My question is the following:
We fix two non-negative integer $n$ and $d$. We can define a natural equivalence relation on $( \mathbb{N}\cup\{0\})^n$, the relation induced by the action of the group of permutation $\Pi_n$ of $\{1,\dots , n\}$ :
$\pi \cdot (a_1,\dots a_n):=(a_{\pi(1)}, \dots , a_{\pi(n)})$.
It is possible to induce $||\cdot ||_1$ on the quotient space. In fact
$||\pi \cdot (a_1,\dots , a_n)||_1=|a_{\pi(1)}+\cdots +a_{\pi(n)}|=|a_1+\cdots +a_n|=||(a_1, \cdots, a_n)||_1$
What is the cardinality of the following set?
$\Lambda:=\{[a_1, \cdots, a_n]\in ( \mathbb{N}\cup\{0\})^n/\Pi_n : ||[a_1, \cdots, a_n]||_1\leq d\}$
The question is more simple if one would compute the cardinality of $p^{-1}(\Lambda)$, where $p:( \mathbb{N}\cup\{0\})^n\to ( \mathbb{N}\cup\{0\})^n/\Pi_n$ is the projection map. In fact
$p^{-1}(\Lambda)=\{(a_1, \cdots, a_n)\in ( \mathbb{N}\cup\{0\})^n: ||(a_1, \cdots, a_n)||_1\leq d\}$
and so its cardinality is equal to
$|p^{-1}(\Lambda)|=\sum_{k=0}^{d}\binom{k+n-1}{n-1}$
by stars and bars argument.
In our case we can observe that
$|\Lambda|=\sum_{k=0}^d |\{[a_1, \cdots, a_n]\in ( \mathbb{N}\cup\{0\})^n/\Pi_n : ||[a_1, \cdots, a_n]||_1= k\}|=\sum_{k=0}^dp(k,n)$
where $p(k,n)$ is the partition of the number $k$ as sum of $n$ non-negative integer, up to commutativity.
So the question becomes the following:
There exists a closed form of the sum
$\sum_{k=0}^dp(k,n)$ ?