\documentclass[12pt]{article} \begin{document} \begin{definition}[Numerical semigroup] A \emph{numerical semigroup} is a subset $S\subseteq \mathbb{N}=\{0,1,2,\dots\}$ such that: (i) $0\in S$; (ii) if $a,b\in S$ then $a+b\in S$; and (iii) $\Delta:=\mathbb{N}\setminus S$ is finite. The finite set $\Delta$ is called the set of \emph{gaps}. \end{definition} \begin{definition}[Gap ``moments''] For an integer $r\ge 0$, define the $r$th \emph{gap power sum} (or \emph{genus moment}) \[ G_r(S):=\sum_{g\in \Delta} g^r. \] \end{definition} \subsection{Hilbert series and the numerator polynomial} Fix a numerical semigroup $S=\langle d_1,\dots,d_m\rangle$ with a generating set of positive integers $d_1,\dots,d_m$ and $\gcd(d_1,\dots,d_m)=1$. Define its (ordinary) generating function \[ H_S(z) := \sum_{s\in S} z^s. \] Because $\Delta$ is finite, the \emph{gap series} \[ \Phi_S(z):=\sum_{g\in \Delta} z^g \] is a polynomial in $z$ (finite sum). \begin{lemma}[Semigroup--gap decomposition] As a formal power series identity, \begin{equation}\label{eq:HS-Phi} H_S(z)+\Phi_S(z)=\sum_{n\ge 0}z^n=\frac{1}{1-z}. \end{equation} \end{lemma} A fact is that $H_S$ has the rational form \begin{equation}\label{eq:rational-form} H_S(z)=\frac{Q_S(z)}{\prod_{i=1}^m(1-z^{d_i})}, \end{equation} where $Q_S(z)$ is a polynomial with integer coefficients called the \emph{Hilbert numerator}. \begin{definition}[Product polynomial] Let \[ P_S(z):=\prod_{i=1}^m(1-z^{d_i}),\qquad \pi_m:=\prod_{i=1}^m d_i. \] \end{definition} \begin{lemma}[Numerator identity]\label{lem:Q-identity} Assuming \eqref{eq:rational-form}, we have the exact identity of formal power series \begin{equation}\label{eq:Q-identity} Q_S(z)=\frac{P_S(z)}{1-z}-P_S(z)\,\Phi_S(z). \end{equation} \end{lemma} \begin{proof} Multiply \eqref{eq:HS-Phi} by $P_S(z)$, then substitute $P_S(z)H_S(z)=Q_S(z)$ from \eqref{eq:rational-form}. \end{proof} \begin{definition}[Alternating power sums of syzygy degrees]\label{def:Cn} Write (the constant coefficient is always 1): \[Q_S(z) = 1 - \sum c_j z^j\] Then for $n\ge 0$ define \begin{equation}\label{eq:Cn-def} C_n(S) := \sum c_j j^n \end{equation} \end{definition} For $n=m+p\ge m$ define positive invariants $K_p(S)$ such that \begin{equation}\label{eq:Cn-Kp} C_{m+p}(S)=(-1)^m \frac{(m+p)!}{p!}\,\pi_m\, K_p(S)\qquad (p\ge 0). \end{equation} \begin{definition}[$T_n(\sigma)$ by generating function]\label{def:Tn-sigma} Define the formal power series \begin{equation}\label{eq:A(t)} A(t):=\prod_{i=1}^m \frac{e^{d_i t}-1}{d_i t}\in \mathbb{Q}[[t]]. \end{equation} Define numbers \begin{equation}\label{eq:Tn-sigma-def} T_n(\sigma):=n!\,[t^n]A(t), \end{equation} i.e.\ $A(t)=\sum_{n\ge 0} T_n(\sigma)\,\frac{t^n}{n!}$. \end{definition} \begin{definition}[$T_n(\delta)$ by generating function]\label{def:Tn-delta} Define \begin{equation}\label{eq:B(t)} B(t):=\frac{t}{e^t-1}\,A(t)\in \mathbb{Q}[[t]]. \end{equation} Define \begin{equation}\label{eq:Tn-delta-def} T_n(\delta):=\frac{n!}{2^n}\,[t^n]B(t), \end{equation} i.e.\ $B(t)=\sum_{n\ge 0} 2^n \cdot T_n(\delta)\,\frac{t^n}{n!}$. \end{definition} \subsection{Statement of Fel's conjecture} \begin{conjecture}[Fel]\label{conj:Fel} With the definitions above, for every $p\ge 0$ one has \begin{equation}\label{eq:Fel-conj} K_p(S)=\sum_{r=0}^{p} \binom{p}{r}\,T_{p-r}(\sigma)\,G_r(S)\;+\;\frac{2^{p+1}}{p+1}\,T_{p+1}(\delta). \end{equation} \end{conjecture} \subsection{Low-lying examples: Evidence} \paragraph{Example 1: $S=\langle 3,5\rangle$.} The gaps are $\Delta=\{1,2,4,7\}$, hence $G_0=4$, $G_1=14$, $G_2=70$, etc. The Hilbert numerator is $Q(z)=1-z^{15}$, so $C_n(S)=15^n$ and \eqref{eq:Cn-Kp} gives $K_0=\frac{15}{2}$, $K_1=\frac{75}{2}$, $K_2=\frac{1125}{4}$, etc. Plugging $\sigma_1=8$, $\sigma_2=34$, $\delta_1=(8-1)/2=7/2$, etc.\ into \eqref{eq:Fel-conj} recovers the same values. \paragraph{Example 2: $S=\langle 4,5,6\rangle$.} One finds $Q(z)=1-z^{10}-z^{12}+z^{22}$ and (using \eqref{eq:Cn-Kp}) $K_0=11$, $K_1=212/3$, $K_2=2002/3$, \dots. The conjectural right side \eqref{eq:Fel-conj} agrees with these values. \end{document}