[seqfan] Re: CF for Constant A071873

[Neil Sloane]

Paul,  I hope you will submit that remarkable continued fraction for
A071873 as a new sequence!

Best regards
Neil

Neil J. A. Sloane, President, OEIS Foundation.
11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
Phone: 732 828 6098; home page: http://NeilSloane.com
Email: njasloane at gmail.com


On Sun, May 10, 2015 at 5:51 AM, C Boyd <cb1 at gmx.co.uk> wrote:

> > contfrac(x) = [0, 1, 1, 48, 1, 9182736455463727, 4, 1, 2, 1, 3, 1,
> > 16413861141941053151166388889231063606316227031696978138434,
> > 9, 9, 2, 1, 3, 8, 1, 1, 19, 1, 1, 2, 3, 1, 7, 1, 1, 4, 1, 1, 1,
> > 3, 3, 1, 1, 2, 2, 2, 1, 5, 2, 1, 1, 1, 1, 5, 1, 78, 1, 21, 1, 1, 5, 3, 2]
> >
> > but 50,000 digits of x will not yield more partial quotients!
>
> The convergent implied by the CF (up to the last known term) has an
> interesting denominator: in Pari,
>
>
> contfrac(56116722783389450057351290684624017957363636363636363636363759820426487093153761054994388327721660/((10^99-1)/9))
> %1 = [0, 1, 1, 48, 1, 9182736455463727, 4, 1, 2, 1, 3, 1,
> 16413861141941053151166388889231063606316227031696978138434, 9, 9, 2, 1, 3,
> 8, 1, 1, 19, 1, 1, 2, 3, 1, 7, 1, 1, 4, 1, 1, 1, 3, 3, 1, 1, 2, 2, 2, 1, 5,
> 2, 1, 1, 1, 1, 5, 1, 78, 1, 21, 1, 1, 5, 3, 2]
>
> The numerator is less interesting (apart from the run of 36's):
>
>
> factor(56116722783389450057351290684624017957363636363636363636363759820426487093153761054994388327721660)
> %2 =
> [2 2]
>
> [5 1]
>
> [2805836139169472502867564534231200897868181818181818181818187991021324354657688052749719416386083
> 1]
>
> The corresponding convergents implied by the CF up to the terms before the
> known massive terms, 50/99 and 41322314049586776860/81818181818181818181,
> also have suggestive denominators.
>
> Best wishes,
>
> CB
>
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