Let $\{a_n\}$ be a sequence defined recursively by $a_1=1,a_{n+1}=a_n^2+6$, find an explicit formula for $a_n$.
I tried some of the usual mathods but none of them led to a solution. It seems that there just isn't such a formula (with only elementary functions). So is there? If there isn't, how do you prove that there isn't such a formula?
There isn't any such formula [EDIT: for arbitrary choice of $a_0$]. In fact, that's the starting point of a field called discrete dynamics where you try to understand qualitatively the behaviour of sequences defined by induction. We study them qualitatively because there is no explicit formula in general. The sequences of the form $z_{n+1}=z_n^2+c$ where $c$ is a constant are especially studied as they are the simplest ones for which there is no explicit formula (except when $c=0$, obviously, or when $c=-2$ because then you have a Chebyshev polynomial and you can use a clever trick). Google Mandelbrot set to learn more.
In your case ($c=6$), what happens is that for every $z_0$ outside of a Cantor set (which is closed set with empty interior) the sequence goes to infinity quite fast, as noted in the other answers (with a speed of order $G(z_0)^{2^n}$ for some constant $G(z_0)>1$ depending on $z_0$.)
For $z_0$ in that Cantor set (google Julia set), the sequence remains bounded.
This is much too long to fit into the comment section beneath Yves Daoust's answer:
Define $\{c_n\}$ by $c_n^{2^{n-1}}=a_n$. Then we have $c=\lim_{n\to\infty}c_n$ and $$ c_{n+1}^{2^n}=a_{n+1}=a_n^2+6=a_n^2\left(1+\tfrac 6{a_n^2}\right) $$ and so $$ \begin{align} 2^n\log(c_{n+1})&=2\log(a_n)+\log(1+6/a_n^2)\\ &=2^n\log(c_n)+\log(1+6/a_n^2) \end{align} $$ therefore $$ \log(c_{n+1})=\log(c_n)+\frac1{2^n}\cdot\log(1+6/a_n^2)\leq\log(c_n)+\frac{6}{2^n\cdot a_n^2} $$ and inductively we see that $$ \begin{align} \log(c)&\leq\log(c_n)+\sum_{k=n}^\infty\frac{6}{2^k\cdot a_k^2}\\ &\leq\log(c_n)+\frac6{a_n^2}\sum_{k=n}^\infty\frac{1}{2^k}\\ &=\log(c_n)+\frac{6}{2^{n-1}\cdot a_n^2} \end{align} $$ and we know that $a_n=\left\lfloor c^{2^{n-1}}\right\rfloor$ will work if and only if $$ a_n\leq c^{2^{n-1}}< a_n+1 $$ We already know that $c>c_n$ so the first inequality holds. The second is equivalent to $2^{n-1}\log(c)<\log(a_n+1)$. Since $a_n+1=a_n(1+1/a_n)$ and $c_n^{2^{n-1}}=a_n$ the latter can be rewritten as $$ \log(a_n+1)=2^{n-1}\log(c_n)+\log(1+1/a_n) $$ and after dividing by $2^{n-1}$ the inequality becomes $$ \log(c)<\log(c_n)+\frac{1}{2^{n-1}}\cdot\log(1+1/a_n) $$ Furthermore we have the first order Taylor approximation $\log(1+x)\approx x$ which is quite accurate for small values of $x$ and already for $x\leq1/a_2=1/7$ we get $\log(1+x)>\frac67x$ and thus for $a_n\geq 7$ $$ \begin{align} \log(c)&\leq\log(c_n)+\frac{6}{2^{n-1}\cdot a_n^2}\\ &\leq\log(c_n)+\frac{6}{2^{n-1}\cdot 7a_n}\\ &<\log(c_n)+\frac{1}{2^{n-1}}\cdot\log(1+1/a_n) \end{align} $$ as desired.
BTW this analysis is easily generalized to $a_{n+1}=a_n^2+b$ for positive numbers other than $b=6$, namely using the same definition of $c_n,c$ then for $a_n\geq b+1$ we have $a_n=\left\lfloor c^{2^{n-1}}\right\rfloor$.
One huge drawback of this method in it's current form, is that it requires one to already compute $a_n$ in order to approximate $c$ well enough to be able to compute $a_n$ with it. That could potentially be fixed by using some analysis like the one I gave above to speed up the convergence $c_n\to c$.
Hint:
$a_n$ grows quickly, so that after a few iterations the term $6$ becomes neglectible and you have
$$a_{n+1}\approx a_n^2$$ of which the solution is
$$a_n=c^{2^n}.$$
The value of $c$ can be determined as the limit of $\sqrt[2^n]{a_n}$ of which here are the first values
$$1,\\ \sqrt7=2.645751\cdots\\ \sqrt[4]{55}=2.723270\cdots\\ \sqrt[8]{3031}=2.723944\cdots\\ \sqrt[16]{9186967}=2.723945\dots\\ $$