Number of partitions of a number on a combinatorial bracelet Suppose we have a
combinatorial bracelet 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)
 A: 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 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. 
A: This is a transform which tightens the problem and asks the reader to consult the bracelet literature.
Consider first the question as specified with the additional proviso that the number of beads is fixed at k.
The number of such bracelets is the same as the number of bracelets with k black and n white beads, if 0 is a color on the original bracelet, otherwise k black and n-k white beads.  I assume 0 is not a color and let someone else deal with the case that 0 is a color.  I also assume k is at most n and at least 1.
Since many of the black and white bracelets do not map to themselves under rotation and reflection, we get a lower bound of (n choose k)/2k.  For more precise values, one needs to look at cyclic bracelets with a period p dividing gcd(k,n)  as well as for each p considering those invariant under reflection.  This analysis should be part of the bracelet literature.
Finally, as the problem did not fix k, one needs presumably to sum over all considered k to get the answer, which should be greater than the number of partitions of n.
Gerhard "Email Me About System Design" Paseman,     2011.06.23 
