Polya Coloring

## 2828924

https://math.stackexchange.com/questions/2828924/coloring-cube-sides-and-vertices/

A mixed cycle index for faces and vertices combined can be of use in
this problem. We can then apply Burnside or Polya as desired.

We proceed to enumerate the permutations of this group. [...]

It follows that the cycle index of $F$ is given by

Z
(
H
# )

1
24

(

a

1

6

b

1

8

6

a

1

2

a

4

b

4

2

3

a

1

2

a

2

2

b

2

4

8

a

3

2

b

1

2

b

3

2

6

a

2

3

b

2

4

)

{\displaystyle Z(H)={\frac {1}{24}}\left(a\_{1}^{6}b\_{1}^{8}+6a\_{1}^{2}a\_{4}b\_{4}^{2}+3a\_{1}^{2}a\_{2}^{2}b\_{2}^{4}+8a\_{3}^{2}b\_{1}^{2}b\_{3}^{2}+6a\_{2}^{3}b\_{2}^{4}\right)}

Using at most $N$ colors for the faces and $M$ for the vertices we get for the number of colorings by Burnside:

H
(
N
,
M
# )

1
24

(

N

6

M

8

6

N

3

M

2

3

N

4

M

4

8

N

2

M

4

6

N

3

M

4

)

{\displaystyle H(N,M)={\frac {1}{24}}(N^{6}M^{8}+6N^{3}M^{2}+3N^{4}M^{4}+8N^{2}M^{4}+6N^{3}M^{4})}

Setting $M=1$ here we should get face colorings. We obtain

1
,
10
,
57
,
240
,
800
,
2226
,
5390
,
11712
,
…

{\displaystyle 1,10,57,240,800,2226,5390,11712,\ldots }

and we encounter [OEIS A047780](https://oeis.org/A047780) where we see that
we have the right values. Setting $N=1$ yields vertex colorings:

$$1, 23, 333, 2916, 16725, 70911, 241913, 701968, \ldots$$

which points to [OEIS A000543](https://oeis.org/A000543) which is
correct as well.

Continuing with the question of colorings that use exactly $N$ colors
for the faces and exactly $M$ for the vertices we find using Stirling numbers
for set partitions

H

X

(
N
,
M
# )

N
!
×
M
!

24

{\displaystyle H\_{X}(N,M)={\frac {N!\times M!}{24}}}

×

(

{

6

N

}

{

8

M

}

6

{

3

N

}

{

2

M

}

3

{

4

N

}

{

4

M

}

8

{

2

N

}

{

4

M

}

6

{

3

N

}

{

4

M

}

)

{\displaystyle \times \left(\left\{{\begin{array}{ll}\ 6\\N\end{array}}\right\}\left\{{\begin{array}{ll}\ 8\\M\end{array}}\right\}+6\left\{{\begin{array}{ll}\ 3\\N\end{array}}\right\}\left\{{\begin{array}{ll}\ 2\\M\end{array}}\right\}+3\left\{{\begin{array}{ll}\ 4\\N\end{array}}\right\}\left\{{\begin{array}{ll}\ 4\\M\end{array}}\right\}+8\left\{{\begin{array}{ll}\ 2\\N\end{array}}\right\}\left\{{\begin{array}{ll}\ 4\\M\end{array}}\right\}+6\left\{{\begin{array}{ll}\ 3\\N\end{array}}\right\}\left\{{\begin{array}{ll}\ 4\\M\end{array}}\right\}\right)}

{

6

N

}

{

8

M

}

6

{

3

N

}

{

2

M

}

{\displaystyle {\begin{Bmatrix}6\\N\end{Bmatrix}}{\begin{Bmatrix}8\\M\end{Bmatrix}}+6{\begin{Bmatrix}3\\N\end{Bmatrix}}{\begin{Bmatrix}2\\M\end{Bmatrix}}}

Setting $M=1$ here we get the count of face colorings with exactly $N$
colors:

$$1, 8, 30, 68, 75, 30, 0, \ldots$$

Note that for six colors, which is the maximum, the orbits have size
$24$ because all the colors are distinct and indeed $6!/24 = 30.$
Similarly with $N=1$ we get vertex colorings:

$$1, 21, 267, 1718, 5250, 7980, 5880, 1680, 0, \ldots$$

and once more for eight colors, the maximum possible, we find that
$8!/24 = 1680.$

Concluding we get for at most two vertex colors and at most four face
colors

$$H(4,2) = 44608$$

and for exactly two vertex colors and four face colors

$$H\_X(4,2) = 16552.$$

## 3827211

https://math.stackexchange.com/questions/3827211

How many necklaces can be made with two red beads, two green beads, and four violet beads?(8 total)

We may as well deploy PET here since we need the cycle index $Z(D\_8)$ of the dihedral group $D\_8$ of order $16$ to apply Burnside. We compute and average the number of assignments of colors to the eight slots fixed by permutations from each conjugacy class in $D\_8$, taking into account the order of the class. This means the assignment is constant on the cycles, so we can place exactly one color in the slots on a given cycle, substituing $a\_d$ from the index with $R^d + G^d + V^d,$ which is PET.

