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COMMENTS
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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,2n-1)+a(k,2n-2)
and a(k,2n+1)=(2k)*a(k,2n)+a(k,2n-1);
b(k,0) = 1, b(k,1) = 2k+1; for n>0, b(k,2n)=2*b(k,2n-1)+b(k,2n-2)
and b(k,2n+1)=(2k)*b(k,2n)+b(k,2n-1).
For example, the convergents to sqrt(4/3) start 1/1, 9/8, 19/17,
161/144, 341/305.
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)^2-a(k,2n-1)*a(k,2n+1)=k=k*a(k,2n-2)*a(k,2n)-a(k,2n-1)^2 and
b(k,2n-1)*b(k,2n+1)-k*b(k,2n)^2=k+1=b(k,2n-1)^2-k*b(k,2n-2)*b(k,2n);
for example, if k=4 and n=3, then b(4,n)=a(n) and
4*a(4,6)^2-a(4,5)*a(4,7)=4*5473^2-2584*46368=4;
4*a(4,4)*a(4,6)-a(4,5)^2=4*305*5473-2584^2=4;
b(4,5)*b(4,7)-4*b(4,6)^2=2889*51841-4*6119^2=5;
b(4,5)^2-4*b(4,4)*b(4,6)=2889^2-4*341*6119=5.
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