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This sequence of distributions interests me and I am looking for an expression in closed form. We are looking at populations of size n!. For n=2, we divide the population in half, and assign to each half the property 2 and not-2 (I write 2'), respectively. So S_2 is

$S_{2} = \frac{1}{2!}[{1[2],1[2']}]$.

The 2, 2' are simply attributes of the elements 1,1. For the second,

$S_{3} = \frac{1}{3!}[{1[2,3],2[2,3'],2[2'3'],1[2'3]}]$

For the fourth,

$S_{4} =\frac{1}{4!}[1[2,3,4],3[2,3,4'],2[2,3'4],6[2,3'4'],6[2',3',4'],2[2',3',4],3[2',3,4'],1[2',3,4]$...

While the elements ([2,3,4], etc.) can be used to construct the next element of the sequence, I hope the inductive process here is clear.

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It's not clear to me how you find $2[2,3']$ and $2[2',3']$ can you explain a bit more? $S_4$ is not divided by $4!$ as the other sequences? –  GarouDan Nov 4 '11 at 21:45
@GarouDan.Sorry. Edited. Will explain more as soon if still unclear. Thanks for pointing out typo. –  daniel Nov 4 '11 at 21:49
@GarouDan: At stage $n+1$ each of the elements from stage $n$ is split into $n+1$ copies; one copy gets the attribute $n+1$, and the other $n$ get the complementary attribute. Thus, $1[2,3,4]$ from stage $4$ will give rise to $1[2,3,4,5]$ and $4[2,3,4,5']$, while $3[2,3,4']$ will produce $3[2,3,4',5]$ and $12[2,3,4',5']$. (At least, this is consistent with what’s displayed.) –  Brian M. Scott Nov 4 '11 at 22:12
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1 Answer

up vote 1 down vote accepted

The coefficient of $[p_2,p_3,\dots,p_n]$, where $p_k$ is either $k$ or $k'$, can be determined as follows. Let $$c_k=\begin{cases}1,&p_k=k\\k-1,&p_k=k'\;;\end{cases}$$ then the coefficient of $[p_2,p_3,\dots,p_n]$ is $$\frac1{n!}\prod_{k=2}^n c_i\;.$$ This is easily proved by induction from the construction of the $S_n$. Is this enough of a closed form?

If you relabel the properties to start at $1$ instead of $2$, the entities in $S_n$ will have the form $[p_1,\dots,p_{n-1}]$, where $p_k$ is either $k$ or $k'$, and the coefficient of $[p_1,\dots,p_{n-1}]$ will be simply $$\frac1{n!}\prod\bigg\{k\in\{1,\dots,n-1\}:p_k=k'\bigg\}=\left(n\prod\bigg\{k\in\{1,\dots,n-1\};p_k=k\bigg\}\right)^{-1}.$$

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I will have to answer this later. It looks promising. Thanks. –  daniel Nov 4 '11 at 22:26
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