Suppose we have a
[combinatorial bracelet](http://en.wikipedia.org/wiki/Bracelet_(combinatorics)) composed of natural numbers. 


(Two bracelets are equivalent if you can get from one to the other via rotation or reflection.)


What is the number of different bracelets whose elements sum up to a previously fixed natural number N?


Also, are there any results if we add a constraint that the number of beads on the bracelet is always odd?


P.S. Any good upper bounds are also helpful.


(**EDITED** in the light of the comments below)



Number of partitions of a number on a combinatorial bracelet
There is almost a bijection between your partition bracelets adding to $n$ and bracelets of length $n$ with $2$ colors. Let the colors be pluses "+" and commas "," and put a $1$ between each two beads. Then the bracelet $++,$ corresponds to the partition bracelet $1+1+1,$ or $(3)$. The bracelet 
$+,+,$ corresponds to $1+1,1+1,$ or $(2~2)$. The bracelet $,,,,,$ corresponds to $(1~1~1~1~1)$. The exception is that there is no partition bracelet which corresponds to $+++...+$, so there is one more bracelet of length $n$ than there are partition bracelets summing to $n$. So, use [the formula for the count of bracelets](http://en.wikipedia.org/wiki/Bracelet_%28combinatorics%29) and subtract $1$. 


To restrict to the case where there are an odd number of terms, you restrict to an odd number of commas. I don't know whether the formula is as simple as the previous. 


