OEIS A229580 mini challenge (RPL)

[Valentin Albillo]

Gerson W. Barbosa Wrote:However I think you were right in the first place, a recursive formula could have indeed been of help for my particular application. In the program below I have used a closed-form formula to compute the last denominator of the continued fraction, but I might as well have computed it using a recursive formula, every term being computed at each step of the first FOR-NEXT loop

A few questions if you don't mind:

What's the origin of the hybrid formula (series + continued fraction) you're evaluating ? Is it original research on your part ? Does it actually converge to Pi\(^2\)/8 and do the coefficients of the continued fraction part actually belong to the OEIS series (divided by 2) or it's just a tentative conjecture ?

I could check the convergence myself but if you already know I'd rather ask and avoid wasting time rediscovering your knowledge. Also, you begin evaluating the continued fraction part starting with the last denominator and working iteratively upwards to the first. I'd do it the opposite way, starting at the top and going downwards to the last denominator because this way it's easier to visualize the alleged convergence as each new denominator is considered.

Regards.
V.
.