Painting a circle that is divided into $6$ pieces Given is a circle that is divided into $6$ identical pieces and we have $4$ different colours. How many ways can we paint it under the condition that no adjacent pieces have the same colour?
You can use all $4$ colours:

But you don't have to:

However, the second one is NOT considered as a different case because it's only rotated:

I thought the answer was $4\cdot3\cdot3\cdot3\cdot3\cdot2$ at first but it should be less since we can rotate it.
 A: Call $C_6$ the number of circles painted with four colors where we do not identify circles linked by rotation.
If you want to regard rotated circles as identical, you should distinguish cases according to the symmetries of your circles:
No circle stays invariant under rotation by one wedge-length because it is forbidden to paint the circle with one color.
A circle stays invariant under rotation by two wedge-lengths if two colors alternate, so there are six such circles (just choose the two colors) that correspond to 12 unrotated ones.
A circle stays invariant under rotation by three wedge-lengths if three colors repeat their pattern, there are eight such choices (choose the three colors and then choose one of the two cyclic orders) that correspond to 24 unrotated ones.
These two cases are mutually exclusive and other rotations do not give as anything new.
The remaining circles are grouped into groups of six, so the remaining circles are $(C_6-12-24)/6$.
This give $C_6/6+8$ painted circles.
So, how do you determine $C_6$? Look at circles divided into $k$ pieces and set up a recursion: Color one wedge after another. Either it works out or you find something which is essentially a coloring of $k-1$ wedges. We obtain
$$C_k=4\cdot3^{k-1} - C_{k-1}.$$
Therefore, $C_k=4\cdot 3^{k-1} - 4 \cdot 3^{k-2} \pm \dots (-1)^{k-1} 4\cdot 3^2 -(-1)^{k-1}C_2 = 4\cdot 3^{k-1} - 4 \cdot 3^{k-2} \pm \dots (-1)^{k-1} 4\cdot 3^2 +(-1)^{k}4\cdot 3 = 4\cdot 3 \dfrac{3^{k-1}-(-1)^{k-1}}{3-(-1)}=  3( 3^{k-1}-(-1)^{k-1})$ .
So, we get $C_6 = 3(3^5+1)=3(243+1)=3\cdot 244$.
The final answer is $C_6/6+8 = 122+8 =130$.
A: There is a wonderful theory, called Polya enumeration theory, that enables precise counting  of coloring patterns when a symmetry group is present. In the  case at stake the involved numbers are so small that an ad hoc procedure seems more efficient than learning this theory in toto.
Consider admissible $4$-colorings $x=(x_0,x_1,\ldots, x_n)$ of the linear graph with  vertices $v_k$ $\>(0\leq k\leq n)$. Denote by $f_n$ the number of such $x$ having $x_n\ne x_0$, and by $g_n$ the number of such $x$ having $x_n=x_0$. Then $g_n$ is at the same time the number of admissible $4$-colorings of the cyclic graph with $n$ vertices. It is easily seen that one has the following recursion scheme:
$$f_1=12, \quad g_1=0,\qquad f_{n+1}=2 f_n+ 3 g_n,\quad g_{n+1}=f_n\qquad(n\geq1)\ .\tag{1}$$
Computation then shows that $g_2=12$, $g_3=24$, and $g_6=732$. (Solving $(1)$ with the "Master Theorem" would give $g_n=3^n+3(-1)^n$.)
In a second step we have to "quotient out" the rotations. This means that we have to descend from individual colorings to types of colorings. It turns out that in this process the symmetries of the individual colorings have to be taken into account.
There are $g_2$ colorings having period $2$. Since two colorings differing only by a rotational shift are considered equivalent we deduce that there are $c_2={g_2\over2}=6$ types of such colorings (this corresponds to the six ways of choosing two colors out of four).
There are $g_3$ colorings having period $3$, and  we deduce that there are $c_3={g_3\over3}=8$ types of such colorings (this corresponds to the four ways of choosing three colors out of four and two ways of arranging them cyclically).
All other colorings $x$ have no rotational symmetry. Therefore each rotational shift equivalence class of such $x$ has exactly $6$ elements. It follows that there are
$$c_6={1\over6}(g_6-g_2-g_3)=116$$
types of such colorings.
The total number of types is therefore given by
$$c_2+c_3+c_6=130\ ,$$
in accordance with the number found by Mark McClure.
