TITLE: how do I find the general term here?
QUESTION [7 upvotes]: I am getting crazy on this series! I found this in a handwritten old book without a reference. I could not figure out how it is built but the series numerically seems to converge to $\pi$.
\begin{align}
a_1&=\frac{16}{3}\\
a_2&=\frac{56}{15}\\
a_3&=\frac{362}{105}\\
a_4&=\frac{1051}{315}\\
a_5&=\frac{90913}{27720}\\
a_6&=\frac{2339483}{720720}\\
a_7&=\frac{9294869}{2882880}\\
a_8&=\frac{314539061}{98017920}\\
a_9&=\frac{95291361359}{29797447680}\\
a_{10}&=\frac{27155335099}{8513556480}\\
a_{11}&=\frac{2493237983453}{783247196160}\\
a_{12}&=\frac{24892232679053}{7832471961600}\\
a_{13}&=\frac{596632945162997}{187979327078400}\\
a_{14}&=\frac{34567420288501151}{10902800970547200}\\
a_{15}&=\frac{4282497882211187099}{1351947320347852800}\\
a_{16}&=\frac{8558465078579558323}{2703894640695705600}\\
...\\
a_{\infty}&=\pi
\end{align}
I have observed that the denominators include multiplication of odd numbers while $2^j$ is also always around. Sometimes the odd numbers appear in a row sometimes they are not in order. For the numerator I do not see much of a pattern!

REPLY [5 votes]: I can give a recursion which seems to fit at least the early values:
$$a_n = a_{n-1} - \dfrac{(2n-4)!}{(n-2)!\,(n-1)!\,(2n+1)\,2^{2n-7}}$$ with a suggestion of Catalan numbers or double factorials in there.
As far as I can tell this gives $a_{513} \approx 3.141722$ so it may well be converging on $\pi$ from above.