<s>[INST] sequence $U_k=a_0U_{k-1}+a_1U_{k-2}+\cdots+a_{k-1}U_0$ Is there a general formula for $U_k$ defined by
$$U_k=a_0U_{k-1}+a_1U_{k-2}+\cdots+a_{k-1}U_0$$
where the $a_i$ are in arithmetic progression and $U_0=1$? Do there always exist $c,d$ such that $U_k\to cd^k$ as $k\to\infty$?
If $a_n=n+1$, 
$U_k=F_{2k}$ where $F_n$ is the $n^{th}$ Fibonacci number.
If $a_n=bn-1 \ \land \ b>1$,
$U_k=b^2(b+1)^{k-2} \ \forall \  k\geq2$ 
 [/INST] The recurrence equation is written as 
$$
   U_k = \sum_{m=0}^{k-1} a_{k-1-m} U_m   \tag{1}
$$
Let's form the generating function $f(x) = \sum_{k=0}^\infty x^k U_k$. Then, summing eq. 1, multiplied on both sides with $x^k$, from $k=1$ to $\infty$: 
$$
  f(x) - 1 = x \cdot f(x) \cdot \sum_{k=0}^\infty a_k x^k = x f(x) g(x)
$$
Since $a_k = a_0 + (a_1-a_0) k$, $g(x) = \sum_{k=0}^\infty a_k x^k = a_0 \frac{1-2x}{(1-x)^2} + a_1 \frac{x}{(1-x)^2}$. Hence
$$
   f(x) = \frac{1}{1-x g(x)} = \frac{(1-x)^2}{(1-x)^2 - a_0 x + x^2 (2a_0 - a_1)}
$$
The case of $a_n = n+1$, that  corresponds to  $a_0=1$ and $a_1 = 2$, we recover the generating function of $F_{2k}$:
$$
   f(x) = \frac{(1-x)^2}{1-3 x + x^2} = 1 + x + 3 x^2 + 8 x^3 + \mathcal{o}(x^3) 
$$
The case of $a_n = b n -1$ maps to $a_0 =-1$ and $a_1 = b-1$ and generating function:
$$
   f(x) = \frac{(1-x)^2}{1-x- b x^2}
$$
The above generating function does not correspond to $U_k = b^2 (b+1)^{k-2}$ for $k \geqslant 2$.
In[44]:= u[0] = 1;

In[45]:= u[k_Integer?Positive] := 
 u[k] = Expand[Sum[a[k - 1 - m] u[m], {m, 0, k - 1}]]

In[49]:= (Sum[u[k] x^k, {k, 0, 12}] /. {a[m_] :> b m - 1}) - 
 Series[-((-1 + x)^2/(-1 + x + b x^2)), {x, 0, 12}]

Out[49]= SeriesData[x, 0, {}, 13, 13, 1]

The asymptotic growth of $U_k$ is determined by poles of $f(x)$.
</s>