TITLE: $3$-colourings of a $3×3$ table with one of $3$ colors up to symmetries
QUESTION [5 upvotes]: Color each cell of a $3×3$ table with one of $3$ colors. What is the number of ways to do so if adjacent cells have different colors?

Of course we consider two paintings the same (equivalent) if exist reflection or rotation which take one to another. So 
$$
\begin{array} {|r|r|r|}
\hline
 \color{blue}{B}& \color{yellow}{Y} &\color{red}{R}  \\
\hline
\color{red}{R}& \color{red}{R}&\color{red}{R}\\
\hline
\color{red}{R}& \color{red}{R}& \color{red}{R} \\
\hline
\end{array}
\;\;\;\;\;{\rm and} \;\;\;\;\;
\begin{array} {|r|r|r|}
\hline
  \color{red}{R}& \color{red}{R}& \color{blue}{B} \\
\hline
 \color{red}{R}& \color{red}{R}& \color{yellow}{Y} \\
\hline
\color{red}{R}& \color{red}{R}& \color{red}{R} \\
\hline
\end{array}
$$
are the same colorings.
Since marked cells are ''independent'' we can color them at random but not with all 3 colors.
\begin{array} {|r|r|r|}
\hline
 & X &  \\
\hline
X & &X \\
\hline
 & X&  \\
\hline
\end{array}
Case 1: If all $X$ are colored with the same color, then for each unmarked cell we have 2 posibilites. So in this case we have $3\cdot 2^{5}$ possibile colorings. But clearly some of them are equivalent. What should I do? Divide this with 4? Or 16? Something else?
Case 2: $Y$ is of different color then $X$. Now we have $3$ colors for $Y$ and $2$ for $X$. Rest of the places we can color $1^3\cdot 2^2$ so we have $6\cdot 2^{2}$ possibile colorings. But again reflections across midlle colum give us equivalent colorings so we should divide this by $2$?
 \begin{array} {|r|r|r|}
\hline
 & Y &  \\
\hline
X & &X \\
\hline
 & X&  \\
\hline
\end{array}
Case 3: ...
 \begin{array} {|r|r|r|}
\hline
 & Y &  \\
\hline
Y & &X \\
\hline
 & X&  \\
\hline
\end{array}
Is there more elegant aproach?

REPLY [4 votes]: Let     me     post     a     pointer.      The     following     MSE
link  from  eleven
months  ago  features  orbital chromatic  polynomials,  which  count
proper   colorings  of   a   graph  under   the   symmetries  of   its
automorphisms. There is extensive documentation at that link. The code
that was posted there is easy to apply here: the underlying graph with
edges for adjacency this the  three-by-three grid graph.  We encode it
as follows:

SQUARE3BY3 :=
proc()
option remember;

    return
    [9,
     {{1, 2}, {2, 3},
      {4, 5}, {5, 6},
      {7, 8}, {8, 9},
      {1, 4}, {2, 5}, {3, 6},
      {4, 7}, {5, 8}, {6, 9}},

     [[1,2,3,4,5,6,7,8,9], # identity
      [3,6,9,2,5,8,1,4,7], # 90 degrees
      [7,4,1,8,5,2,9,6,3], # -90 degrees
      [9,8,7,6,5,4,3,2,1], # 180 degrees

      [7,8,9,4,5,6,1,2,3], # horizontal flip
      [3,2,1,6,5,4,9,8,7], # vertical flip

      [1,4,7,2,5,8,3,6,9],    # falling diagonal
      [9,6,3,8,5,2,7,4,1]]];  # rising diagonal

end;

The Maple  command OCP(SQUARE3BY3()); then immediately  yields the
OCP:
$$P(k) = 1/8\,{k}^{9}+8\,k-{\frac {133\,{k}^{2}}{4}}-3/2\,{k}^{8}
+{\frac {33\,{k}^{7}}{4}}-{\frac {53\,{k}^{6}}{2}}
+{\frac {217\,{k}^{5}}{4}}-{\frac {291\,{k}^{4}}{4}}
+{\frac {507\,{k}^{3}}{8}}.$$
This yields for up to twelve colors the sequence
$$0, 2, 69, 1572, 19865, 153480, 830802, 3476144, 12003462,
\\35757630, 94780235, 228579252, \ldots$$
which confirms the value for three colors that was first to appear.
 Remark, as  per comments. The value $P(k)$ of  this OCP counts
the number  of colorings  using at  most $k$  colors.  We  can compute
$P'(k)$ which gives  the proper colorings using exactly  $k$ colors by
inclusion-exclusion. Here  the nodes of  the poset are subsets  $Q$ of
$[k]$ representing proper colorings using some subset of the colors in
$Q$, which is counted by $P(|Q|).$ A coloring using exactly the colors
from  some  set $R$  is  represented  by  all nodes  corresponding  to
supersets  $Q$ of  $R.$  With the  weight  being $(-1)^{k-|Q|}$  those
colorings using exactly  $k$ colors only occur at  $Q=[k]$ with weight
$(-1)^{k-|Q|} = (-1)^{0} = 1.$ A coloring using exactly $R\subset [k]$
colors is  represented by all $Q$  such that $R \subseteq  Q \subseteq
[k]$, with total weight
$$\sum_{R'\subseteq [k] \setminus R} (-1)^{k-|R\cup R'|}
= (-1)^{k-|R|}
\sum_{r=0}^{|[k]\setminus R|} {|[k]\setminus R| \choose r} (-1)^{-r}
= 0.$$
Hence only  the colorings  with exactly $k$  colors contribute  and we
find
$$P'(k) = \sum_{Q\subseteq [k]} (-1)^{k-|Q|} P(|Q|) =
\sum_{q=0}^k {k\choose q} (-1)^{k-q} P(q).$$
This gives the finite sequence
$$0, 2, 63, 1308, 12675, 56520, 120960, 120960, 45360, 0, \ldots$$
because it  is clearly impossible  to color  the grid using  more than
nine distinct colors.  Observe also  the entry for three colors, which
is $P(3) - {3\choose 2} P(2) = 69 - 3\times 2$ i.e. we have subtracted
the colorings using two colors (there  are no coloring using one color
and hence $P(2)$ counts colorings  with exactly two colors). Also note
that with nine colors all orbits have the same size, namely eight, and
indeed we  obtain $9!/8 =  45360.$ As to  what happens when  there are
more than nine colors we can recover $P(k)$ as follows:
$$\sum_{q=0}^9 {k\choose q} P'(q).$$
Addendum.  The reader  might be  interested  to know  that we  can
compute  the  OCP  for  larger  grids, using  the  following  code  in
conjunction with the quoted link:

