defines the lemniscate cotangent = (X/Y) = T

where X = T/(T + 1) ]]>

You may want to see this post by Robert Israel who answered my question at http://math.stackexchange.com/questions/153504.

]]>The reduced form of the sextic you mentioned does not have a concise form. You have to go through TWO Tschirnhausen transformations, a quadratic and quartic one. I described an easy approach in “A New Way to Derive the Bring-Jerrard Quintic”, the 18th file in http://sites.google.com/site/tpiezas/ramanujan. (While I discuss there the quintic, I trust you can extrapolate it for the sextic.)

]]>in general

\begin{equation}

A_s(z)=\sum^{\infty}_{m=1}\frac{z^m}{\left(^{sm}_m\right)}=

\frac{1}{z}\sum^{s}_{k=1}\frac{\log\left(\frac{1+p_{s,k}(z)}{p_{s,k}(z)}\right)}{p_{s,k}(z)^{s-2}\left(s-1+sp_{s,k}(z)\right)}

\end{equation} ]]>

How you get it?

I personally searching such formulas for years. Can you find $R(q)^{-5}-11-R(q)^5$ in hypergeometric functions? This may lead to the solution of a sextic equation

$b^2/(20a)+bx+ax^2=c x^{5/3}$. I want to ask you, can help me find the reduced form of the above equation $x^6+ax^2+bx+c=0$? ]]>