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Jan 24, 2014 at 10:45 comment added Mark Wildon Here is an involutive proof of the cancellation. As in the proof, let $(j_1,j_2,\ldots)$ encode a partition with $j_i$ parts of size $2^{i-1}$. If all the non-zero multiplicities are odd, do nothing. Otherwise, the sequence is of the form $(\ldots, 2a, a, \ldots, a, c, \ldots)$ where $2a$ is the first even multiplicity and $c\not =a$. Replace this subsequence with $(\ldots, 2c, c, \ldots, c, a, \ldots)$. If the $c$ in the first sequence is in position $r$ then both subsequences contribute $s_a^{2^{r-1}}s_c^{2^{r-1}}$.
Jan 22, 2014 at 1:25 vote accept Richard Stanley
Jan 22, 2014 at 0:07 history edited Gjergji Zaimi CC BY-SA 3.0
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Jan 21, 2014 at 23:55 comment added Gjergji Zaimi @Will, yes, precisely.
Jan 21, 2014 at 23:06 comment added Will Sawin So the terms that appear are exactly the terms where the exponents of the $s_n$, $n$ odd, add up to $2^k-1$ for some $k$ without cancellation.
Jan 21, 2014 at 22:30 history answered Gjergji Zaimi CC BY-SA 3.0