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COMMENTS
| sqrt(3/2) = 1.224744871... = 2/2 + 2/9 + 2/(9*89) + 2/(89*881) + 2/(881*8721) + 2/(8721*86329), + ... [From /Gary W. Adamson (qntmpkt(AT)yahoo.com), Oct 08 2008]
Contribution from Charlie Marion (charliemath(AT)optonline.net), Jan 07 2009: (Start)
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(3/2) start 1/1, 5/4, 11/9,
49/40, 109/89.
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=2 and n=3, then a(2,n)=a(n) and
2*a(2,6)^2-a(2,5)*a(2,7)=2*881^2-396*3920=2;
2*a(2,4)*a(2,6)-a(2,5)^2=2*89*881-396^2=2;
b(2,5)*b(2,7)-2*b(2,6)^2=485*4801-2*1079^2=3;
b(2,5)^2-2*b(2,4)*b(2,6)=485^2-2*109*1079=3.
Cf. A000129, A001333, A142238, A153313-153318.
[From Charlie Marion (charliemath(AT)optonline.net), Jan 07 2009]
(End)
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