Consulting the following [fact sheet on necklaces and
bracelets](https://math.stackexchange.com/questions/3216181/)
we get for the cycle index of the dihedral group

D
8

{\displaystyle D\_8}

Z
(

D

8

# )

1
16

a

1

8

1
4

a

1

2

a

2

3

5
16

a

2

4

1
8

a

4

2

1
4

a

8

.

{\displaystyle Z(D\_{8})={\frac {1}{16}}a\_{1}^{8}+{\frac {1}{4}}a\_{1}^{2}a\_{2}^{3}+{\frac {5}{16}}a\_{2}^{4}+{\frac {1}{8}}a\_{4}^{2}+{\frac {1}{4}}a\_{8}.}

We seek

[

R

2

G

2

V

4

]
Z
(

D

8

;
R
G
V
)
.

{\displaystyle [R^{2}G^{2}V^{4}]Z(D\_{8};R+G+V).}

Here we assume that the OP asks for the full symmetry i.e. dihedral,
which means that the label to use is \*bracelet.\* Working through the five
terns in the cycle index we obtain

1
16

a

1

8

{\displaystyle {\frac {1}{16}}a\_{1}^{8}}

 fills with

[

R

2

G

2

V

4

]

1
16

(
R
G
V

)

8

=

1
16

(

8

2
,
2
,
4

)

{\displaystyle [R^{2}G^{2}V^{4}]{\frac {1}{16}}(R+G+V)^{8}={\frac {1}{16}}{8 \choose 2,2,4}}

1
4

a

1

2

a

2

3

{\displaystyle {\frac {1}{4}}a\_{1}^{2}a\_{2}^{3}}

[

R

2

G

2

V

4

]

1
4

(
R
G
V

)

2

(

R

2

G

2

V

2

)

3

=
[

R

2

G

2

V

4

]

1
4

(

R

2

G

2

V

2

)
(

R

2

G

2

V

2

)

3

=
[

R

2

G

2

V

4

]

1
4

(

R

2

G

2

V

2

)

4

=
[
R
G

V

2

]

1
4

(
R
G
V

)

4

{\displaystyle [R^{2}G^{2}V^{4}]{\frac {1}{4}}(R+G+V)^{2}(R^{2}+G^{2}+V^{2})^{3}=[R^{2}G^{2}V^{4}]{\frac {1}{4}}(R^{2}+G^{2}+V^{2})(R^{2}+G^{2}+V^{2})^{3}=[R^{2}G^{2}V^{4}]{\frac {1}{4}}(R^{2}+G^{2}+V^{2})^{4}=[RGV^{2}]{\frac {1}{4}}(R+G+V)^{4}}

=

1
4

(

4

2
,
1
,
1

)

.

{\displaystyle ={\frac {1}{4}}{4 \choose 2,1,1}.}

5
16

a

2

4

{\displaystyle {\frac {5}{16}}a\_{2}^{4}}

[

R

2

G

2

V

4

]

5
16

(

R

2

G

2

V

2

)

4

=

5
16

(

4

2
,
1
,
1

)

.

{\displaystyle [R^{2}G^{2}V^{4}]{\frac {5}{16}}(R^{2}+G^{2}+V^{2})^{4}={\frac {5}{16}}{4 \choose 2,1,1}.}

1
8

a

4

2

{\displaystyle {\frac {1}{8}}a\_{4}^{2}}

[

R

2

G

2

V

4

]

1
8

(

R

4

G

4

V

4

)

2

=

{\displaystyle [R^{2}G^{2}V^{4}]{\frac {1}{8}}(R^{4}+G^{4}+V^{4})^{2}=0.}

1
4

a

8

{\displaystyle {\frac {1}{4}}a\_{8}}

[

R

2

G

2

V

4

]

1
4

(

R

8

G

8

V

8

# )

{\displaystyle [R^{2}G^{2}V^{4}]{\frac {1}{4}}(R^{8}+G^{8}+V^{8})=0.}

We thus get for our answer

1
16

(

8

2
,
2
,
4

)

9
16

(

4

2
,
1
,
1

)

=

{\displaystyle {\frac {1}{16}}{8 \choose 2,2,4}+{\frac {9}{16}}{4 \choose 2,1,1}={33.}}

## 2016732

https://math.stackexchange.com/questions/2016732

How many necklaces can be made with two red beads, two green beads, and two violet beads?

I upvoted the first answer but I would like to show how to compute the cycle index $Z(D\_6)$ of the dihedral group $D\_6$ and apply the Burnside lemma and the Polya Enumeration Theorem to this problem.

Observe that the OEIS uses the convention of refering to rotational symmetry as a necklace and as a bracelet when reflectional symmetry is included, so we are working with a bracelet here.

