Find $a_{n}$ from a convolution formula Suppose that $c_{n}$ satisfies the recurrence formula below: $c_2=\alpha$, and
$$c_{2n}+c_{2n-2}=\frac{(\alpha)_n}{n!},n\geq2.$$
were $(\alpha)_n = \alpha(\alpha-1)·\cdots·(\alpha-n+1)$ and $\alpha$  is a real number. If $c_n$ is also a convolution of $a_n$, i.e.
$$c_{2n}=\sum_{m=1}^n a_m·a_{n+1-m}$$
So how can we find the analytical formulas of $c_n$ and $a_n$? (PS: $c_n = 0$ if $n$ is an odd integer.) $$$$
Here below are the first few values:
$$c_2=\alpha, c_4=\frac{\alpha(\alpha-3)}{2}$$
$$a_1=\sqrt{\alpha}, a_2=\frac{\sqrt{\alpha}(\alpha-3)}{4}$$
 A: Hint.
Let's set $d_k:=c_{2k}$, $k \geq 1$, with $d_1=\alpha$. The identity
$$c_{2k}+c_{2k-2}=\frac{(\alpha)_k}{k!},k\geq2,$$
were $\color{blue}{(\alpha)_k = \alpha(\alpha-1)·\cdots·(\alpha-k+1)}$  may be rewritten as
$$
d_{k}+d_{k-1}=\frac{(\alpha)_k}{k!},k\geq2,
$$ equivalently
$$
(-1)^k d_{k} - (-1)^{k-1} d_{k-1}=(-1)^k\frac{(\alpha)_k}{k!},
$$ summing from $k=2$ to $k=n$, by telescoping we get
$$
(-1)^n d_{n} - \alpha=\sum_{k=2}^{n}(-1)^k\frac{(\alpha)_k}{k!}
$$ and we obtain the analytic form
$$
c_{2n}=(-1)^n \sum_{k=1}^{n}(-1)^k\frac{(\alpha)_k}{k!}, \quad n=1,2,\ldots,
$$ 
which may be simplified to 

$$
\color{blue}{c_{2n}=(-1)^n \left(\frac{\Gamma(n+1-\alpha)}{\Gamma(1-\alpha)}\cdot\frac{1}{n!}-1\right), \quad n=1,2,\ldots. \tag1}
$$ 

Observe that, by the binomial theorem, we have
$$
\sum_{k=0}^{\infty}(-1)^k\frac{(\alpha)_k}{k!} x^k =(1-x)^{\alpha},
$$ then, setting $\displaystyle \color{blue}{f(x)=\sum_{k=1}^{\infty} a_k x^k}$, by the Cauchy product we have
$$
\left(f(x)\right)^2 =\frac{x}{1+x}\left((1+x)^{\alpha}-1\right)
$$
giving

$$
\color{blue}{f(x)= \sqrt{\frac{x}{1+x}}\sqrt{(1+x)^{\alpha}-1} , \quad  \tag2}
$$ 

with a more involved analytic form for $a_n$.

You may check that
$$\color{blue}{
\begin{align}
c_{2} &=\alpha\\
c_{4} &=\frac{\alpha}{2}(\alpha-3)\\
c_{6} &=\frac{\alpha}{6}(\alpha^2-6\alpha+11)\\
c_{8} &=\frac{\alpha(\alpha-5)}{24}(\alpha^2-5\alpha+10)\\
\ldots &=\ldots\\
\end{align}}
$$
and
$$
\color{purple}{
\begin{align}
a_{2} &=\sqrt{\alpha}\\
a_{4} &=\frac{\sqrt{\alpha}}{4}(\alpha-3)\\
a_{6} &=\frac{\sqrt{\alpha}}{96}(5\alpha^2-30\alpha+61)\\
a_{8} &=\frac{\sqrt{\alpha}(\alpha-7)}{384}(3\alpha^2-14\alpha+31)\\
\ldots &=\ldots\\
\end{align}}
$$
Hoping this helps you!
