Solve Advanced Recurrence Relations

generating-functionshypergeometric functionrecurrence-relationsreference-request

For some research I’m doing, I had to solve recurrences of the form: $a(n)=n \cdot a(n-1)-(n-3) \cdot a(n-2)$.

WolframAlpha gave the solution:

$$ a(n) = 4c_2 \Gamma (n-1) ( \Gamma (n-1) F(1,n-2;n,n;1) -(c_1-4c_2(e-3))(n-2) )$$
Where F is the regularized generalized hypergeometric function.

I was taught a little bit about creating closed forms from recurrence relations, but have no clue to derive something such as this. It looks really similar to examples where you use exponential generating functions, but I can’t see exactly how it would work in this case. So my question is, how would a human get this solution, and where could I go to learn this and similar advanced techniques.

Best Answer

Here are two references which might be useful when learning recurrence relations.

Hint: Sometimes we are lucky and can find a solution also by elementary means. Assuming initial values $a_0, a_1$ are given, we derive from \begin{align*} a_n=na_{n-1}-(n-3)a_{n-2}\qquad n\geq 2 \end{align*} successively \begin{align*} \color{blue}{1}a_2&=2a_1+a_0\\ a_3&=\color{blue}{3}a_2\\ a_4&=4a_3-a_2=\color{blue}{11}a_2\\ a_5&=5a_4-2a_3=\color{blue}{49}a_2\\ a_6&=6a_5-3a_4=\color{blue}{261}a_2\\ &\cdots\\ \end{align*}

Looking for the values $1,3,11,49,261$ in OEIS we find the sequence A001339 with representation \begin{align*} a_{n+2}=\sum_{k=0}^n\binom{n}{k}(k+1)!\qquad\qquad n\geq 0 \end{align*}

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