0,2

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.

For n>0, a(2n)=2a(2n-1)+a(2n-2) and a(2n+1)=8a(2n)+a(2n-1).

The initial convergents are 1, 9/8, 19/17, 161/144,

341/305, 2889/2584, 6119/5473, 51841/46368,

109801/98209, 930249/832040, 1970299/1762289,

Cf. A000129, A001333, A142238-A142239, A153313-153318.

Sequence in context: A068174 A165247 A041677 this_sequence A041160 A089565 A001154

Adjacent sequences: A153313 A153314 A153315 this_sequence A153317 A153318 A153319

nonn

Charlie Marion (charliemath(AT)optonline.net), Jan 07 2009