Introduction: We have a question of how to calculate the number of $n$-variables monomials of degree at most $d$ (which are known as Laurent monomials).

For example: If $n=2$, $d=2$ then we have 19 monomials, which are: $x^{-2}$, $x^{-2}y$, $x^{-2}y^2$, $x^{-1}y^{-1}$, $x^{-1}$, $x^{-1}y$, $x^{-1}y^2$, $y^{-2}$, $y^{-1}$, 1, $y$, $y^{2}$, $xy^{-2}$, $xy^{-1}$, $x$, $xy$, $x^2y^{-2}$, $x^2y^{-1}$, $x^2$. They are corresponding to the following 19 pairs of powers: (-2,0), (-2,1), (-2,2), (-1,-1), (-1,0), (-1,1), (-1,2), (0,-2), (0,-1), (0,0), (0,1), (0,2), (1,-2), (1,-1), (1,0), (1,1), (2,-2), (2,-1), (2,0).

So how about, e.g., $n=4$, $d=4$ ?

Formulation: Let $n,d\in\mathbb N$. Let \begin{align*} \Omega(n,d) &= \{\alpha\in\mathbb N^n: \alpha_1+\ldots+\alpha_n \leq d\},\\ \Gamma(n,d) &= \Omega(n,d) + \Omega(n,d),\\ \Delta(n,d) &= \Omega(n,d) - \Omega(n,d). \end{align*} Then, the elements of $\Delta(n,d)$ corresponds to the tuples of powers of the monomials.

The cardinality of $\Omega(n,d)$ is well-known to be $\binom{n+d}{n}$. The cardinality of $\Gamma(n,d)$ can be shown to be $\binom{n+2d}{n}$. However, is there a formula to compute the cardinality of $\Delta(n,d)$ ?

Some results: We did some attempts and got that \begin{align*} |\Delta(n,1)| &= n(n+1)+1,\\ |\Delta(1,d)| &= 2d+1,\\ |\Delta(2,d)| &= (2d+1)^2 - d(d+1),\\ |\Delta(3,d)| &= (2d+1)^3 - \frac{7}{3}d(d+1)(2d+1),\\ |\Delta(4,d)| &= (2d+1)^4 - \frac{1}{12}d(d+1)(157d^2+157d+46). \end{align*} We still haven't found a way to calculate $|\Delta(n,d)|$ in general. By definition of $\Delta(n,d)$ as above, I think we may have a geometric approach. An orientable suggestion would be nice.

Introduction: We have a question of how to calculate the number of $n$-variables Laurent monomials of degree at most $d$.

For example: If $n=2$, $d=2$ then we have 19 monomials, which are: $x^{-2}$, $x^{-2}y$, $x^{-2}y^2$, $x^{-1}y^{-1}$, $x^{-1}$, $x^{-1}y$, $x^{-1}y^2$, $y^{-2}$, $y^{-1}$, 1, $y$, $y^{2}$, $xy^{-2}$, $xy^{-1}$, $x$, $xy$, $x^2y^{-2}$, $x^2y^{-1}$, $x^2$. They correspond to the following 19 pairs of powers: $(-2,0)$, $(-2,1)$, $(-2,2)$, $(-1,-1)$, $(-1,0)$, $(-1,1)$, $(-1,2)$, $(0,-2)$, $(0,-1)$, $(0,0)$, $(0,1)$, $(0,2)$, $(1,-2)$, $(1,-1)$, $(1,0)$, $(1,1)$, $(2,-2)$, $(2,-1)$, $(2,0)$.

So how about, e.g., $n=4$, $d=4$ ?

Formulation: Let $n,d\in\mathbb N$. Let \begin{align*} \Omega(n,d) &= \{\alpha\in\mathbb N^n: \alpha_1+\dotsb+\alpha_n \leq d\},\\ \Gamma(n,d) &= \Omega(n,d) + \Omega(n,d),\\ \Delta(n,d) &= \Omega(n,d) - \Omega(n,d). \end{align*} Then, the elements of $\Delta(n,d)$ correspond to the tuples of powers of the monomials.

The cardinality of $\Omega(n,d)$ is well-known to be $\binom{n+d}{n}$. The cardinality of $\Gamma(n,d)$ can be shown to be $\binom{n+2d}{n}$. However, is there a formula to compute the cardinality of $\Delta(n,d)$ ?

Some results: We did some attempts and got that \begin{align*} |\Delta(n,1)| &= n(n+1)+1,\\ |\Delta(1,d)| &= 2d+1,\\ |\Delta(2,d)| &= (2d+1)^2 - d(d+1),\\ |\Delta(3,d)| &= (2d+1)^3 - \frac{7}{3}d(d+1)(2d+1),\\ |\Delta(4,d)| &= (2d+1)^4 - \frac{1}{12}d(d+1)(157d^2+157d+46). \end{align*} We still haven't found a way to calculate $|\Delta(n,d)|$ in general. By definition of $\Delta(n,d)$ as above, I think we may have a geometric approach. An orientable suggestion would be nice.

