|
%I #14 Jul 13 2021 06:23:58
%S 8,125,216,1728
%N Numbers k such that (period of the continued fraction for sqrt(k))^3 = k.
%C Next term, if it exists, exceeds 6600000. - Sam Handler (sam_5_5_5_0(AT)yahoo.com), Nov 14 2004
%t Do[ If[ Length[ Last[ ContinuedFraction[ Sqrt[n]]]]^3 == n, Print[n]], {n, 1, 10^5} ]
%o (PARI) f(n) = {if (issquare(n), 0, my(p=100); while (! isokq(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(m) = f(m)^3 == m; \\ _Michel Marcus_, Jul 12 2021
%Y Cf. A003285.
%K nonn,more
%O 1,1
%A _Naohiro Nomoto_, Nov 09 2001
|
