I’ve actually been struggling with this:

For $n>0$, $\lbrace s_i\rbrace$ is said to simply solve $SRS(n)$ if for all $k=0,1,...,n-1$, it is true that $s_k
$s_k = 1 + \sum_{i=0}^{n-1} \text{Ch}_{k}(s_i)$ where $\text{Ch}_k$ is the indicator function for the singleton set containing $k$.

Prove or disprove that the only simple solution that exists for $n>5$, which aren’t multiples of 3, is

$s_0=s_{n-2}=s_{n-1}=1$

$s_1=n-3$

$s_2=3$

$s_i=2$ for $2
And that there are none for $n$ which are multiples of $3$