Solution of the algebraic equation

Question

I used to think about the following physics question. In quantum mechanics one has a quantity know as spin of a particle. A particle with total spin value $S$ (here $S\in\{0, 1/2, 1, 3/2, ...\}$ corresponds to an eigenvalue of the casimir operator of $SO(3)$ for a given irreducible representation) is described by a so called state vector, which belongs to an $2S+1$ (which is a dimension of space to which a given irreducible representation belongs) dimensional Hilbert space $\mathcal{H}$. When one studies a system of $N$ particles of a given spin $S$, the Hilbert space on which a state vector lies is a tensor product one $\mathcal{H}^{\otimes{N}}$ thus having a dimension $(2S+1)^{N}$. The question is the following, for which values of $N$ for fixed $S$ one can construct vectors belonging to $\mathcal{H}^{\otimes{N}}$ which are eigenvectors to the permutation operator which acts on the label of a given single particle. To me it seems that a reasonable argument is as follows. For $N$ particles there are $N!$ permutations, thus one has to look at the intersection of the dimension of space and $N!$, so we have to solve the following equation $$(2S+1)^{N}=N!$$ For $S=1/2$, which is the first non-trivial case, the solution is evident $N=2$. However, I'd like to know about the case for general $S$. Unfortunately they don't train us physicists to solve such equations, that is why I'm here asking this question.... Great thanks anyways.

Answer 1

I'd suggest using the Stirling approximation

$$ \ln N!\approx N\ln N-N$$

which if you're going to use integer $N$ is really good even at low $N$

thus

$$ N\ln (2S+1)=N \ln N-N\\ \ln(2S+1)=\ln N-1 $$

Which is easily solvable for $N$