SQUARE :=
proc(n)
    option remember;
    local src, rot, automs, edges, v2n;

    src := [seq(seq([p, q], q=0..n-1), p=0..n-1)];

    edges :=
    {seq(seq({[p, q], [p+1, q]},
             p=0..n-2), q=0..n-1),
     seq(seq({[p, q], [p, q+1]},
             p=0..n-1), q=0..n-2)};

    rot := v -> [v[2], n-1-v[1]];

    automs :=
    [src, # identity
     map(rot, src),                    #  90 degrees
     map(v -> rot(rot(v)), src),       # 180 degrees
     map(v -> rot(rot(rot(v))), src),  # 270 degrees

     map(v -> [n-1-v[1], v[2]], src),  # horizontal flip
     map(v -> [v[1], n-1-v[2]], src),  # vertical flip

     map(v -> rot([n-1-v[1], v[2]]),
         src),  # rising diagonal
     map(v -> rot(rot(rot([n-1-v[1], v[2]]))),
         src)]; # falling diagonal

     v2n :=
     [seq(seq([p, q] = 1 + p*n + q, q=0..n-1), p=0..n-1)];

    [n*n, subs(v2n, edges), subs(v2n, automs)];
end;

We obtain for a four-by-four the OCP
$$1/8\,{k}^{16}-3\,{k}^{15}+{\frac {69\,{k}^{14}}{2}}
-{\frac {2015\,{k}^{13}}{8}}+{\frac {10437\,{k}^{12}}{8}}
\\-{\frac {20307\,{k}^{11}}{4}}+15333\,{k}^{10}-{\frac {292907\,{k}^{9}}{8}}
-{\frac {848501\,{k}^{7}}{8}}+{\frac {1023195\,{k}^{6}}{8}}
\\-{\frac {240539\,{k}^{5}}{2}}+{\frac {557915\,{k}^{8}}{8}}
-{\frac {8807\,k}{4}}+{\frac {112831\,{k}^{2}}{8}}
+{\frac {683997\,{k}^{4}}{8}}-{\frac {347485\,{k}^{3}}{8}}$$
with the sequence
$$0, 1, 1155, 759759, 103786510, 4767856260, 107118740001, \ldots$$
We get for a five-by-five the OCP
$$1/8\,{k}^{25}+{\frac {69997383\,{k}^{17}}{8}}-5\,{k}^{24}
+{\frac {195\,{k}^{23}}{2}}-1233\,{k}^{22}+{\frac {45399\,{k}^{21}}{4}}
\\-80919\,{k}^{20}+{\frac {928545\,{k}^{19}}{2}}
-{\frac {17590911\,{k}^{18}}{8}}-{\frac {118477969\,{k}^{16}}{4}}
+{\frac {172111059\,{k}^{15}}{2}}
\\-{\frac {1726958987\,{k}^{14}}{8}}
+{\frac {3754019329\,{k}^{13}}{8}}-{\frac {1770719251\,{k}^{12}}{2}}
\\+{\frac {5797425049\,{k}^{11}}{4}}-2053661272\,{k}^{10}
+{\frac {20055169857\,{k}^{9}}{8}}+{\frac {9236896437\,{k}^{7}}{4}}
\\-{\frac {6780818949\,{k}^{6}}{4}}+{\frac {8083053959\,{k}^{5}}{8}}
-{\frac {20932696169\,{k}^{8}}{8}}+4017958\,k
\\-{\frac {145271789\,{k}^{2}}{4}}-{\frac {3768579695\,{k}^{4}}{8}}
+{\frac {1292510453\,{k}^{3}}{8}}$$
with the sequence
$$0, 2, 76332, 2557101612, 6352711134515, 2747239197568620,
\\378972203462839707, \ldots$$