Black and white beads on a circle There are $n$ beads placed on a circle, $n\ge 3$. They are numbered in random order as viewed clockwise. Beads for which the number of the previous bead is less than the number of a next bead are painted in white color,and others - in black.
Two colourations that can be made equal by rotation are considered identical. How many different colourations can occur?
I've write a programm and for $n=3...11$ I've got answers $2 , 1 , 6 , 7 ,  18 , 25 , 58 , 93 , 186$ 
 A: Let $a_n$ be the number of configurations for this problem
and let $b_n$ be the number of two-colored necklaces with $n$ beads (no flips allowed).
It is well-known that $b_n=\frac 1n\sum_{d\mid n}\phi(d)2^\frac nd$ (this is OEIS A000031).
Computer runs suggest that for odd $n$ all these patterns also occur for this problem,
except the two monochromatic ones, so $a_n=b_n-2$.
They also suggest that for even $n$ all patterns occur except the ones where one color occurs every second spot.
In this case the number of necklaces where black occurs every second spot is $b_{\frac n2}$.
We get the same number for the patterns where white occurs every second spot,
and then we have double counted the one pattern where black and white alternate all around the necklace,
so we get $a_n=b_n-2b_{\frac n2}+1$.
For the proof of these observations we look at the cases of even $n$ and odd $n$ separately.
First we prove the case for even $n$. Note that in this case the odd beads and the even beads
form two disjoint circular sequences. The colors on the odd beads prescribe the ordering on the even beads (and v.v.),
but there is a huge amount of freedom in assigning the values, and what is more important:
the value assignments on the subsequences are independent: any value assignment on the
odd beads consistent with the colors on the even beads can be combined with any value assignment
on the even beads that is consistent with the colors on the odd beads (as long as no values are
duplicated).
We first show that for any non-monochromatic color sequence on the odd beads we can find a value assignment
for the even beads that is consistent with it. Let $n=2t$ and let $p_1,\ldots,p_t$ be the odd beads
and $q_1,\ldots,q_t$ the even beads, so that the complete necklace is $p_1,q_1,p_2,q_2,\ldots$.
Because the color sequence is not monochromatic and because we have cyclic arrangements
we may assume that $p_1$ is black and $p_2$ is white.
We give $q_1$ the value 1. Now assume we have assigned a value to $q_i$.
If $p_{i+1}$ is white, we assign value $q_i+n$ to $q_{i+1}$, otherwise
we assign value $q_i-1$.
Because we have less than $n$ values to assign this we end up in $q_t$ having a larger value than $q_1$,
which shows that indeed all value assignments on the even beads are consistent with the colors on the odd beads.
We can similarly find a value assignment for the odd beads consistent with the colors on the even beads
(just make sure to use a different range of numbers).
Combining these value assignments and 'flattening' the values to make the valueset equal to $1,\ldots,n$
finishes the proof that for even $n$ any color sequence that is not monochromatic on the odd beads
or the even beads is realizable.
Now we turn to the case that $n$ is odd. In this case we take $n=2t-1$ and we consider the
beads to be numbered $p_1,p_{t+2},p_2,p_{t+3},\ldots,p_n,p_{t+1}$. Since we have excluded
monochromatic patterns we may again assume that $p_1$ is black and $p_2$ is white.
We use the same procedure. We start with value 1 on $p_{t+1}$ and assuming we have
assigned a value to $p_i$ we assign value $p_i+n$ to $p_{i+2}$ if the bead after $p_i$ is white
and value $p_i-1$ if the bead after $p_i$ is black.
This will eventually assign a value to $p_n$ that is larger than 1, so again we have managed
a value assignment consistent with the colors.
Btw, the values found by computer simulation were (starting with $n=3$)
2, 1, 6, 7, 18, 25, 58, 93, 186, 325
which coincides with the given formulas and with the values reported in the question. 
A: The facts observed in the answer by Leen Droogendijk can easily be explained. I recall that it says that for odd$~n$ all colour patters except the monochromatic (all-white or all-black) ones can be obtained, and for even $n$ all patters except those where either the odd-position or the even-position subsequences are monochromatic (and possibly both simultaneously).
Since one is not counting the actual numberings of the beads, let us start with a given colouring and try to see if the conditions it gives on the associated permutation (which are a collection of inequalities among pairs of permutation entries) are contradictory (in which case the colouring is rejected) or not. It is easy to see that a collection of inequalities is contradictory if it contains any oriented cycles, and consistent otherwise. (Formally, in the latter case the inequalities define by transitivity a partial ordering among the positions of the entries, and partial orderings can always be extended to a total ordering; this prescribes a permutation.)
Now starting at the first bead, write down the colour of every other bead (so starting with positions $1$, $3$, $5$,...) wrapping back to the beginning of the necklace at the end, and continuing until bead $1$ is encountered again. If $n$ is odd this happens when all beads have been seen (and two tours of the necklace are completed) while if $n$ is even it happens after the first tour, after which all odd-position beads have been seen. In the latter case similarly make a separate tour for the even-position beads. So we have one or two cyclic chains of colours (according as $n$ is odd or even), and what we wish to show is that the colouring is contradictory if and only if at least one of those chains is monotonic.
Between each pair of successive colours (in cyclic order) of one chain we can place the permutation entry (position) whose value is involved in the both inequalities for those colours. (For instance if the first chain starts BW... then between the B and the W is placed permutation entry $p_2$, which is involved in the inequality $p_n>p_2$ for the black bead coming from position$~1$ and in the inequality $p_2<p_4$ for the white bead coming from position$~3$.) Thus the permutation entries are arranged in one or two cyclic chains linked by inequalities, and all required inequalities are thus taken into account. As observed above the inequalities are contradictory if they are all of the same kind (all "$<$" or all "$>$"; either gives an oriented cycle), and else non-contradictory.
A: Let $a_n$ be the number of configurations for this problem
and let $b_n$ be the number of two-colored necklaces with $n$ beads (no flips allowed).
It is well-known that $b_n=\frac 1n\sum_{d\mid n}\phi(d)2^\frac nd$ (this is OEIS A000031).
Computer runs suggest that for odd $n$ all these patterns also occur for this problem,
except the two monochromatic ones, so $a_n=b_n-2$.
They also suggest that for even $n$ all patterns occur except the ones where one color occurs every second spot.
In this case the number of necklaces where black occurs every second spot is $b_{\frac n2}$.
We get the same number for the patterns where white occurs every second spot,
and then we have double counted the one pattern where black and white alternate all around the necklace,
so we get $a_n=b_n-2b_{\frac n2}+1$.
