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I'm trying to solve a second order differential equation and I got a recurrence. Can someone help to solve $$n(n-1+q)a_{n}-a_{n-3}+e\cdot a_{n-2}=0$$ where $q$, $e$, and $a_{0}$ are some real numbers with $a_{1}=0$ and $a_{2}=-e\cdot a_{0}/(2(1+q))$.

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Perhaps you should edit your question so that it looks a little cleaner. Some $a_0$'s just floating in there – Patrick Da Silva Feb 11 '12 at 14:51
Recurrences like this don't always have neat solutions. One approach is to calculate the next few terms to see whether there is a pattern. – Gerry Myerson Feb 12 '12 at 0:59
I actually calculate the first six terms, the dependence on n and q clears up but it remains unclear in "e" – mohamed benbitour Feb 12 '12 at 10:13
Maybe if you edit into your question what you have found out about $n$ and $q$, someone will be able to help you with $e$. – Gerry Myerson Feb 13 '12 at 3:55
I'm very hapy becouse I found a solution in "Combinatorics Function technics" – mohamed benbitour Feb 15 '12 at 13:49

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