<s>[INST] Number of possible unique patterns of $5$ points on an $8\times 8$ grid. I'm after a specific answer and workings/formula for the number of possible unique patterns that $5$ points can have on an $8\times 8$ grid. The pattern must be unique and not match another pattern when it's rotated or mirrored. Can the workings be explained in laymans terms? Thanks
 [/INST] We solve the case of the points going into the slots between the grid,
so there are  $64$ slots. We put  at most one point into  one slot. We
use  the Polya  Enumeration Theorem  (PET), which  requires the  cycle
index of the action of the symmetries on the slots. Supposing the grid
is $n\times  n$ with  $n$ even  we get for  the rotations,  first, the
identity:
$$a_1^{n^2}.$$
Next, two rotations by $90$ degrees and $270$ degrees:
$$2 a_4^{n^2/4}.$$
The rotation by $180$ degrees:
$$a_2^{n^2/2}.$$
For reflections there is the horizontal and the vertical reflection:
$$2 a_2^{n^2/2}$$
and the reflection in the two diagonals:
$$2 a_1^n a_2^{(n^2-n)/2}.$$
We get the cycle index
$$Z(G) = \frac{1}{8}
\left(a_1^{n^2} + 2 a_4^{n^2/4} + a_2^{n^2/2}
+ 2 a_2^{n^2/2} + 2 a_1^n a_2^{(n^2-n)/2}\right)$$
or
$$\bbox[5px,border:2px solid #00A000]{Z(G) = \frac{1}{8}
\left(a_1^{n^2} + 2 a_4^{n^2/4}
+ 3 a_2^{n^2/2} + 2 a_1^n a_2^{(n^2-n)/2}\right).}$$
For an odd number of points call is $m$ we thus have
$$[z^m] Z(G; 1+z) \\=
[z^m] \frac{1}{8}
\left((1+z)^{n^2} + 2 (1+z^4)^{n^2/4}
+ 3 (1+z^2)^{n^2/2} + 2 (1+z)^n (1+z^2)^{(n^2-n)/2}\right)
\\ = [z^m] \frac{1}{8}
\left((1+z)^{n^2} + 2 (1+z)^n (1+z^2)^{(n^2-n)/2}\right).$$
This is
$$\bbox[5px,border:2px solid #00A000]{\frac{1}{8} {n^2\choose m}
+ \frac{1}{4} \sum_{k=0}^{n/2-1}
{n\choose 2k+1} {(n^2-n)/2\choose (m-2k-1)/2}}$$
where $n$ is even and $m$ is odd.
 We get for $n=8$ and $m=5$ the answer
$$\bbox[5px,border:2px solid #00A000]{954226.}$$
</s>