
COMMENTS

In general, denominators, a(k,n) and numerators, b(k,n), of continued
fraction convergents to sqrt((k+1)/k) may be found as follows:
a(k,0) = 1, a(k,1) = 2k; for n>0, a(k,2n)=2*a(k,2n1)+a(k,2n2)
and a(k,2n+1)=(2k)*a(k,2n)+a(k,2n1);
b(k,0) = 1, b(k,1) = 2k+1; for n>0, b(k,2n)=2*b(k,2n1)+b(k,2n2)
and b(k,2n+1)=(2k)*b(k,2n)+b(k,2n1).
For example, the convergents to sqrt(4/3) start 1/1, 11/10, 23/21,
241/220, 505/461.
In general, if a(k,n) and b(k,n) are the denominators and numerators,
respectively, of continued fraction convergents to sqrt((k+1)/k)
as defined above, then
k*a(k,2n)^2a(k,2n1)*a(k,2n+1)=k=k*a(k,2n2)*a(k,2n)a(k,2n1)^2 and
b(k,2n1)*b(k,2n+1)k*b(k,2n)^2=k+1=b(k,2n1)^2k*b(k,2n2)*b(k,2n);
for example, if k=5 and n=3, then b(5,n)=a(n) and
5*a(5,6)^2a(5,5)*a(5,7)=5*10121^24830*106040=5;
5*a(5,4)*a(5,6)a(5,5)^2=5*461*101214830^2=5;
b(5,5)*b(5,7)5*b(5,6)^2=5291*1161615*11087^2=6;
b(5,5)^25*b(5,4)*b(5,6)=5291^25*505*11087=6.