We need to enumerate and factor the twelve permutations that
contribute to $Z(D\_6).$

There is the identity, which contributes
$$a\_1^6.$$
The two rotations by a distance of one and five contribute
$$2 a\_6.$$
The two rotations by a distance of two or four contribute
$$2 a\_3^2.$$
Finally the rotation by a distance of three contributes
$$a\_2^3.$$

There are three reflections about an axis passing through opposite
vertices, giving
$$3 a\_1^2 a\_2^2$$

and three reflections about an axis passing through the midpoints of
opposite edges, giving
$$3 a\_2^3.$$

This finally yields the cycle index
$$Z(D\_6) = \frac{1}{12}
\left(a\_1^6 + 2a\_6 + 2a\_3^2 + 3a\_1^2 a\_2^2 + 4a\_2^3\right).$$

Do the Burnside calculation first. We have three colors and two instances of each. The colors must be constant on the cycles. We now proceed to count these. We get for $a\_1^6$ the contribution ${6\choose 2,2,2}.$ There are no candidates for $a\_6$ (we do not have six instances of a color). We do not have three instances either, so zero for $a\_3^2.$ For $a\_1^2 a\_2^2$ we must choose a pair of colors for the two-cycles, giving $3\times 2\times {3\choose 2}.$

Finally for $a\_2^3$ we get six permutations of the three colors. This yields

1
12

(

(

6

2
,
2
,
2

)

3
×
2
×

(

3
2

)

4
×
6

)

=

1
12

(
90
18
24
# )

{\displaystyle {\frac {1}{12}}\left({6 \choose 2,2,2}+3\times 2\times {3 \choose 2}+4\times 6\right)={\frac {1}{12}}(90+18+24)=11.}

On the other hand Polya says that we need the coefficient

[

A

2

B

2

C

2

]
Z
(

D

6

)
(
A
B
C
)

{\displaystyle [A^{2}B^{2}C^{2}]Z(D\_{6})(A+B+C)}

 which is

[

A

2

B

2

C

2

]

1
12

(

(
A
B
C

)

6

2
(

A

6

B

6

C

6

)
2
(

A

3

B

3

C

3

)

2

3
(
A
B
C

)

2

(

A

2

B

2

C

2

)

2

4
(

A

2

B

2

C

2

)

3

)

.

{\displaystyle [A^{2}B^{2}C^{2}]{\frac {1}{12}}\left((A+B+C)^{6}+2(A^{6}+B^{6}+C^{6})+2(A^{3}+B^{3}+C^{3})^{2}+3(A+B+C)^{2}(A^{2}+B^{2}+C^{2})^{2}+4(A^{2}+B^{2}+C^{2})^{3}\right).}

Now we may drop the terms that cannot possibly produce multiples of
$A^2 B^2 C^2$ which leaves

[

A

2

B

2

C

2

]

1
12

(

(
A
B
C

)

6

3
(
A
B
C

)

2

(

A

2

B

2

C

2

)

2

4
(

A

2

B

2

C

2

)

3

)

.

{\displaystyle [A^{2}B^{2}C^{2}]{\frac {1}{12}}\left((A+B+C)^{6}+3(A+B+C)^{2}(A^{2}+B^{2}+C^{2})^{2}+4(A^{2}+B^{2}+C^{2})^{3}\right).}

Doing the coefficient extraction then yields

1
12

(

(

6

2
,
2
,
2

)

3
×
3
×
2
4

(

3

1
,
1
,
1

)

)

=

{\displaystyle {\frac {1}{12}}\left({6 \choose 2,2,2}+3\times 3\times 2+4{3 \choose 1,1,1}\right)=11.}

Here we have used the observation that we must choose a square from the first term of $(A+B+C)^2 (A^2+B^2+C^2)^2$ and then choose the two remaining squares from the second factor, which may be done in two ways.

## 1099238

https://math.stackexchange.com/questions/1099238/how-many-ways-can-you-put-8-red-6-green-and-7-blue-balls-in-4-indistinguishable

How many ways can you put 8 red, 6 green and 7 blue balls in 4 indistinguishable bins?

Partitions with no equal parts

Here is a simpler example to explain the mechanism above.

Suppose we have a rectangle marked in every corner with several points, such that the sum of points is 7 and each distribution fixes the rectangle, otherwise being acted by the Klein group.

The \*\*asymmetry cycle index\*\* is

$Z(rectangle) = 1/4 \ a\_1^3 - 3/4 \,a\_2^2 + 1/2 \ a\_4$

The asymmetry cycle index goes back to Rota 1964 and, totally counterintuitive, it involves negative terms as well as 4-cycles.

We can securely use the pet\_varinto\_cind procedure delivered above.

```
pet_varinto_cind (poly, Z(rectangle))

```

The polynomial would be $poly = (p+p^2+p^3+...p^{10} + ..+p^m)$. In fact, power 10 is more than enough to follow the example, and there is no need to sum too many terms.

We then extract the desired coefficient representing the sum of points over the four corners.

sum of points 5 -----> 1 fixed rectangle

sum of point 6 -----> 1 fixed rectangle

sum of points 7 -----> 5 fixed rectangles

sum of points 8 -----> 7 fixed rectangles

for 7 points, the fixed rectangles are

4111, 3211, 3121, 3112 and 2221

Counting partitions without equal parts follows the same mechanism.

Note the asymmetry series/polynomial is not a function of cycle index only, because it depends on some lattice of subgroups, not uniquely characterized by the cycle index.