Only few strongly regular graphs with parameters $\lambda=0$ (triangle-free) and $\mu=2$ (any two non-adjacent vertices
have exactly two common neighbors) are known, see the [wikipedia page](https://en.wikipedia.org/wiki/Strongly_regular_graph): the 4-cycle, the Clebsch graph and the Sims-Gewirtz graph.


I am looking for any information about the potential existence of more such graphs. For which values of $n$ and $k$ are they known *not* to exist?



What is known about the non-existence of strongly regular graphs srg(n,k,0,2)?
[Example 1 in A.Neumaier paper](https://doi.org/10.1016/S0304-0208(08)73275-4) says in partcular that the vertex degree in this case must be $k=t^2+1$, for $t$ not divisible by 4. As well, the number of vertices is $v=1+k+\binom{k}{2}$. The examples you list correspond to $t=2,3$. The next possible parameter set corresponds to $t=5$, so you have $v=352$, $k=26$. [A.Brouwer's database](https://www.win.tue.nl/~aeb/graphs/srg/srgtab351-400.html) lists this tuple of parameters as feasible, but no examples known. Similarly for $t=6,7$ you have feasible sets of parameters $v=704,1276$, resp. $k=37,50$, but no examples known.




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To see that $k=t^2+1$, note that the [2nd eigenvalue of the adjacency matrix](https://en.wikipedia.org/wiki/Strongly_regular_graph#Eigenvalues) is
$$
r:=\frac{1}{2}\left[(\lambda-\mu)+\sqrt{(\lambda-\mu)^2 + 4(k-\mu)}\right]=-1+\sqrt{k-1},\quad \text{i.e. $t^2:=(r+1)^2=k-1.$}
$$
Similarly, the 3rd eigenvalue is $s:=-1-\sqrt{k-1}$, and one can compute their multiplicites, see e.g. [Brouwer-van Lint, p.87](https://pure.tue.nl/ws/portalfiles/portal/2394798/595248.pdf)
to rule out the case $t$ divisible by 4.
Namely, the multiplicity of $r$ is given by
$$
-\frac{k(s+1)(k-s)}{(k+rs)(r-s)}=\frac{k\sqrt{k-1}(k+1+\sqrt{k-1})}{4\sqrt{k-1}}=\frac{(t^2+1)(t^2+2+t)}{4},
$$
which cannot be an integer if $4|t$.


