Can we solve this recurrence relation? Consider the following recurrence relation
$$x_1=1\\ x_2=a\\ x_{n+2}=ad^nx_{n+1}-dx_n$$
where $a,x_n\in\mathbb{C}$ and $d\in\mathbb{R},d>1$.

Is there any $a\in\mathbb{C}$ such that $\lim x_n=0$ ? or at least can
  we find a formula for $x_n$ depending only on $a$ and $d$?

 A: 
Let $x_n$ be a complex sequence satisfying $x_0=0$, $x_1=1$ and
  $$x_{n+2}=ad^nx_{n+1}-dx_n$$
  for all $n\in\Bbb N$ with $d>1$ and $a\in\Bbb C\setminus\{0\}$.
  Then
  \begin{align}
&x_n=O(a^nd^{n^2/2-n/2})&
&(n\to\infty)\tag 1
\end{align}
  Moreover, we have $x_n=o(a^nd^{n^2/2-n/2})$ if and only if $x_n\xrightarrow{n\to\infty}0$.

Proof. First note that the sequence $u_n=a^nd^{n^2/2-n/2-1}$ satisfy $u_{n+2}=ad^nu_{n+1}$, hence if we put $w_n=x_n/u_n$, the we have
$$w_{n+2}=w_{n+1}-d\frac{u_n}{u_{n+2}}w_n$$
Since $d\frac{u_n}{u_{n+2}}=\frac{d^2}{a^2d^{2n}}$ we get
$$w_{n+2}=w_1-\frac{d^2}{a^2}\sum_{k=0}^n\frac{w_k}{d^{2k}}\tag 3$$
We claim that this sequence converges.
For all $n\in\Bbb N$, we have
\begin{align}
&|w_n|\leq c^n&
&\text{where } c=1+\frac{d^4}{|a|^2(d^2-1)}>1
\end{align}
This follows by induction on $n$, for
\begin{align}
|w_{n+2}|
&\leq|w_1|+\frac{d^2}{|a|^2}\sum_{k=0}^n\frac{|w_k|}{d^{2k}}\\
&\leq c+\frac{d^2}{|a|^2}c^n\sum_{k=0}^nd^{-2k}\\
&\leq c^{n+1}+\frac{d^2}{|a|^2}c^{n+1}\frac 1{1-d^{-2}}\\
&\leq c^{n+2}
\end{align}
Consequently, $\sqrt[n]{|w_n|}$ is bounded and let $\ell=\limsup\sqrt[n]{|w_n|}$.
If $\ell<d^2$ then the power series in $(3)$ is convergent, hence $w_n$ is convergent as well.
Assume $\ell\geq d^2>1$.
Then for all $\varepsilon>0$, we have $|w_n|\leq(\ell+\varepsilon)^n$ eventually, hence there exists a constant $C>0$ satisfying
\begin{align}
\ell
&=\limsup\sqrt[n]{|w_n|}\\
&\leq\limsup\sqrt[n]{|w_1|+C\sum_{k=0}^n\left(\frac{\ell+\varepsilon}{d^2}\right)^k}\\
&\leq\lim
\sqrt[n]{|w_1|+C\frac{\left(\frac{\ell+\varepsilon}{d^2}\right)^{n+1}-1}{\frac{\ell+\varepsilon}{d^2}-1}}\\
&=\frac{\ell+\varepsilon}{d^2}
\end{align}
from which $\ell\leq\frac\ell{d^2}$ which implies $d^2\leq 1$, a contradiction which proves $w_n$ to be convergent and hence proves $(1)$.
Finally, from $(3)$, we get $w_n\xrightarrow{n\to\infty}0$ if and only if
$$w_1=\frac{d^2}{a^2}\sum_{k=0}^\infty\frac{w_k}{d^{2k}}$$ from which
$$w_{n+2}=\frac{d^2}{a^2}\sum_{k=n+1}^\infty\frac{w_k}{d^{2k}}$$
Let $$s_n=\sup_{k\geq n}|w_k|$$
Then $s_n\downarrow 0$ and for all $k\geq n$ we have
\begin{align}
|w_{k+2}|
&\leq\frac{d^2}{|a|^2}\frac{s_{k+1}}{d^{2k+2}}\\
&\leq\frac{d^2}{|a|^2}\frac{s_{n+1}}{d^{2n+2}}
\end{align}
from which
$$s_{n+2}\leq\frac{d^2}{|a|^2}\frac{s_{n+1}}{d^{2(n+1)}}$$
Consequently
\begin{align}
s_n
&\leq s_0\prod_{k=0}^{n-1}\frac{d^2}{|a|^2 d^{2k}}\\
&\leq s_0\frac{d^{2n}}{|a|^{2n}d^{n^2-n}}
\end{align}
Consequently,
\begin{align}
|x_n|
&=|u_n||w_n|\\
&\leq |u_n| s_n\\
&\leq s_0\frac{|u_n|}{|a|^{2n}d^{n^2-n}}\\
&=\frac{|a|^nd^{n^2/2-n/2-1}}{|a|^{2n}d^{n^2-3n}}\\
&\xrightarrow{n\to\infty}0
\end{align}
$\square$

Assuming $x_n\neq 0$ eventually, we have $x_n\xrightarrow{n\to\infty}0$ if and only if $|x_{n+1}/x_n|$ is bounded.

