Intuiting factorials in recurrence relations

Question

I'm trying to solve differential equations via series, and I understand the underlying concepts and processes to get to a recurrence relation. For example, from a homogeneous differential equation, I can reach an relation such as $$ a_n = - \frac{a_{n-2}}{n(3n +1)} \ \text{where} \ a_{3, 5, 7...} = 0 $$

My method to try and find the answer so far has been to write out large-ish terms without simplifying and look for a pattern by inspection $$ a_6 = \frac{-1}{2(7)*4(13)*6(19)} $$ I haven't had much luck. I can find some terms that appear in the final answer, but there always seems to be at the least a factorial function that goes unaccounted for. My textbook says that the answer to this recurrence relation is $$ a_m = \frac{(-1)^m}{2^m m!7*13* ... (6m+1)} $$

I can evaluate this and show that it is equivalent to my recurrence relation, but I don't know what steps to take to get to that point. I'm especially baffeled by how the factorial $m!$ and $2^m$ fit into the answer. Any clarification would be appreciated.

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It was suggested in the comments that I think of the problem in terms of $ a_n = (something)a_{n-2} = (something)a_{n-4} $
That makes the function
$$ a_n = - \frac{a_{n-6}}{n(n-2)(n-4)(3n+1)(3n-5)(3n - 11)} $$
I can see a likely candidate for a factorial function in the $ n(n-2)(n-4) $ sequence, and the other three terms in the denominator decrease by 6 in each term, but I don't have any ideas for how to turn these patterns into an actual function.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Lets $\ds{b_{n} \equiv a_{2n} = -\,{a_{2n - 2} \over 2n\pars{6n + 1}} \implies b_{n} = -\,{b_{n - 1} \over 2n\pars{6n + 1}}}$

\begin{align} b_{n} & = -\,{1 \over 2}\,{b_{n - 1} \over n\pars{6n + 1}} = {\pars{-1}^{2} \over 2^{2}} \,{b_{n - 2} \over n\pars{6n + 1}\pars{n - 1}\bracks{6\pars{n - 1} + 1}} \\[5mm] & = {\pars{-1}^{3} \over 2^{3}} \,{b_{n - 3} \over n\pars{6n + 1}\pars{n - 1}\bracks{6\pars{n - 1} + 1} \pars{n - 2}\bracks{6\pars{n - 2} + 1}} \\[5mm] & = \cdots = {\pars{-1}^{n} \over 2^{n}}\,{1 \over n\pars{n - 1}\ldots 1}\, {1 \over \prod_{k = 1}^{n}\pars{6k + 1}}\,b_{0} = {\pars{-1}^{n} \over 2^{n}}\,{1 \over n!}\, {1 \over 6^{n}\prod_{k = 1}^{n}\pars{k + 1/6}}\,b_{0} \\[5mm] & = {\pars{-1}^{n} \over 12^{n}}\,{1 \over n!}\, {1 \over {\vphantom{\Large A}\pars{7/6}^{\,\large \overline{n}}}}\,b_{0} = {\pars{-1}^{n} \over 12^{n}}\,{1 \over n!}\, {1 \over \Gamma\pars{7/6 +n}/\Gamma\pars{7/6}}\,b_{0}\qquad \pars{~\Gamma: Gamma\ Function~} \\[5mm] & = \bbx{{\pars{-1}^{n} \over 12^{n}}\,{\pars{1/6}! \over n!\pars{1/6 + n}!}\,b_{0}} \qquad\mbox{where}\ z! = \Gamma\pars{z + 1}. \end{align}

Note that $\ds{a^{\overline{m}} = a\pars{a + 1}\pars{a + m - 1}}$, with $\ds{m \in \mathbb{N}_{\ >\ 0}}$ and $\ds{a^{\overline{0}} = 1}$, is the Rising Factorial which can be expressed in terms of the Gamma Function as follows: $\ds{a^{\overline{m}} = {\Gamma\pars{a + m} \over \Gamma\pars{a}}}$. 'Extension' of the raising factorial to any $\ds{m \in \mathbb{Z}}$ can be seen in the above cited link.

Answer 2

I suppose you are trying to derive $$a_{2m}=\frac{(-1)^m}{2^m m!7*13* ... (6m+1)}a_0$$ since $a_n=0$ for odd indices $n$. The patterns that develop are pretty clear:

$$a_{2m}=\color{blue}{\frac{-1}{2m(6m+1)}a_{2m-2}}$$

$$=\frac{-1}{2m(6m+1)}\cdot \frac{-1}{2(m-1)(6(m-1)+1)}a_{2(m-1)-2}=\color{blue}{\frac{(-1)^2}{2^2[m(m-1)]\cdot[(6m+1)(6m-5)]}a_{2m-2.2}}$$

$$=\frac{(-1)^2}{2^2[m(m-1)]\cdot[(6m+1)(6m-5)]}\cdot\frac{-1}{2(m-2)(6(m-2)+1)}a_{2(m-2)-2}=\color{blue}{\frac{(-1)^3}{2^3[m(m-1)(m-2)]\cdot[(6m+1)(6m-5)(6m-11)]}a_{2m-2.3}}$$

$$\vdots$$

$$=\color{blue}{\frac{(-1)^m}{2^m[m(m-1)\cdots(m-(m-1))]\cdot[(6m+1)(6m-5)\cdots(6(m-(m-1))+1)]}a_{2m-2m}}$$

$$=\frac{(-1)^m}{2^m[m(m-1)(m-2)\cdots3\cdot2\cdot1]\cdot[(6m+1)(6m-5)(6m-11)\cdots13\cdot7]}a_0$$

$$=\frac{(-1)^m}{2^mm!(6m+1)(6m-5)(6m-11)\cdots13\cdot7}a_0$$