Sequence for which no closed form can exist

Inevitably, the answer will depend on what one means by closed form. We consider computable sequences of non-negative integers, that is, computable functions $f(x)$ from the non-negative integers to the non-negative integers. We will allow the least number operator $\mu$, as well as operations of ordinary arithmetic.

The least number operator may not be familiar, so we define it. Let $R(y,x_1,\dots,x_n)$ be a relation such that for all $x_1,\dots,x_n$ there is a $y$ such that $R(y, x_1,\dots,x_n)$ holds. Then $\mu y R(y,x_1,\dots,x_n)$ is the least such $y$.

As a consequence of Matijasevic's solution of Hilbert's Tenth Problem, there is a fixed polynomial $P(e, x, u_1,\cdots, u_k)$ with the following property.

For any computable function $f$, there is a non-negative integer $e=e(f)$ such that for any $x$, $$f(x)=[\mu y( P(e, x, [y]_1,[y]_2, \dots, [y]_k)=0)]_0.$$ Here by $[w]_i$ we mean the exponent of the $i$-th prime $p_i$ in the prime power decomposition of $w$. (The $0$-th prime is $2$.)

This gives what I would consider a positive answer to the closed form question: Every computable sequence has a closed form. Many theorems of the same general kind were known long before the work of Matijasevic, except that instead of a polynomial $P$, one had a more complicated function.

If the $\mu$-operator is not allowed, there are quite a few different workarounds.

The most famous and interesting one is probably the sequence of primes (if you mean a closed form in terms of elementary functions).

Goldbach proved that no polynomial with integer coefficients can give a prime for all integer values. However it is not fully clear that there isn't some elementary function that generates all primes.

There is a whole wikipedia article on formulas for primes.