Recurrence relations that do not have a closed form solution

A recurrence relation computes value $d_n$ with the $\{d_{i},d_{j}\ldots\}$ previous digits of the sequence

So for example, the Fibonacci sequence is defined as follows:
$$
F_n = F_{n-1}+F_{n-2},\ \ \ F_0=0, F_1=1
$$
The classic closed form solution is given by the Binet formula:
$$
F_n = \frac{\phi^n-\psi^n}{\sqrt{5}}
$$
where $\phi=\frac{1+\sqrt{5}}{2}$ and $\psi=1-\phi$.

The question is: are there such sequence of digits defined through some arbitrary recurrence relation that do not have a closed form solution, that is, that provably do not have a closed form solution?

I am asking this question in CS stack exchange because the context of this question is Random Number Generation... If such iterated sequences exist, this could be (could be) helpful in designing RNG's...

Yes, there are. Consider the recurrence $a_{n+1}=(7a_n/4 + 1/2) - (5a_n/4 + 1/2)(-1)^{a_n}$ where $a_0$ is a given integer value. It is well known that this gives the Collatz sequence. There is no "known" closed form solution to the recurrence (and if you find it, you would be quite famous in mathematical circles). Furthermore, it is also well known that there are recurrent sequences that can't be solved in closed form. For instance, Conway shows that a natural generalization of the Collatz problem is undecidable. On the other hand, there are large classes of recurrence relations for which methods are known to determine closed form solutions in terms of sometimes multiple finite or infinite sums. Hypergeometric functions can be used to solve large classes of them. For instance, in theory, all linear recurrences with polynomial coefficients can be solved. More information can be found in Solving Linear Recurrence Equations with
Polynomial Coefficients by Petkovsek and Zakrajsek (PDF).

StackExchange Computer Science Q#27598, answer score: 6