In this post, I will introduce the partition \(L\) function.
Definitions
Let \(n\) be a positive integer. Consider the symmetric group \(S_n\). Let \(c\) be a conjugacy class of the symmetric group \(S_n\). We can define the sign of \(c\).
【Definition 1】Let \(n\) be a positive integer and let \(\lambda\) be a partition of \(n\). We define the sign of \(\lambda\) as the sign of \(c\) where \(c\) is a conjugacy class of \(S_n\) corresponds to \(\lambda\). We denote the sign of \(\lambda\) as \(\textrm{sgn}(\lambda)\).
【Example 1】Let \(n=4\).
The partition \([4]\) corresponds to the conjugacy class of \((1, 2, 3, 4) \in S_4\). Therefore, \(\textrm{sgn}([4]) = \textrm{sgn}(1, 2, 3, 4) = -1\).
The partition \([3, 1]\) corresponds to the conjugacy class of \((1, 2, 3)(4) \in S_4\). Therefore, \(\textrm{sgn}([3, 1]) = \textrm{sgn}((1, 2, 3)(4)) = 1\).
【Definition 2】Let \(n\) be a positive integer and let \(s > 0\). The partition \(L\) function is
\[
L_{\textrm{Par}}(n, s) := \sum_{\lambda \ \vdash n} \textrm{sgn}(\lambda) f(\lambda)^s,
\] where \(\lambda\) is any partition of \(n\) and \(f(\lambda)\) is the number of 1’s in \(\lambda\).
【Example 2】
\begin{align}
L_{\textrm{Par}}(4, s) &= -f([4])^s + f([3, 1])^s +f([2, 2])^s-f([2, 1, 1])^s + f([1, 1, 1, 1])^s \\
&= – \ 0^s + 1^s + 0^s-2^s+4^s \\
&= 1-2^s+4^s.
\end{align}
【Example 3】
\begin{align}
L_{\textrm{Par}}(5, s) &= f([5])^s-f([4, 1])^s-f([3, 2])^s + f([3, 1, 1])^s + f([2, 2, 1])^s-f([2, 1, 1, 1])^s + f([1, 1, 1, 1, 1])^s \\
&= 0^s-1^s-0^s+2^s+1^s-3^s+5^s \\
&= 2^s-3^s+5^s.
\end{align}
Theorem
【Theorem】Let \(m\) and \(n\) be positive integers. It holds that
\[
L_{\textrm{Par}}(n, m) = \sum_{k = 0}^{n-1} p_{\textrm{sc}}(n-k-1) \{(k+1)^m-k^m\},
\] where \(p_{\textrm{sc}}(l)\) is the number of self-conjugate partitions of \(l\).
Proof
【Proof】We can prove the theorem similarly to this post. The proof is by induction on \(n\). If \(n = 1\), it is clear that the equation holds. Suppose that the equation holds when \(n = N\), i.e. suppose that
\[
L_{\textrm{Par}}(N, a) = \sum_{k = 0}^{N-1} p_{\textrm{sc}}(N-k-1) \{(k+1)^a-k^a\},
\] for all positive integer \(a\). Since this post, it holds that
\[
\sum_{\lambda \ \vdash N} \textrm{sgn}(\lambda) = p_{\textrm{sc}}(N).
\] Therefore, if \(n = N+1\), then
\begin{align}
L_{\textrm{Par}}(N+1, m) &= \sum_{\lambda \ \vdash N+1} \textrm{sgn}(\lambda) f(\lambda)^m \\
&= \sum_{\lambda \ \vdash N} \textrm{sgn}(\lambda) \{f(\lambda) + 1\}^m \\
&= \sum_{\lambda \ \vdash N} \textrm{sgn}(\lambda) \left\{1 + \sum_{a=1}^m \binom{m}{a} f(\lambda)^a \right\} \\
&= \sum_{\lambda \ \vdash N} \textrm{sgn}(\lambda) + \sum_{\lambda \ \vdash N} \sum_{a=1}^m \binom{m}{a} \textrm{sgn}(\lambda) f(\lambda)^a \\
&= p_{\textrm{sc}}(N) + \sum_{a=1}^m \binom{m}{a} \sum_{\lambda \ \vdash N} \textrm{sgn}(\lambda) f(\lambda)^a \\
&= p_{\textrm{sc}}(N) + \sum_{a=1}^m \binom{m}{a} L_{\textrm{Par}}(N, a) \\
&= p_{\textrm{sc}}(N) + \sum_{a=1}^m \binom{m}{a} \sum_{k = 0}^{N-1} p_{\textrm{sc}}(N-k-1) \{(k+1)^a-k^a\} \\
&= p_{\textrm{sc}}(N) + \sum_{k = 0}^{N-1} p_{\textrm{sc}}(N-k-1) \sum_{a=1}^m \binom{m}{a} \{(k+1)^a-k^a\} \\
&= p_{\textrm{sc}}(N) + \sum_{k = 0}^{N-1} p_{\textrm{sc}}(N-k-1) \{\{(k+1)+1\}^m-(k+1)^m\} \\
&= p_{\textrm{sc}}(N) + \sum_{l = 1}^N p_{\textrm{sc}}(N-l) \{(l+1)^m-l^m\} \\
&= \sum_{l = 0}^N p_{\textrm{sc}}(N-l) \{(l+1)^m-l^m\}.
\end{align} This completes the proof.
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