Proof.
The sequence $z_n=\sqrt d x_{n+1}/x_n$ satisfy
$$z_{n+1}+\frac 1{z_n}=\frac a{\sqrt d}d^n$$
If $|z_n|$ is bounded, then $|\frac 1{z_n}|\geq\frac{|a|}{\sqrt d}d^n-|z_{n+1}|\to\infty$.
Thus $|z_n|\to 0$.
In particular, $|z_n|$ is bounded by $1/2$, hence $|x_n|$ is bounded $2^{-n}$, hence $x_n\to 0$.
$\square$

We have $x_n\xrightarrow{n\to\infty}0$ if and only if $\sqrt[n]{|x_n|}$ is bounded.

Proof.
Let $\sqrt[n]{|x_n|}$ be bounded and
$$\ell=\limsup_{n\to\infty}\sqrt[n]{|x_n|}$$
Then for all $\varepsilon>0$, we have $|x_n|<(\ell+\varepsilon)^n$ eventually, hence
\begin{align}
\ell
&=\limsup_{n\to\infty}\sqrt[n+1]{|x_{n+1}|}\\
&\leq\limsup_{n\to\infty}\sqrt[n+1]{\left|\frac{x_{n+2}}{ad^n}\right|+\left|\frac{dx_n}{ad^n}\right|}\\
&\leq\lim_{n\to\infty}\sqrt[n+1]{\frac{(\ell+\varepsilon)^{n+2}}
{|a|d^n}+\frac{d(\ell+\varepsilon)^{n}}{|a|d^n}}\\
&=\lim_{n\to\infty}\sqrt[n+1]{\frac{(\ell+\varepsilon)^{n}}{|a|d^n}((\ell+\varepsilon)^2+d)}\\
&=\frac{\ell+\varepsilon}d
\end{align}
Thus $\ell\leq\frac{\ell+\varepsilon}d$ for all $\varepsilon>0$, hence $\ell\leq\ell/d$ from which $\ell(d-1)\leq 0$.
Since $d>1$ and $\ell\geq 0$, this implies $\ell=0$, hence $x_n\to 0$.
$\square$
A generating function.

We have $x_n\xrightarrow{n\to\infty}0$ if and only if there exists an entire function $G:\Bbb C\to\Bbb C$ satisfying the functional equation
  $$(1+dz^2)G(z)=1+azG(dz)$$

Proof. Let $x_0=0$ and $x_1=1$ and let $F$ be its generating function
$$F(z)=\sum_{n=0}^\infty x_n z^n=zG(z)$$
Then $G$ satisfy the functional equation
$$(1+dz^2)G(z)=1+azG(dz)$$
In particular, this gives $G(0)=1$ and for $z=\pm i\sqrt d$
$$G(\pm i\sqrt d)=\pm i\frac{\sqrt d}a$$
A: I am not convinced, yet, that what follows below is right (I'm sure there's a flaw in there somewhere), but I feel it's worth posting so that others can check for errors...

Suppose that $y_n=x_{n+1}-A_n x_n$. Now, we can observe that
$$\begin{align}
y_{n+1}=x_{n+2}-A_{n+1}x_{n+1} &= (ad^n-A_{n+1}) x_{n+1} - dx_n\\
&=(ad^n-A_{n+1})\left(x_{n+1}-\frac{d}{ad^n-A_{n+1}}x_n\right)
\end{align}$$
To keep our expression independent of $x_n$, we require that
$$
A_n=\frac{d}{ad^n-A_{n+1}}
$$
or
$$
A_{n+1}=ad^n-\frac{d}{A_n}
$$
Note that this expression does not depend on $x_n$ in any way, and we can choose $A_0$ as we please (so long as $A_n$ never takes the value of 0).
Now, if we have an $A_n$ satisfying this expression, then we have
$$
y_{n+1}=(ad^n-A_{n+1})y_n = \frac{d}{A_n}y_n
$$
Also note that $y_0=a-A_0$. If $|A_n|$ is small, then $y_n$ will grow. If $|A_n|>d$, then $y_n$ will shrink.
If $A_n$ remains bounded, then for $x_n$ to converge to zero, it is necessary and sufficient that $y_n$ converge to zero.
From these two facts, we conclude that, if there exists an $A_0\neq a$ for which $|A_n|$ is bounded above by $d$ for sufficiently large $n$, then $x_n$ cannot converge to zero.
This, however, causes a problem. Returning to the original recurrence, dividing by $x_{n+1}$ and letting $B_n=\frac{x_{n+1}}{x_n}$, we obtain
$$
B_{n+1}=ad^n-\frac{d}{B_n}
$$
Now, if $x_n\to 0$, then $|B_n|$ is bounded above by 1 for sufficiently large $n$, at least in terms of geometric mean. Of course, B_0=a$, and thus this fact itself does not indicate that convergence cannot be obtained.
However, consider $A_0=a+\epsilon$ for a small value, $\epsilon$. We can easily see that
$$
\frac{dA_{n+1}}{d\epsilon} = \frac{d}{A_n^2}\frac{dA_n}{d\epsilon}
$$
and therefore, $\frac{dA_n}{d\epsilon}$ will stabilise when $A_n\to \sqrt{d}$, limiting the impact for sufficiently small $\epsilon$. And as $\sqrt{d}<d$ (because $d>1$), we therefore conclude that $\exists\epsilon$ such that $|A_n|$ remains bounded above by $d$.
From this, we can conclude (by contradiction) that there cannot be a value of $a$ for which $x_n$ converges to zero, so long as $d>1$.
