Suppose one starts with an (infinite) multiset of positive integers $\mathcal{A} = \{a_i\}_{i\geq 0}$ such that:

$1=a_0\leq a_1\leq a_2\leq\ldots$

Can one always find a (necessarily infinite) group $G$ such that the set of degrees of its finite dimensional irreducible complex representations is exactly $\mathcal{A}$?

Clearly the answer to the above question is no for arbitrary $\mathcal{A}$; for example if $a_1 = N>1$ (i.e. the trivial representation is the only 1-dimensional representation of the potential group), then a simple argument involving the decomposition of the tensor square of this $N$-dimensional representation shows that $N\leq a_2\leq \frac{N^2+N}{2}$. Using more general forms of such arguments involving tensor powers of decompositions of $GL_n(\mathbb{C})$-representations, one can find other restrictions ruling out certain potential degree sequences.

$\bf{Question:}$ Given $\mathcal{A}$ such that there are no obstructions arising from tensor power considerations, is it always possible to find/construct a group $G$ whose irreducible degree sequence is exactly $\mathcal{A}$?

I assume that in general this is still a hard question, so I am also interested in partial results. I would also be interested in negative results along the lines "Even though this sequence $\mathcal{A}$ has no obstructions, it still cannot be the degree sequence of any group for some deeper reason."