It is left to the reader to confirm that we can start a Greek ladder for \(\sqrt{k}\) with any nonnegative integers \(x_1\) and \(y_1\) (as long as they are not both zero) and the sequence of approximations produced by the ladder will still converge to \(\sqrt{k}\). Thus, a natural question to ask is, "If we don't start the Greek ladder with initial values \(x_1=1\) and \(y_1=1\), can we still find a continued fraction with convergents that match the Greek ladder at every term?" We leave this question as an opportunity for further investigation by students who wish to engage in mathematical research on their own. We offer the following examples as enticement.
Example 1: If we start with \(x_1 = 1\) and \(y_1=2\), the ladder for \(\sqrt{k}\) looks like:


\(x_n\) 
\(y_n\) 
\({y_n}/{x_n}\) 
\(n=1\) 

\(1\) 
\(2\) 
\(2\) 
\(n=2\) 

\(3\) 
\(k+2\) 
\(\frac{k+2}{3}\) 
\(n=3\) 

\(k+5\) 
\(4k+2\) 
\(\frac{4k+2}{k+5}\) 
\(n=4\) 

\(5k+7\) 
\(k^2 +9k+2\) 
\(\frac{k^2 +9k+2}{5k+7}\) 
\(\vdots\) 

\(\vdots\) 
\(\vdots\) 
\(\vdots\) 
The continued fraction that matches it is
\[[2;k4,3;k1,2;k1,2;k1,2,\dots].\]
Example 2: If we start with \(x_1 = 2\) and \(y_1=3\), the ladder for \(\sqrt{k}\) looks like:


\(x_n\) 
\(y_n\) 
\({y_n}/{x_n}\) 
\(n=1\) 

\(2\) 
\(3\) 
\(\frac{3}{2}\) 
\(n=2\) 

\(5\) 
\(2k+3\) 
\(\frac{2k+3}{5}\) 
\(n=3\) 

\(2k+8\) 
\(7k+3\) 
\(\frac{7k+3}{2k+8}\) 
\(n=4\) 

\(9k+11\) 
\(2k^2 +15k+3\) 
\(\frac{2k^2 +15k+3}{9k+11}\) 
\(\vdots\) 

\(\vdots\) 
\(\vdots\) 
\(\vdots\) 
The continued fraction that matches it is
\[[3/2;4k9,10;2k2,1;k1,4;k1,1;k1,4;k1,1; \dots].\]
We say that a continued fraction of this type is eventually periodic of period 2.
We believe there are additional patterns and general statements that can be proved concerning the starting values of the ladder and its corresponding continued fraction.