# Factorial coefficient simplification

1. Mar 28, 2012

### physicsjock

Hey,

I've been trying to find a way to simplify these two series coefficients,

\begin{align} & {{a}_{4n}}={{a}_{o}}\frac{2!6!....(n-6)!(n-2)}{4!8!....(n-4)!n!} \\ & {{a}_{4n+1}}={{a}_{1}}\frac{3!7!....(n-6)!(n-2)}{5!9!....(n-4)!n!} \\ & E.g. \\ & {{a}_{12}}={{a}_{o}}\frac{2!6!10!}{4!8!12!}\,= \frac{{{a}_{o}}}{3\cdot 4\cdot 7\cdot 8\cdot 11\cdot 12}\,\,\,\,\,\,and \\ & {{a}_{13}}={{a}_{1}}\frac{3!7!11!}{5!9!13!}\,=\,\,\frac{{{a}_{1}}}{4\cdot 5\cdot 8\cdot 9\cdot 12\cdot 13} \\ \end{align}

This is the best way I could find but I was wondering if there were any ways to avoid the use of ....'s

Is there any way to simplify these coefficients?