I'm trying to solve the combinatorics problem. It is the following :
"For each integer $n > 2$ determine a self-conjugate partition of $n$ that has at least two parts."
In my study group, we are discussing, for odd $n = 2k + 1$, the partition $1^k(k + 1)^1$ is a self-conjugate partition, on the other hand, for even $n = 2k$, the partition $1^{k-1}2^1(k-1)^1$, where $a^b$ means a repeated b times.
But, I don't quite understand this argument. Does anyone explain about this?
Take $k=4$ as an example. For $n=2k+1=9$ the partition $1^45^1$ has the following Ferrers diagram:
$$\begin{array}{ccc} \bullet&\bullet&\bullet&\bullet&\bullet\\ \bullet\\ \bullet\\ \bullet\\ \bullet \end{array}$$
Since conjugation just reflects the Ferrers diagram in its upper-left-to-lower-right diagonal, this partition of $9$ is clearly self-conjugate. Changing $k$ just changes the lengths of the horizontal and vertical arms: each has length $k+1$.
Now take $n=2k=8$. Ferrers diagrams like the one above always have an odd number of dots, so we need a slightly different trick: we tuck one extra dot in between the two arms, like this:
$$\begin{array}{ccc} \bullet&\bullet&\bullet&\bullet\\ \bullet&\bullet\\ \bullet\\ \bullet \end{array}$$
This corresponds to the partition $1^22^14^1$, and in general you’ll have $k-2$ rows of $1$ at the bottom, one row of $2$ just above them, and a row of $k$ at the top, so you’ll have the partition $1^{k-2}2^1k^1$.
