How many distinct patterns exist for a 5x5 grid by filling 3 colors? Using 3 colors to fill in a $5\times5$ grid (you don't have to use all colors), then how many distinct patterns exist? The "distinct" means we have to consider the symmetry.
Any effective approach is appreciated.
 A: Supposing that we want non-isomorphic patterns we need the cycle index
$Z(G)$ of the automorphism group $G$ acting on the cells of the grid.
We assume full symmetry, which includes rotations and reflections.
We  now compute  this  cycle  index. There  is  the identity  which
contributes $$a_1^{25}.$$
The  $90$  degree and  the  $270$  degree  rotation contribute  $$2a_1
a_4^6.$$
The $180$ degree rotation contributes $$a_1 a_2^{12}.$$
The horizontal and vertical reflections contribute
$$2a_1^5 a_2^{10}.$$
The reflections in the diagonals contribute
$$2a_1^5 a_2^{10}.$$
This yields the cycle index
$$Z(G) = \frac{1}{8}
\left(a_1^{25} + 2a_1 a_4^6 + a_1a_2^{12} + 4 a_1^5 a_2^{10}\right).$$
With at most  $N$ colors we thus have (by Burnside  the colors must be
constant on each cycle and there are $N$ colors available)
$$Q_N = \frac{1}{8} 
(N^{25} + 2N^7 + N^{13}+ 4N^{15}).$$
We get for at most three colors
$$105918450471.$$
Remark.  If we  seek the  number  of colorings  using an  exact
number of  distinct colors  like three colors  and call  the colorings
with at most $N$ colors $Q_N$ we have by inclusion-exclusion
$$\sum_{q=1}^N {N\choose q} (-1)^{N-q} Q_q.$$
This  produces a  finite sequence  that starts  at $N=1$  and  ends at
$N=25$ (we cannot place more than $25$ distinct colors on the grid).
We obtain
$$1, 4211742, 105905815242, 140314385087520, 36550287370308180,\ldots,
\\ 1938901255416373248000000, 0,\ldots$$
We have  $$1938901255416373248000000 = \frac{25!}{8}$$  when $N=25$ as
all orbits are the same size in this case.
