Find $a_{n}$ from a convolution formula

Question

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}$$

Answer 1

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 $a_n$.

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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!