OEIS A229580 mini challenge (RPL)

[Gerson W. Barbosa]

(05-02-2018 02:59 PM)Valentin Albillo Wrote:  

(05-02-2018 12:14 PM)Gerson W. Barbosa Wrote:  I arrived at that OEIS sequence when testing a continued fraction with consecutive denominators 3, 20 and 112. Since a continued fraction is evaluated backwards, a recursive formula wouldn’t help.

Would you please elaborate on this ? What's the particular, complete continued fraction ? What makes you think a recursive formula wouldn't help ?

Well, I've noticed that

\(\frac{\pi ^{2}}{8} \approx 1+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\cdots +\frac{1}{\left ( 2n-1 \right )^{2}}+\frac{1}{4n+\frac{1}{3n+\frac{2^{4}}{20n+\frac{3^{4}}{112n}}}}\)

If I wanted to manually test the continuous fraction with the next few possible denominators, like 576, 2816, 13312... (A229580/2), then a recursive formula would surely have been of help, but if I decided to do it programmatically, with many more terms, then it would not. Anyway, it turns out that apparently the next term is not 256/576... but I had to check that out.

Thanks again for your interest!

Gerson.