Number of Laurent monomials of n variables with degree at most d 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.
 A: One such formula is
$$\sum_{p=0}^n \binom{n}{p} \binom{d}{p} \binom{d+n-p}{n-p}.$$
To derive this, let $P \subseteq [n]$ be the set of variables with positive exponents and let $p = |P|$. There are $\binom{n}{p}$ ways to choose $P$. After choosing $P$, we must choose a monomial in $\{ x_i \}_{i \in P}$ of degree $\leq d$ where each variable has degree $\geq 1$; there are $\binom{d}{p}$ ways to do this. And we must choose a monomial in $\{ x_i \}_{i \in [n] \setminus P}$ of degree $\leq d$ where each variable has degree $\geq 0$; there are $\binom{d+n-p}{n-p}$ ways to do this. I think this is equivalent to Ira Gessel's formula in the comment below.
My previous answer is preserved below the line:

One such formula is 
$$\sum_{\substack{0 \leq p,q \leq d\\ p+q \leq n}} \binom{n}{p,q} \binom{d}{p} \binom{d}{q}$$
where $\binom{n}{p,q}$ is the trinomial coefficient $\tfrac{n!}{p! q! (n-p-q)!}$.
To see this, let $P$ and $Q \subset [n]$ be the sets of variables whose exponents are positive and negative. To choose a monomial in your set, first choose the index sets $P$ and $Q$, which we can do in $\binom{n}{|P|,|Q|}$ ways. Then choose two monomials, one in the variables $\{ x_i \}_{i \in P}$, and one in the variables $\{ x_j \}_{j \in Q}$, where each variable occurs to degree $\geq 1$ and the total degree is $\leq d$. The number of monomials in $p$ variables where each variable occurs to degree $\geq 1$ and the total degree is $\leq d$ is $\binom{d}{p}$. (Note that this is even correct when $d<p$.)
This isn't a particularly nice formula, but I don't see how to simplify it more. (EDIT: But Ira Gessel did; see his comment below.) Note that it is manifestly a polynomial in $d$, for fixed $n$.
