isok(k) = isprime(f(k)); \\ _Michel Marcus_, Jan 05 2022
%Y Cf. A003285, A206586, A003814.
%K nonn
%O 1,1
%A _Giorgos Kalogeropoulos_, Jan 04 2022
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