 Solution

With $a=0$ and $b>0$, the recurrence is $p_n=(b/n)p_{n-1}$. Iteration gives $p_n=p_0{b}^n/n!$. Since ${\sum }_{n\ge 0}{b}^n/n!=e^b$, normalization givesHence the claim count distribution is a Poisson distribution of mean $b$, and the Panjer recursion begins at $g_0=e^{-b}$.