Sequence & Its Numerators & Denominators

[Leroy Quet]

a[1] should be a POSITIVE rational to ensure no divide-by-zero problems.

As for someone computing these examples, I am only really interested in 
the a[1] = 1 example.



I wrote:

>I just submitted this to sci.math. I already sent the m=1 example to 
>seq.fan a little while back. Perhaps someone could calculate the 
>numerator/denominator sequences for the first couple of m's and post them 
>to the EIS.
>
>Thanks,
>Leroy Quet
>
>---------------
>
>
>This is a generalization of the sequence discussed at:
>http://mathforum.org/discuss/sci.math/t/482979
>
>Given a fixed positive integer m;
>
>Let a[1] = any nonzero rational where its denominator is relatively
>prime to m.
>
>Let:
>                     1
>a[k+1] = a[k] + -------------  , for all positive integers k.
>                 m + 1 /a[k]
>
>In linear mode:
>
>a[k+1] = a[k] + 1/(m + 1 /a[k]) , for all positive integers k.
>
>If n[k] = numerator of a[k] in reduced form,
>
>and d[k] = denominator of a[k] in reduced form,
>
>then, for k = all positive integers,
>
>
>n[k+2] =
>
> n[k+1]*(m*n[k+1] + n[k+1]^2/(2n[k]^2) -m^2 *n[k]^2/2),
> 
>
>and d[k+2] = 
>
> d[k+1]*(d[k+1] + d[k+1]^2/d[k]^2 -d[k]^2)
>
>
>
>The above are from these recursions:
>
>n[k+1] = n[k] *(m*n[k] + 2*d[k]),
>
>d[k+1] = d[k] *(m*n[k] + d[k])
>
>
>Noteworthy facts:
>
>*d[2] involves m. But the recursion, for d[k+2] in terms of d[k] and
>d[k+1], itself is independent of m.
>
>*Each d[k+1] is divisible by d[k], and each n[k+1] is divisible by
>n[k].
>
>*(d[k+1]/d[k]) is relatively prime to (n[j+1]/n[j]) for every positive
>integer k and every positive integer j.
>
>Thanks,
>Leroy Quet