In this post, I will introduce the partition zeta function.
Definition
【Definition】Let \(n\) be a positive integer and \(s > 0\). The partition zeta function is
\[
\zeta_{\textrm{Par}}(n, s) := \sum_{\lambda \ \vdash \ n} f(\lambda)^s,
\] where \(\lambda\) is any partition of \(n\) and \(f(\lambda)\) is the number of 1’s in \(\lambda\).
【Example 1】
\begin{align}
\zeta_{\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 2】
\begin{align}
\zeta_{\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 + 2^s + 3^s + 5^s.
\end{align}
Theorem
【Theorem】Let \(m\) and \(n\) be positive integers. It holds that
\[
\zeta_{\textrm{Par}}(n, m) = \sum_{k = 0}^{n-1} p(n-k-1) \{(k+1)^m-k^m\}.
\]
Proof
【Proof】The proof is by induction on \(n\). When \(n = 1\), it is clear that the theorem holds for all positive integer \(m\). Suppose that the theorem is true when \(n = N\), i.e. suppose that
\[
\zeta_{\textrm{Par}}(N, a) = \sum_{k = 0}^{N-1} p(N-k-1)\{(k+1)^a-k^a\},
\] for all positive integer \(a\). When \(n = N+1\), then
\begin{align}
\zeta_{\textrm{Par}}(N+1, m) &= \sum_{\lambda \ \vdash N+1} f(\lambda)^m \\
&= \sum_{\lambda \ \vdash N} \{f(\lambda) + 1\}^m \\
&= \sum_{\lambda \ \vdash N} \left\{1 + \sum_{a=1}^m \binom{m}{a} f(\lambda)^a \right\} \\
&= \sum_{\lambda \ \vdash N} 1 + \sum_{\lambda \ \vdash N} \sum_{a=1}^m \binom{m}{a} f(\lambda)^a \\
&= p(N) + \sum_{a=1}^m \binom{m}{a} \sum_{\lambda \ \vdash N} f(\lambda)^a \\
&= p(N) + \sum_{a=1}^m \binom{m}{a} \zeta_{\textrm{Par}}(N, a) \\
&= p(N) + \sum_{a=1}^m \binom{m}{a} \sum_{k = 0}^{N-1} p(N-k-1) \{(k+1)^a-k^a\} \\
&= p(N) + \sum_{k = 0}^{N-1} p(N-k-1) \sum_{a=1}^m \binom{m}{a} \{(k+1)^a-k^a\} \\
&= p(N) + \sum_{k = 0}^{N-1} p(N-k-1) \{\{(k+1)+1\}^m-(k+1)^m\} \\
&= p(N) + \sum_{l = 1}^N p(N-l) \{(l+1)^m-l^m\} \\
&= \sum_{l = 0}^N p(N-l) \{(l+1)^m-l^m\}.
\end{align} This completes the proof.
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