I have observed that the number of triangles $\frac{vk}{6}$ of a strongly regular graph with parameters $(v,k,1,2)$ is given by the coefficient $2(k-1)$ in the molien series of the "4-D extraspecial group $2^{1+2\cdot 2}$": For example the beginning of the molien series is: $1 + t^2 + 5t^4 + 6t^6 + 15t^8 + 19t^{10} + 35t^{12} + 44t^{14} + 69t^{16} + 85t^{18}\\ + 121t^{20} + 146t^{22} + 195t^{24} + 231t^{26} + 295t^{28} + 344t^{30} + 425t^{32} + O(t^{34})$

We know that for $srg(9,4,1,2)$ we have $\frac{9\cdot4}{6}=6$ triangles and the coeffient $6 \cdot t^6$ tells us that at $6 = 2\cdot(4-1)$ we have $6$ triangles.

We know that for $srg(99,14,1,2)$ we have $\frac{99\cdot14}{6}=231$ triangles and the coeffient $231 \cdot t^{26}$ tells us that at $26 = 2\cdot(14-1)$ we have $231$ triangles.

And so on ...

I have checked this for all known parameters for such hypothetical graphs: The parameters of all such graphs are: $(9,4,1,2),(99,14,1,2),(243,22,1,2),(6273,112,1,2),(494019,994,1,2)$ of which only the $n=9$ and $n=243$ are known. The Conway $99$-graph problem asks if there exists such graph with 99 vertices. The number of triangles in such a (hypthetical) graph is: $6,231,891,117096,81842481$

(1) My (naive) question is, if there is a (deeper) connection, and maybe one can find unified way to construct all such graphs using the group above and the invariant homogenous polynomials (if this is not too much to ask)?

(2) The 6 homogenous invariants of degree 6 are:

$\frac14x_0^2x_1^2x_2^2 + \frac14x_0^2x_1^2x_3^2 + \frac14x_0^2x_2^2x_3^2 + \frac14x_1^2x_2^2x_3^2,$

$ \frac14x_0^4x_2^2 + \frac14x_0^2x_2^4 + \frac14x_1^4x_3^2 + \frac14x_1^2x_3^4,$

$ \frac14x_0^6 + \frac14x_1^6 + \frac14x_2^6 + \frac14x_3^6,$

$ \frac14x_0^3x_1x_2x_3 + \frac14x_0x_1^3x_2x_3 + \frac14x_0x_1x_2^3x_3 + \frac14x_0x_1x_2x_3^3,$

$\frac14x_1^4x_2^2 + \frac14x_1^2x_2^4 + \frac14x_0^4x_3^2 + \frac14x_0^2x_3^4,$

$ \frac14x_0^4x_1^2 + \frac14x_0^2x_1^4 + \frac14x_2^4x_3^2 + \frac14x_2^2x_3^4$

My question is, if one can find a way to associate to each such a homogenous invariant a unique triangle. I am struggling to find out what the set of vertices $V$ with $|V|=9$ is.

**Edit**:
I found some way for the case $\deg=6$ with a brute force attack in SAGEMATH.
But this attack doesn't work for $\deg=26$ since there I have to compute the common zeros of the 231 homogenous polynomials over some finite field. The problem I am facing is that the ideal generated by the homogenous invariants has dimension 2. But for a finite field there should be at most finitely many zeros. Any idea how to solve this in SAGEMATH (Computing the common zeros over some finite field?)

J is the ideal of interest:

```
sage: [I.gens() for I in J.minimal_associated_primes()]
[[x1 + x2, x0 + x3], [x2 + x3, x0 + x1], [x1 + x3, x0 + x2]]
sage: J.dimension()
2
```

https://pure.tue.nl/ws/portalfiles/portal/2157517

https://people.maths.bris.ac.uk/~matyd/GroupNames/1/ES+(2,2).html

that, I'd be convinced something is going on. $\endgroup$ – Gro-Tsen Jun 12 '18 at 9:47