In [MM1], P. A. MacMahon introduced his "Omega Calculus" as a tool for solving problems in Diophantine equations and inequalities. This computational method turns out to useful in attacking certain kinds of problems involving the generating functions for various types of partition functions. MacMahon's Omega operator, \(\Omega_\ge\), is defined by: $$\mathop{\Omega}_\ge \sum_{s_1=-\infty}^\infty \cdots \sum_{s_r=-\infty}^{\infty} A_{s_1,\dots ,s_r} \lambda_1^{s_1} \cdots \lambda_r^{s_r} := \sum_{s_1=0}^\infty \cdots \sum_{s_r=0}^{\infty} A_{s_1,\dots ,s_r},$$ where the domain of the \(A_{s_1,\dots ,s_r}\) is the field of rational functions over \(\mathbb{C}\) in several complex variables and \{\lambda_i\) are restricted to a neighborhood of the circle \(|\lambda_i|=1.\)
Note that the effect of the operator \(\Omega_\ge\) is to set any term which has at least one factor \(\lambda_i^{s_i} \) with \( s_i < 0 \) equal to 0, and to set any term \(A_{s_1,\dots ,s_r}\lambda_1^{s_1} \cdots \lambda_r^{s_r}\) with all \(s_i \geq 0\) equal to \(A_{s_1,\dots ,s_r}\) (or alternatively, in the case all \(s_i \geq 0\) , each \(\lambda_i\) is set equal to 1) .

As an example of an application of the operator \(\Omega_\ge\), the generating function for the sequence \(\{p_m(n)\}\) will be derived, where \(m\) is a fixed positive integer for any positive integer \(n\), \(p_m(n)\) denotes the number of partitions of \(n\) into at most \(m\) parts.
First, observe that \begin{align} \Omega_\ge \frac{1}{(1-\lambda x)(1-y/\lambda)} = \frac{1}{(1- x)(1-xy)}. &&(1.1) \end{align} To see this \begin{align*} \Omega_\ge \frac{1}{(1-\lambda x)(1-y/\lambda)} &= \Omega_\ge \sum_{ n_1\geq 0}(\lambda x)^{n_1} \sum_{ n_2\geq 0}\left(\frac{y}{\lambda}\right)^{n_2}\\ &\\ &=\Omega_\ge \sum_{ n_1\geq 0}\sum_{ n_2\geq 0} x^{n_1} y^{n_2}\lambda^{n_1-n_2}\\ &\\ &=\sum_{ n_1\geq n_2\geq 0} x^{n_1} y^{n_2}\\ &\\ &=\sum_{ k\geq 0, n_2\geq 0} x^{k+n_2} y^{n_2} \,\,\,\, (\text{after setting } n_1=k+n_2)\\ &\\ &=\sum_{ k\geq 0}x^k \sum_{n_2\geq 0} (xy)^{n_2}\\ &\\ &=\frac{1}{(1- x)(1-xy)}. \end{align*} Next, the number of partitions of \(n\) into at most \(m\) is the number of solutions to the equation $$ n_1+n_2+\dots + n_m = n, \text{ with } n_1\geq n_2 \geq \dots \geq n_m \geq 0, $$ and so the generating function for the sequence \(\{p_m(n)\}\) is \begin{align*} \sum_{n=0}^{\infty}&p_m(n)q^n=\sum_{ n_1\geq n_2 \geq \dots \geq n_m \geq 0} q^{n_1+n_2+\dots + n_m}\\ &\\ &=\Omega_\ge \sum_{ n_1\geq 0, n_2 \geq 0, \dots , n_m \geq 0} q^{n_1+n_2+\dots + n_m}\lambda_1^{n_1 - n_2}\lambda_2^{n_2 - n_3}\dots \lambda_{m-1}^{n_{m-1} - n_{m}}\\ &\\ &=\Omega_\ge \sum_{ n_1\geq 0}(q\lambda_1)^{n_1} \sum_{ n_2\geq 0}\left(\frac{q\lambda_2}{\lambda_1}\right)^{n_2} \sum_{ n_3\geq 0}\left(\frac{q\lambda_3}{\lambda_2}\right)^{n_3} \dots \sum_{ n_m\geq 0}^{\infty}\left(\frac{q}{\lambda_{m-1}}\right)^{n_m}\\ &\\ &=\Omega_\ge \frac{1}{1-q\lambda_1} \frac{1}{1-\frac{q\lambda_2}{\lambda_1}} \frac{1}{1-\frac{q\lambda_3}{\lambda_2}} \dots \frac{1}{1-\frac{q}{\lambda_{m-1}}}\\ &\\ &=\Omega_\ge \frac{1}{1-q} \frac{1}{1-q^2\lambda_2} \frac{1}{1-\frac{q\lambda_3}{\lambda_2}} \dots \frac{1}{1-\frac{q}{\lambda_{m-1}}}, \end{align*} where the last equality is by (1.1) with \(\lambda=\lambda_1\), \(x=q\) and \(y=q\lambda_2\). Now apply (1.1) again with \(\lambda=\lambda_2\), \(x=q^2\) and \(y=q\lambda_3\) to get $$ \Omega_\ge \frac{1}{1-q} \frac{1}{1-q^2} \frac{1}{1-q^3\lambda_3} \dots \frac{1}{1-\frac{q}{\lambda_{m-1}}}. $$ Continue like this to eventually get that the generating function is $$ \sum_{n=0}^{\infty}p_m(n)q^n = \frac{1}{1-q} \frac{1}{1-q^2} \frac{1}{1-q^3} \dots \frac{1}{1-q^m}. $$

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