If I understand this correctly these partitions are colorings of the
faces under rotational symmetry with the colors representing
membership in a partition and the colors being swappable (symmetric
group acting on them). This enumeration problem can be solved by
Power Group Enumeration and indeed this was done for the twelve
edges rather than the six faces at the following MSE link
I. To obtain an
answer to the present query the only change is to replace the cycle
index of the edges under rotational symmetries by the one of the
faces, which was computed at this MSE link
II and found to
be:
$$Z(G) = \frac{1}{24}
(a_1^6 + 8 a_3^2 + 6 a_1^2 a_4 + 3 a_1^2 a_2^2 + 6 a_2^3).$$
With these ingredients we obtain the classification of face colorings
by the number of colors with at most six colors to be
$${P_{{1}}}^{6}+2\,{P_{{1}}}^{3}{P_{{2}}}^{3}+2\,{P_{
{1}}}^{2}{P_{{2}}}^{4}+4\,{P_{{1}}}^{2}{P_{{2}}}^{2
}{P_{{3}}}^{2}+P_{{1}}{P_{{2}}}^{5}+3\,P_{{1}}{P_{{
2}}}^{2}{P_{{3}}}^{3}\\+2\,P_{{1}}P_{{2}}{P_{{3}}}^{4
}+5\,P_{{1}}P_{{2}}{P_{{3}}}^{2}{P_{{4}}}^{2}+2\,P_
{{1}}P_{{2}}P_{{3}}{P_{{4}}}^{3}\\+2\,P_{{1}}P_{{2}}P
_{{3}}P_{{4}}{P_{{5}}}^{2}+P_{{1}}P_{{2}}P_{{3}}P_{
{4}}P_{{5}}P_{{6}}.$$
We thus obtain e.g. five $2$-partitions and nine $3$-partitions, same
as found by OP. We also get two $5$-partitions which is correct as
well (double color adjacent or not). The Maple code for this is quite
compact and relatively straightforward once the PGE algorithm is
known.
with(combinat);
cube_face_cind :=
1/24(a[1]^6 + 8a[3]^2 + 6a[1]^2a[4]
+ 3a[1]^2a[2]^2 + 6*a[2]^3);
pet_cycleind_symm :=
proc(n)
option remember;
local l;
if n=0 then return 1; fi;
expand(1/n*add(a[l]*pet_cycleind_symm(n-l), l=1..n));
end;
pet_indets2rep :=
proc(ip)
local rep, var, deg, pos, s;
rep := []; pos := 1;
for var in indets(ip) do
for deg to degree(ip, var) do
rep :=
[op(rep), [seq(s, s=pos..pos+op(1, var)-1)]];
pos := pos + op(1, var);
od;
od;
rep;
end;
cube_face_colorings_gf :=
proc(n)
option remember;
local idx_cols, rep, res, term_a, term_b,
v_a, v_b, inst_a, len_a, len_b, p, q,
parts, alldeg, cols, col, v;
if n = 1 then return P[1]^6 fi;
idx_cols := pet_cycleind_symm(n);
res := 0;
for term_b in idx_cols do
rep := pet_indets2rep(term_b);
for term_a in cube_face_cind do
p := 1;
for v_a in indets(term_a) do
len_a := op(1, v_a);
inst_a := degree(term_a, v_a);
q := 0;
for v_b in rep do
len_b := nops(v_b);
if len_a mod len_b = 0 then
q := q + len_b*
mul(P[col], col in v_b)
^(len_a/len_b);
fi;
od;
p := p*q^inst_a;
od;
res := res +
lcoeff(term_a)*lcoeff(term_b)*p;
od;
od;
res;
parts := 0;
for cols in expand(res) do
alldeg :=
sort(map(v -> degree(cols, v),
[op(indets(cols))]));
parts := parts +
lcoeff(cols)*
mul(P[v]^alldeg[v], v=1..nops(alldeg));
od;
parts;
end;
Addendum. OP asks what we can say about the case of the cube having
no symmetries, i.e. being a strip of six slots with the group permuting
the slots being the identity. The code shown above will produce the
following generating function in that case:
$${P_{{1}}}^{6}+10\,{P_{{1}}}^{3}{P_{{2}}}^{3}+15\,{
P_{{1}}}^{2}{P_{{2}}}^{4}+15\,{P_{{1}}}^{2}{P_{{2}
}}^{2}{P_{{3}}}^{2}+6\,P_{{1}}{P_{{2}}}^{5}\\+60\,P_
{{1}}{P_{{2}}}^{2}{P_{{3}}}^{3}+15\,P_{{1}}P_{{2}}
{P_{{3}}}^{4}+45\,P_{{1}}P_{{2}}{P_{{3}}}^{2}{P_{{
4}}}^{2}+20\,P_{{1}}P_{{2}}P_{{3}}{P_{{4}}}^{3}\\+15
\,P_{{1}}P_{{2}}P_{{3}}P_{{4}}{P_{{5}}}^{2}+P_{{1}
}P_{{2}}P_{{3}}P_{{4}}P_{{5}}P_{{6}}.$$
We can verify these by inspection i.e. with five colors we must choose
two slots for the double color, giving ${6\choose 2} = 15$ or with two
colors one of which has two instances we also get ${6\choose 2} = 15$ or
at last with two colors both of which have three instances we get
$\frac{1}{2} {6\choose 3} = 10.$
Now if we are only after the count for $E_m$ being the slots so that
$Z(E_m) = a_1^m$ and $Z(S_n)$ acting on $n$ colors we can actually
compute a closed form. With PGE we must cover the cycles of the slot
permutation $\alpha$ with cycles of the permutation of the colors
$\beta$ but here we only have $m$ fixed points in $\alpha$ to cover.
Therefore if $\beta$ has $q$ fixed points we get a contribution of
$q^m.$ The combinatorial class of permutations with fixed points marked
is
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\textsc{SET}(\mathcal{U}\times \textsc{CYC}_{=1}(\mathcal{Z})
+ \textsc{CYC}_{=2}(\mathcal{Z})
+ \textsc{CYC}_{=3}(\mathcal{Z})
+ \textsc{CYC}_{=4}(\mathcal{Z})
+ \cdots).$$
This gives the EGF
$$G(z, u) = \exp\left(uz
+ \frac{z^2}{2} + \frac{z^3}{3} + \frac{z^4}{4} + \cdots\right)
\\ = \exp\left(uz-z + \log\frac{1}{1-z}\right)
= \frac{\exp(-z)}{1-z} \exp(uz).$$
We seek to turn $u^q z^n/n!$ into $q^m z^n/n!$ and use
$$q^m = \sum_{r=1}^m q^{\underline{r}} {m\brace r}.$$
We thus obtain
$$[z^n] \left. \sum_{r=1}^m {m\brace r}
\frac{\exp(-z)}{1-z} \exp(uz) z^r \right|_{u=1}
= [z^n] \sum_{r=1}^m {m\brace r} \frac{z^r}{1-z}.$$
This gives the closed form
$$\bbox[5px,border:2px solid #00A000]{
\sum_{r=1}^n {m\brace r}.}$$
Here we have confirmed a formula that could have been obtained by
inspection. In particular for $n=m$ we get the sequence of Bell
numbers:
$$1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975, \ldots$$
