enumerating necklaces with certain conditions Suppose I have a necklace with $2k$ regions, and I have $n\geq k$ indistinguishable beads which I can string onto these regions. So each region contains $n_{j}$ (possibly zero) beads such that $n_{1}+\dots + n_{2k} = n$. Moreover, suppose that every second region must contain at least 1 bead, i.e. $n_{2i}\geq1$ for $i = 1,\dots,k$.
How many different necklaces can I make? In other words I want to consider possibilities up to rotations (but not flipping - rotational symmetry but not dihedral symmetry).
I've seen some great answers here on stackexchange such as
Summation with combinations
Primitive binary necklaces
but I have not been successful in tweaking those answers to suit my problem. I have come up with a generating function but it doesn't seem to have the same flavour as the linked solutions that involve the Polya enumeration theorem, and totient functions etc, so I worry I'm working in the wrong direction.
 A: Hint: As you have to place $k$ beads in each alternate region ($n_{2i} = 1$ for $1 \le i \le k$). Your problem really reduces to ($n-k$) beads being placed in $2k$ regions. This is same as possible combinations of $2k$ numbers adding to ($n-k$). Numbers can be zero and can also repeat. Please see this post for help on that - Number of possible combinations of x numbers that sum to y.
But given this is a necklace, you have to also consider the rotational symmetry.
EDIT:
Your problem is more complicated because it combines two problems in one -

*

*Arrange indistinguishable beads in $2k$ regions adding to $n$ beads [or, ($n-k$) beads. Same thing.]

*Arrange $n$ beads of $m$ colors (indistinguishable other than the color) to make different necklaces [I will explain how through an example]

The example of $m$ indistinguishable seats and $n$ indistinguishable people (with $n \le m$) is your problem as some seats can be empty like some regions in the necklace can be empty but it is a much reduced problem as none of the seats can have more than one person. In your case each region can have multiple beads and can widely vary in number.
Let me take the same example we have gone over before -
$k = 3, 2k = 6$ regions, $n = 6$ beads
We place $3$ beads in $3$ alternate regions so we are now left with the problem of placing $3$ beads in $6$ regions. Combination of $6$ numbers adding to $3$ -
a) $\{3,0,0,0,0,0\}$ -> $1$ distinct number and $5$ empty ($0$) places in a circle
= $\dfrac{(6-1)!}{5!} = 1$ (which we anyway know).
b) $\{2,1,0,0,0,0\} -> 2$ distinct numbers and $4$ empty places in a circle
= $\dfrac{(6-1)!}{4!}$ = 5
c) $\{1,1,1,0,0,0\} -> 3$ indistinct numbers and $3$ empty places in a circle
Now this is where the formula used in a) and b) would not work as there is no way to distinguish $1$ from itself. So we go to the problem of $3$ beads each of $2$ colors - "$1$" and "$0$". Given we have $3$ beads of two colors each and $3$ is a prime number, it is easier to solve. Given it is a small number, we can just draw and find out too. Otherwise we have to apply Polya theorem.
Anyway, this case = $4$.
Adding $a), b), c)$, we get the answer $10$.
Now if $n = 8$ and $2k = 6$, we are talking about $n - k = 5$. Let's see how many scenarios are there?
$ i) \{5,0,0,0,0,0\}, ii) \{4,1,0,0,0,0\}, iii) \{3,2,0,0,0,0\}, iv) \{2,2,1,0,0,0\}, v) \{3,1,1,0,0,0\}, vi) \{2,1,1,1,0,0\}, vii) \{1,1,1,1,1,0\}$.
$i), ii), iii)$ are easy to find out - similar to cases a) and b) in our previous example. $iv)$ is a similar problem to c) and we have to apply Polya theorem - the problem is of $6$ beads of $3$ different colors - 3 beads of "$0$" color, $2$ beads of "$2$" color and $1$ beads of "$1$" color. $v)$ and $vi)$ are again beads of $3$ colors and once we know $iv)$, we can just multiply by 3. But $vii)$ is beads of 2 colors.
So I think the way you have to go about it is to have a Polya function that returns the number of arrangements for $n$ beads of $m$ colors with ($\sum_{i=1}^m n_i = n$).
Then the next step is to find all distinct combinations of $2k$ numbers adding to $n$. For cases where $2k$ break-up of $n$ is all distinct numbers (other than empty regions or "$0$"), we can easily find the arrangements. For all other cases of repeating numbers, we need to call the Polya function with $m$ and array $n_i$ as inputs.