Introduction: We have a question of how to calculate the number of $n$-variables monomials of degree at most $d$ (which are known as Laurent monomials).

For example: If $n=2$, $d=2$ then we have 19 monomials, which are: $x^{-2}$, $x^{-2}y$, $x^{-2}y^2$, $x^{-1}y^{-1}$, $x^{-1}$, $x^{-1}y$, $x^{-1}y^2$, $y^{-2}$, $y^{-1}$, 1, $y$, $y^{2}$, $xy^{-2}$, $xy^{-1}$, $x$, $xy$, $x^2y^{-2}$, $x^2y^{-1}$, $x^2$. They are corresponds to the following 19 pairs of powers: (-2,0), (-2,1), (-2,2), (-1,-1), (-1,0), (-1,1), (-1,2), (0,-2), (0,-1), (0,0), (0,1), (0,2), (1,-2), (1,-1), (1,0), (1,1), (2,-2), (2,-1), (2,0).

So how about, e.g., $n=4$, $d=4$ ?

Formulation: Let $n,d\in\mathbb N$. Let \begin{align*} \Omega(n,d) &= \{\alpha\in\mathbb N^n: \alpha_1+\ldots+\alpha_n \leq d\},\\ \Gamma(n,d) &= \Omega(n,d) + \Omega(n,d),\\ \Delta(n,d) &= \Omega(n,d) - \Omega(n,d). \end{align*} Then, the elements of $\Delta(n,d)$ corresponds to the tuples of powers of the monomials.

The cardinality of $\Omega(n,d)$ is well-known to be $\binom{n+d}{n}$. The cardinality of $\Gamma(n,d)$ can be shown to be $\binom{n+2d}{n}$. However, is there a formula to compute the cardinality of $\Delta(n,d)$ ?

Some results: We did some attempts and got that \begin{align*} |\Delta(n,1)| &= n(n+1)+1,\\ |\Delta(1,d)| &= 2d+1,\\ |\Delta(2,d)| &= (2d+1)^2 - d(d+1),\\ |\Delta(3,d)| &= (2d+1)^3 - \frac{7}{3}d(d+1)(2d+1),\\ |\Delta(4,d)| &= (2d+1)^4 - \frac{1}{12}d(d+1)(157d^2+157d+46). \end{align*} We still haven't found a way to calculate $|\Delta(n,d)|$ in genreal. By definition of $\Delta(n,d)$ as above, I think we may have a geometric approach. Orientable suggestion would be nice.

Introduction: We have a question of how to calculate the number of $n$-variables monomials of degree at most $d$ (which are known as Laurent monomials).

For example: If $n=2$, $d=2$ then we have 19 monomials, which are: $x^{-2}$, $x^{-2}y$, $x^{-2}y^2$, $x^{-1}y^{-1}$, $x^{-1}$, $x^{-1}y$, $x^{-1}y^2$, $y^{-2}$, $y^{-1}$, 1, $y$, $y^{2}$, $xy^{-2}$, $xy^{-1}$, $x$, $xy$, $x^2y^{-2}$, $x^2y^{-1}$, $x^2$. They are corresponding to the following 19 pairs of powers: (-2,0), (-2,1), (-2,2), (-1,-1), (-1,0), (-1,1), (-1,2), (0,-2), (0,-1), (0,0), (0,1), (0,2), (1,-2), (1,-1), (1,0), (1,1), (2,-2), (2,-1), (2,0).

So how about, e.g., $n=4$, $d=4$ ?

Formulation: Let $n,d\in\mathbb N$. Let \begin{align*} \Omega(n,d) &= \{\alpha\in\mathbb N^n: \alpha_1+\ldots+\alpha_n \leq d\},\\ \Gamma(n,d) &= \Omega(n,d) + \Omega(n,d),\\ \Delta(n,d) &= \Omega(n,d) - \Omega(n,d). \end{align*} Then, the elements of $\Delta(n,d)$ corresponds to the tuples of powers of the monomials.

The cardinality of $\Omega(n,d)$ is well-known to be $\binom{n+d}{n}$. The cardinality of $\Gamma(n,d)$ can be shown to be $\binom{n+2d}{n}$. However, is there a formula to compute the cardinality of $\Delta(n,d)$ ?

Some results: We did some attempts and got that \begin{align*} |\Delta(n,1)| &= n(n+1)+1,\\ |\Delta(1,d)| &= 2d+1,\\ |\Delta(2,d)| &= (2d+1)^2 - d(d+1),\\ |\Delta(3,d)| &= (2d+1)^3 - \frac{7}{3}d(d+1)(2d+1),\\ |\Delta(4,d)| &= (2d+1)^4 - \frac{1}{12}d(d+1)(157d^2+157d+46). \end{align*} We still haven't found a way to calculate $|\Delta(n,d)|$ in general. By definition of $\Delta(n,d)$ as above, I think we may have a geometric approach. An orientable suggestion would be nice.