TITLE: sequence $U_k=a_0U_{k-1}+a_1U_{k-2}+\cdots+a_{k-1}U_0$
QUESTION [5 upvotes]: 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 $a_n=bn-1 \ \land \ b>1$,
$U_k=b^2(b+1)^{k-2} \ \forall \  k\geq2$

REPLY [3 votes]: Suppose that
$$U_k=\sum_{l=0}^{k-1}\big(a(k-l)+b\big)U_l$$
Then
$$V(x)-1=\sum_{k=1}^\infty U_kx^k=\sum_{k=1}^\infty \left(\sum_{l=0}^{k-1}\big(a(k-l)+b\big)U_l\right)x^k $$
$$=\sum_{l=0}^\infty \left(\sum_{k>l}^\infty \big(a(k-l)+b\big)x^k\right)U_l=\sum_{l=0}^\infty\left(\sum_{\delta=1}^\infty (a\delta+b)x^\delta\right)x^lU_l=\left(\frac{ax}{(1-x)^2}+\frac{bx}{1-x}\right)V(x)$$
whence
$$V(x)=\left(1-\frac{ax}{(1-x)^2}-\frac{bx}{1-x}\right)^{-1}=\frac{(1-x)^2}{1-(a+b+2)x+(b+1)x^2}.$$
As a rational function, its coefficients will be a linear combination of powers of the zeros of the denominator above. Assuming a non-degenerate case, one power will dominate the other and we will indeed have $U_k\sim c d^k$ for some $c,d$, as desired.