Notice
In this post, a continued fraction is represented as follows$$ a + \cfrac{1^2}{b+\cfrac{3^2}{b+\cfrac{5^2}{\ddots}}} = a +K^\infty_{k=1}\frac{(2k-1)^2}{b} $$
When I was checking the continued fractions of pi in wolframalpha, I noticed the following regularity.
\begin{eqnarray*} \frac{4}{\pi} &=& 1 + K^\infty_{k=1}\frac{(2k-1)^2}{2} \\ \pi &=& 3 + K^\infty_{k=1}\frac{(2k-1)^2}{6} \\ \frac{16}{\pi} &=& 5 + K^\infty_{k=1}\frac{(2k-1)^2}{10} \end{eqnarray*}
So I was wondering what the value of $ 7 + K^\infty_{k=1}\frac{(2k-1)^2}{14} $ would be. Some numerical investigation yielded the following speculative results.
The statement from here on out is unproven.
\begin{eqnarray*} \frac{9\pi}{4} = 7 + K^\infty_{k=1}\frac{(2k-1)^2}{14} \\ \frac{256}{9\pi} = 9 + K^\infty_{k=1}\frac{(2k-1)^2}{18} \\ \frac{225\pi}{64} = 11 + K^\infty_{k=1}\frac{(2k-1)^2}{22} \\ \frac{1024}{25\pi} = 13 + K^\infty_{k=1}\frac{(2k-1)^2}{26} \\ \frac{1225\pi}{256} = 15 + K^\infty_{k=1}\frac{(2k-1)^2}{30} \\ \end{eqnarray*}
The numbers appearing on LHS are all square numbers, indicating that there is some regularity. If you know of any proof of this or related information, please let me know. Thank you.
All these continued fractions (and many others..) can be derived from a formula known to Wallis and Euler. I do not know how (and if..) they proved it. It was also proven by Stieltjes and Ramanujan. These Continued fractions can be evaluated using Euler's Differental Method. In this way one can prove the Wallis-Euler formula's which state that:
$4k+1+\cfrac{1^2}{2(4k+1)+\cfrac{3^2}{2(4k+1)+\cfrac{5^2}{2(4k+1)+...}}}=\cfrac{2k+1}{W(k)}\cfrac{4}{\pi}$
and
$4k+3+\cfrac{1^2}{2(4k+3)+\cfrac{3^2}{2(4k+3)+\cfrac{5^2}{2(4k+3)+...}}}=(2k+1)\cdot{W(k)}\cdot{\pi}$
with the Wallis product $W(k)=\cfrac{1\cdot3}{2\cdot2}\cdot\cfrac{3\cdot5}{4\cdot4}\cdot\cfrac{5\cdot7}{4\cdot4}....\cfrac{(2k-1)\cdot)2k+1)}{2k\cdot2k}$