How does one deal with expressions like $$ \sum_\mu s_\mu \delta_{2 \mu-1,j} \delta_{2\mu,k} \quad \text{ or } \quad \sum_\mu a_\mu\delta_{2\mu,j} \delta_{2\mu -1,k} \quad ? $$ The usual rules would tell me set $\mu=\frac{j+1}{2}$ so that the first expression would be equal to $$ \sum_\mu s_\mu \delta_{2 \mu-1,j} \delta_{2\mu,k} \stackrel{?}{=} s_{\frac{j+1}{2}}\delta_{j+1,k} $$ but it should be $$ \sum_\mu s_\mu \delta_{2 \mu-1,j} \delta_{2\mu,k} = s_{(j+1)/2}\delta_{j+1,k}\delta_{j\in \mathbb{O}\ } $$ where $\mathbb{O}$ here denotes the set of (positive) odd numbers.
So, what are the rules to tackle these problems fast?
The key is to keep in mind what the possible indices $\mu$ are in this context. I will assume that in this context, we take $\mu = 1,2,\dots$.
Note that for a given $j,k$, the sum $\sum_{\mu} s_\mu \delta_{2 \mu-1,j} \delta_{2\mu,k}$ can only be non-zero if there exists an integer $\mu \geq 0$ such that $2\mu - 1 = j$ and $2 \mu = k$. Note that such a $\mu$ can only exist if $j$ is odd and $k$ is even. Moreover, we can see that if these equations hold, then $j + 1 = (2\mu - 1) + 1 = 2 \mu = k$. Finally, it is clear that if such a $\mu$ exists for any $j,k$, then there is only one such $\mu$, which is equal to $(j+1)/2$.
Putting that all together, we find that \begin{align} \sum_{\mu} s_\mu \delta_{2 \mu-1,j} \delta_{2\mu,k} &= \delta_{j \in \Bbb O} s_{(j+1)/2}\delta_{j+1,k}. \end{align} Assuming that there is no further context here, this seems to contradict both answers that you have given.