Recursive sequence growth rate Take arbitrary $a>0$, define $\{x_n\}$ recursively by $x_0=1$ and $x_n=a(\sum_{i=1}^n \frac{x_{n-i}}{i})$. Describe the growth of $\{x_n\}$ (is it linear, exponential, etc), and determine the limit of $\sqrt[n]{x_n}$ as $n$ approaches infinity.
some things that I have trying: Due to the first term it seems the sequence grows similar to exponential, but I am unsure of limit because first term does not dominate. For $a=1$ the first few terms are: $b_0=1$, $b_1=1$, $b_2=\frac{3}{2}$, $b_3=\frac{7}{3}$, $b_4=3$, $b_5=\frac{157}{60}$
Edit: A user has shown that once we know function exponential, we can approximate answer. However, how can we rigorously prove its exponential?
 A: This is a nice one, and it was tough. I provide here a conjectured answer.
I have found it with a CAS (mathematica) which gave me numerically an overview of the solution, which then could be transformed into a a conjected analytic expression which in turn was verified numerically by the CAS. A proof of the conjecture is still missing (I thank @jxia1234
for pointing this out).
The conjected answer is
$$L(a) :=\lim_{n\to\infty} (x_n)^{\frac{1}{n}}= \frac{1}{1-e^{-1/a}}\tag{1}$$
The heuristic derivation proceeds in two steps
Step 1: Numericaly solving the recursion
$$x_n = a  \sum_{k=1}^{n} \frac{x_{n-k}}{k}, x_0=1\tag{2}$$
The solution for  some values of the parameter $a$ are shown here (as lines instead of points for better visibility)

We can see that for large enough $n$ we have an exponential behaviour
$$x_n \simeq A e^{B n}\tag{3}$$
where $A$ and $B$ are numbers depending on the parameter $a$.
The quantity we are finally looking for is the $n$th root
$$L(a)=\lim_{n\to\infty} (x_n)^{\frac{1}{n}} = \lim_{n\to\infty} A^{\frac{1}{n}}e^{B} = e^{B(a)}\tag{4}$$
notice that it is independent of $A$.
Step 2: In order to find an approximate analytic expression for $B(a)$ we insert $(3)$ into the recursion $(2)$, giving $A e^{B n} = a  \sum_{k=1}^{n} \frac{A  e^{B (n-k)}}{k}$. Cancelling the common factors $A$ and $e^{B}$ this reads
$$1=a \sum_{k=1}^{n} \frac{e^{-B k}}{k}\tag{5}$$
Extending the range of $n$ to $\infty$ (here enters the approximation) this simplifies to $\frac{1}{a}= -\log(1-e^{-B})$ which can be solved for $B$ leading to the result $(1)$.
