6.2.3.2 step 2

Using quantities calculated in step \(1\), the algorithm now searches for a non-negative integer \(d\) using

\begin{align*} d&=\frac {n}{12} \left ( e_{\infty } - \sum _{c\in \Gamma } e_c \right ) \end{align*}

Where in the above \(e_c \in E_c\), \(e_\infty \in E_\infty \) \(n\) is any of \(\{4,6,12\}\) values. If non-negative \(d\) is found, then

\begin{align*} \theta &= \frac {n}{12} \sum _{c\in \Gamma } \frac {e_c}{x-c} \end{align*}

The sum above is over all families of \(\{e_\infty ,e_c\}\) which generated the non-negative integer \(d\). Next define

\begin{align*} S &= \prod _{c\in \Gamma } (x-c) \end{align*}

The product above is over families of \(\{e_\infty ,e_c\}\) which generated the non-negative integer \(d\). If no non-negative integer \(d\) is found, then no Liouvillian solution exists.