Does there exist a constant $C$ such that $2^n\le a_n \le C^n$ for any $n$? I'm interested in making original sequences and in finding a special property. I've been thinking a sequence $\{a_n\}$.
$\{a_n\}\ (n=0,1,2,\cdots)$ is defined as the following:
$$a_0=1, a_{n+1}=\sum_{j=0}^{n}a_j+\sum_{j=0}^na_ja_{n-j}.$$
By using computer, I expect the following would be proven true:
My expectation: There exists a constant $C\gt2$ such that $2^n\le a_n \le C^n$ for any $n$.
I can't prove this, but it seems true, indeed. Here is my question.
Question: Is this property is true? In other words, can this property be a theorem? If it can be a theorem, please show me how to prove it. If not, again please show me how to prove it.
 A: Let
$$f(z) = \sum_{n\ge 0}^{}a_n z^n.$$
It's straightforward to verify (P.S.: See end of this post.) that
$$z(1-z)f(z)^2+(2z-1)f(z)+1-z=0,$$
and therefore
$$f(z) = \frac{2z-1+\sqrt{1-8z+12z^2-4z^3}}{2z(z-1)}.$$
In this form, there appears to be a singularity at $z=0$, but by rationalizing the numerator we see that there really isn't one:
$$f(z)=\frac{2(1-z)}{1-2z+\sqrt{1-8z+12z^2-4z^3}}.$$
So $f$ is analytic for $|z|<|r|$, where $r$ is the root of $1-8z+12z^2-4z^3=0$ with smallest absolute value.  Numerically, we find that  $r=0.16243...$ .  Therefore the series defining $f(z)$ converges whenever $|z|\le 0.16243$, which implies that
$$\lim_{n\rightarrow\infty}a_n 0.16243^n=0.$$
So, at least for sufficiently large values of $n$,
$$a_n 0.16243^n\le1$$
and
$$a_n\le\frac{1}{0.16243^n}<6.15650^n.$$
Numerically, this appears to be true for all $n$, but I haven't proved that.
In any case, $C$ does exist; in fact we can take $C$ to be the maximum of $6.15650$ and the finitely many values (if any) of $a_n^{1/n}$ for which $a_n 0.16243^n>1.$
P.S.: Proof of functional equation for $f(z)$:
$$f(z)^2=\sum_{n\ge0}\sum_{j=0}^na_j a_{n-j} z^n
=\sum_{n\ge0}\Big(a_{n+1}-\sum_{j=0}^na_j\Big)z^n$$
so
$$\begin{align}
(1-z)f(z)^2&=\sum_{n\ge0}\Big(a_{n+1}-\sum_{j=0}^na_j\Big)z^n -
\sum_{n\ge1}\Big(a_n-\sum_{j=0}^{n-1}a_j\Big)z^n\\
&=a_1-a_0+\sum_{n\ge1}(a_{n+1}-2a_n)z^n\\
&=1+\sum_{n\ge2}a_nz^{n-1}-2\sum_{n\ge1}a_nz^n\\
&=1+z^{-1}(f(z)-a_0-a_1z)-2(f(z)-a_0)\\
&=1-z^{-1}+(z^{-1}-2)f(z).
\end{align}$$
Multiplying by $z$ and rearranging gives the stated equation.
