%I #13 Jan 23 2022 07:29:18
%S 1,2,3,41,58,106,193,337,586,949,1061,1117,1153,1249,1669,2381,3733,
%T 5857,6577,6781,8389,11173,14293,15817,17137,17209,23017,37921,38377,
%U 46261,47293,56929,82561,90121,113173,122401,148957,151057,161149,163729,193873,206209,225769,322513,497473,576529,676129,686893,706621,862921,946489,992281,1032649,1198081,1597033,1655677,1779409,1930021,2299489,2367481,2584081,3209281,3528409,3933073,4068241,4160521,4283689,4726009,4833901
%N Smallest m such that the period of the continued fraction of sqrt(m) is A215485(n).
%C The continued fractions of these numbers have the "hard to get" lengths listed in sequence A215485. They fill the last gaps in the table when computing A013646.
%H Patrick McKinley, <a href="/A215508/a215508.txt">Table of n, a(n) for n=0..253</a>
%e The lengths of the continued fractions of sqrt(1), sqrt(2), sqrt(3) and sqrt(41) are 0, 1, 2 and 3 respectively. The rest of the sequence follows A215485 similarly.
%Y Cf. A013646, A215485.
%K nonn
%O 0,2
%A _Patrick McKinley_, Aug 13 2012