%I #6 Mar 28 2022 21:35:06
%S 2,3,5,6,8,10,11,15,17,18,20,24,26,27,30,35,37,38,39,40,42,48,50,51,
%T 56,63,65,66,68,72,80,82,83,84,87,90
%N Integers k (not perfect squares) such that the continued fraction expansion of the square root of k has period at most 2.
%C In a recent paper, Justin Thomas, Julian Rosen, and I show that this is equivalent to the following criterion: let d be the integer part of the square root. Then sqrt(k) has period at most 2 if and only if 2d/(k - d^2) is an integer.
%D Justin Thomas, Krishnan Shankar, Julian Rosen, "Continued Fractions, Square Roots and the orbit of 1/0 on the boundary of the hyperbolic plane", preprint.
%H K. Shankar, <a href="http://www.math.ou.edu/~shankar/pubs.html">Square roots and continued fractions</a>.
%H K. Shankar, <a href="http://www.math.ou.edu/~shankar/research/cfrac.pdf">SQUARE ROOTS, CONTINUED FRACTIONS AND THE ORBIT OF 1/0 ON dH2</a>
%e The first term is 2 because sqrt(2) is irrational and for k=2, d=1, 2d/(k - d^2) = 1 is an integer.
%K nonn
%O 1,1
%A Krishnan Shankar (shankar(AT)math.ou.edu), Apr 17 2006