Let me contribute some material here to help you get started and
initiate additional research. I must ask you to consult two previous
posts, however, where essentially 100% of the answer to your question
is already documented. These are
Using the notation from the above, for the rotations we must replace
the count $P_d(k)$ by a generating function. By construction the
$P_d(k)$ arise for permutations with $d$ cycles of length $n/d$ which
owing to the coloring being proper requires a properly colored cycle
of length $d$ call it $\beta$ with at most the given number of colors
(the $d$ cycles are adjacent and monochrome by Burnside). This cycle
is repeated around the bracelet. Therefore we have solved the problem
if we can construct a generating function by colors of properly
colored cycles, where symmetries are not taken into account. This is
done recursively using a memoized algorithm that classifies generating
functions of paths by the first and last color of the paths they
represent, which are then used to build cycles. Now if a color
appears on one of these cycles we must replace its variable $C$ in the
generating function by $C^{n/d}$ because the cycle $\beta$ induces the
colors of the $d$ cycles which make up the permutation. With the
reflections which only contribute when $n$ is even where the axis of
reflection passes through opposite vertices we need a generating
function of properly colored paths with no symmetry and we take one of
these to constitute the colors on one side and induce the ones on the
other. This means we replace each variable from the generating
function by its square, compensating for the two colors which appear
on the fixed points, which contribute only once. This is basically
all. Here is the generating function for proper non-isomorphic
colorings of a nine-bracelet using at most three colors as requested
by OP:
$${C_{{1}}}^{4}{C_{{2}}}^{4}C_{{3}}+3\,{C_{{1}}}^{4}{C_{{2}
}}^{3}{C_{{3}}}^{2}+3\,{C_{{1}}}^{4}{C_{{2}}}^{2}{C_{{3}}
}^{3}+{C_{{1}}}^{4}C_{{2}}{C_{{3}}}^{4}\\+3\,{C_{{1}}}^{3}{
C_{{2}}}^{4}{C_{{3}}}^{2}+8\,{C_{{1}}}^{3}{C_{{3}}}^{3}{C
_{{2}}}^{3}+3\,{C_{{1}}}^{3}{C_{{2}}}^{2}{C_{{3}}}^{4}+3
\,{C_{{1}}}^{2}{C_{{2}}}^{4}{C_{{3}}}^{3}\\+3\,{C_{{1}}}^{2
}{C_{{2}}}^{3}{C_{{3}}}^{4}+C_{{1}}{C_{{2}}}^{4}{C_{{3}}}
^{4}.$$
Therefore the answer to the query is
$$\bbox[5px,border:2px solid #00A000]{8.}$$
Here is an excerpt for the case of a twelve-bracelet with at most $5$
colors:
$$\ldots +16\,C_{{1}}C_{{2}}{C_{{3}}}^{6}
{C_{{4}}}^{2}{C_{{5}}}^{2}+10\,C_{{1}}C_{{2}}{C_{{3}}}^{6
}C_{{4}}{C_{{5}}}^{3}+3\,C_{{1}}C_{{2}}{C_{{3}}}^{6}{C_{{
5}}}^{4}\\+15\,C_{{1}}C_{{2}}{C_{{3}}}^{5}{C_{{4}}}^{5}+153
\,C_{{1}}C_{{2}}{C_{{3}}}^{5}{C_{{4}}}^{4}C_{{5}}+408\,C_
{{1}}C_{{2}}{C_{{3}}}^{5}{C_{{4}}}^{3}{C_{{5}}}^{2}\\+408\,
C_{{1}}C_{{2}}{C_{{3}}}^{5}{C_{{4}}}^{2}{C_{{5}}}^{3}+153
\,C_{{1}}C_{{2}}{C_{{3}}}^{5}C_{{4}}{C_{{5}}}^{4}+15\,C_{
{1}}C_{{2}}{C_{{3}}}^{5}{C_{{5}}}^{5}\\+3\,C_{{1}}C_{{2}}{C
_{{3}}}^{4}{C_{{4}}}^{6}+153\,C_{{1}}C_{{2}}{C_{{3}}}^{4}
{C_{{4}}}^{5}C_{{5}}+1014\,C_{{1}}C_{{2}}{C_{{3}}}^{4}{C_
{{4}}}^{4}{C_{{5}}}^{2}\\+1783\,C_{{1}}C_{{2}}{C_{{3}}}^{4}
{C_{{4}}}^{3}{C_{{5}}}^{3}+\ldots$$
The Maple code for this is shown below. It includes a very basic
enumeration routine that confirmed the results from Burnside for all
examples that were examined.
