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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Jan 22, 2014 at 1:25 vote accept Richard Stanley
Jan 22, 2014 at 0:33 answer added Per Alexandersson timeline score: 1
Jan 21, 2014 at 22:30 answer added Gjergji Zaimi timeline score: 11
Jan 21, 2014 at 15:20 answer added James Cranch timeline score: 1
Jan 21, 2014 at 7:44 comment added Greg Martin Still considering the "special" partitions of $2n-1$ as given by the $p_n(s_1,\dots)$: let $N_{n,k}$ $(n\ge1,k\ge0)$ denote the number of "special" partitions of $2n-1$ into $2^{k+1}-1$ parts. Then $N_{n,0}=1$ for all $n\ge1$, and $N_{n,k}=0$ when $n<2^k$. Emperically, it seems that $N_{n,k} = \sum_{j=0}^k N_{n-2^k,j}$. If I'm not mistaken, the same recursion/initial values are satisfied by the number of ways to write $n$ as a sum of powers of $2$ where the largest element of the sum equals $2^k$. This might be a clue to the desired bijection....
Jan 21, 2014 at 7:32 comment added Greg Martin Random remark: the partition $(2n-5) + 3 + 1$ never seems to occur.
Jan 21, 2014 at 7:26 comment added Greg Martin @JamesCranch: emperically, the partitions of $2n-1$ into odd parts that occur all have a number of parts which is one less than a power of $2$. For example, when $n=5$, the corresponding partitions of $9$ are $9$, $7+1+1$, $3+3+3$, and $3+1+1+1+1+1+1$. However, not all such partitions appear: $5+3+1$ is missing, and $7+3+1$ and $5+3+3$ are missing when $n=6$, while $9+3+1$ and $5+3+1+1+1+1+1$ are missing when $n=7$, for example.
Jan 20, 2014 at 23:17 comment added James Cranch Interesting! If we define $|s_n| = n$, then the polynomial $p_n$ is homogeneous of degree $2n-1$. In particular, if the conjecture is true, then there is some natural subset of the partitions of $2n-1$ into odd parts which is in bijection with the partitions of $n$ into powers of $2$.
Jan 20, 2014 at 21:14 history asked Richard Stanley CC BY-SA 3.0