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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, 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)^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=5 and n=3, then b(5,n)=a(n) and
5*a(5,6)^2-a(5,5)*a(5,7)=5*10121^2-4830*106040=5;
5*a(5,4)*a(5,6)-a(5,5)^2=5*461*10121-4830^2=5;
b(5,5)*b(5,7)-5*b(5,6)^2=5291*116161-5*11087^2=6;
b(5,5)^2-5*b(5,4)*b(5,6)=5291^2-5*505*11087=6.
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