with(combinat);
with(numtheory);
P := (d,k) -> (k-1)^d + (-1)^d*(k-1);
chr_bracelet_uniq :=
proc(n, k)
local res, d;
res := 1/2/n*add(phi(n/d)*P(d,k),
d in divisors(n));
if type(n, even) then
res := res +
1/4*k*(k-1)^(n/2);
fi;
res;
end;
chr_count :=
proc(gf)
local vars, v, sl;
vars := indets(gf);
sl := [seq(v=1, v in vars)];
subs(sl, gf);
end;
chr_gf_rec :=
proc(len, cols, first, last)
option remember;
local c1, c2, res;
if len = 1 then return 0 fi;
if len = 2 then
if first <> last then
return C[first]*C[last]
else
return 0;
fi;
fi;
res := 0;
if len = 3 then
for c1 to cols do
if c1 <> first and c1 <> last then
res := res +
C[first]*C[c1]*C[last];
fi;
od;
return res;
fi;
for c1 to cols do
if c1 <> first then
for c2 to cols do
if c2 <> last then
res := res +
C[first]*C[last] *
chr_gf_rec(len-2, cols, c1, c2);
fi;
od;
fi;
od;
expand(res);
end;
chr_gf_path :=
proc(len, cols)
local c1, c2;
if len = 1 then
return add(C[c1], c1=1..cols);
fi;
add(add(chr_gf_rec(len, cols, c1, c2),
c2 = 1..cols), c1=1..cols);
end;
chr_gf_cycle :=
proc(len, cols)
local c1, c2;
if len=1 then return 0 fi;
add(add(`if`(c1 <> c2,
chr_gf_rec(len, cols, c1, c2), 0),
c2 = 1..cols), c1=1..cols);
end;
chr_gf_bracelet_uniq :=
proc(n, k)
option remember;
local res, d, sl, q, c1, c2;
res := 0;
for d in divisors(n) minus {1} do
sl := [seq(C[q] = C[q]^(n/d), q=1..k)];
res := res +
phi(n/d)*subs(sl, chr_gf_cycle(d, k))/2/n;
od;
if type(n, even) then
sl := [seq(C[q] = C[q]^2, q=1..k)];
for c1 to k do
for c2 to k do
res := res +
expand(subs(sl, chr_gf_rec(n/2+1, k, c1, c2))
/C[c1]/C[c2]/4);
od;
od;
fi;
res;
end;
ENUM :=
proc(n, k)
option remember;
local orbits, rec, col, gf, term;
orbits := table();
rec :=
proc(sofar)
local orbit, rot, rseq, q, c;
if nops(sofar) < n then
for c to k do
if sofar[-1] <> c then
rec([op(sofar), c]);
fi;
od;
return;
fi;
if sofar[1] = sofar[-1] then
return;
fi;
orbit := [];
for rot from 0 to n-1 do
rseq :=
[seq(sofar[1+((q+rot) mod n)],
q=0..n-1)];
orbit := [op(orbit), rseq];
rseq :=
[seq(rseq[n+1-q], q=1..n)];
orbit := [op(orbit), rseq];
end;
orbits[sort(orbit)[1]] := 1;
end;
for col to k do rec([col]) od;
gf := 0;
for term in [indices(orbits, 'nolist')] do
gf := gf +
mul(C[col], col in term);
od;
gf;
end;
Best Answer
There are $10$ ways to arrange the red balls. Of these:
two arrangements leave three adjacent free spots and one solo free spot: $$ R\,\_\,\_\,\_R\,\_R\\ R\,\_R\,\_\,\_\,\_R $$For each of these: Whichever colour is in the middle of the three free spots must also be in the solo spot, so 2 possibilities for each of these arrangements
six arrangements leave two adjacent spots and two solo spots: $$ \,\_\,\_R\,\_R\,\_R\\ \,\_R\,\_\,\_R\,\_R\\ \,\_R\,\_R\,\_\,\_R\\ R\,\_\,\_R\,\_R\,\_\\ R\,\_R\,\_\,\_R\,\_\\ R\,\_R\,\_R\,\_\,\_ $$ and one leaves two pairs of adjacent spots: $$ R\,\_\,\_R\,\_\,\_R $$For each of these: the colours must differ in a pair of adjacent spots, but they can be swapped around, so 4 possibilities for each of these arrangements
one arrangement leaves no adjacent spots: $$ \,\_R\,\_R\,\_R\,\_ $$ Here there are six possible arrangements of the remaining $4$ balls
In total: $$ 2\cdot 2 + 7\cdot 4 + 1\cdot 6 = 38 $$ I don't think there is a more "elegant" way to do this than simply splitting into cases and count. The interactions are too complicated to make inclusion-exclusion viable, and there isn't a simple formula for "non-adjacent" that we can apply. Of course, how you decide to split into cases changes how easy the calculations are and how easily you miss a case or double count.