This question is related to another question posed on this site.
Let me recall the construction: Let $A:=k[x,y]/I$ with $k$ the complex numbers (or any algebraically closed field) and $\dim_k(A)< \infty$. It follows $A$ is artinian and hence there is an isomorphism
$$\phi:A \cong A_1 \oplus \cdots \oplus A_d$$
where $A_i$ is an artinian local ring for every $i$ with maximal ideal $\mathfrak{m}_i$. Since $\dim_k(A_i)< \infty$ it follows $\mathfrak{m}_i^{l_i}=0$ for some integer $l_i \geq 1$.
Let $\mathfrak{p}_i:=A_1\oplus \cdots \oplus \mathfrak{m}_i \oplus \cdots \oplus A_d$. It follows $A/\mathfrak{p}_i \cong A_i/\mathfrak{m}_i\cong k$ and $A/\mathfrak{p}_i^{l_i} \cong A_i$. It follows
$$\mathfrak{p}_1^{l_1}\cdots \mathfrak{p}_d^{l_d} =0.$$
By the chinese remainder theorem there is an isomorphism
$$A \cong A/(0) \cong A/\mathfrak{p}_1^{l_1}\cdots \mathfrak{p}_d^{l_d} \cong A_1 \oplus \cdots \oplus A_d$$
and since $k$ is algebraically closed it follows $\mathfrak{p}_i \cong(x-a_i,y-b_i)$ for complex numbers $(a_i,b_i)\in k^2$.
Let us lift the ideals $\mathfrak{p}_i$ to ideals $\mathfrak{q}_i \subseteq B:=k[x,y]$ with $B/\mathfrak{q}_i \cong A/\mathfrak{p}_i \cong k$. It follows the ideals $\mathfrak{q}_i \subseteq B$ are maximal ideals with
$$\mathfrak{q}_1^{l_1}\cdots \mathfrak{q}_d^{l_d} \subseteq I \subseteq \mathfrak{q}_i$$
for all $i=1,\ldots,d$.
Let us assume there is an equality of ideals
$$I:=(x-a_1,y-b_1)^{l_1}\cdots (x-a_d,y-b_d)^{l_d}$$
in $k[x,y]$, where $l_1,...,l_d$ is a set of integers $\geq 1$ satisfying a certain condition (see $C_1$ below).
Hence if our aim is to study the Hilbert scheme, we want to parametrize all length $n$ ideals, in particular we want to study the set of products of maximal ideals
$$I:=\mathfrak{p}_1^{l_1}\cdots \mathfrak{p}_d^{l_d} \subseteq k[x,y]$$
with $1 \leq d \leq n$ and $l_i \geq 1$. We find the formula
$$\dim_k(k[x,y]/(x,y)^i)=\binom{i+1}{2}$$
hence if $I:=(x-a_1,y-b_1)^{l_1}\cdots (x-a_d,y-b_d)^{l_d}$
it follows
$$\dim_k(k[x,y]/I)= \sum_{j=1}^d \binom{l_j+1}{2}.$$
Note that $\dim_k(k[x,y]/(x-a,y-b)^l)=\dim_k(k[x,y]/(x,y)^l$ hence
$$\dim_k(A) = \sum_j \dim_k(k[x,y]/(x-a_j,y-b_j)^{l_j}=\sum_j \binom{l_j+1}{2}.$$
Hence when studying the Hilbert scheme we want to parametrize ideals $I=\mathfrak{p}_1^{l_1}\cdots \mathfrak{p}_d^{l_d}$ with $1 \leq d \leq n $ and
$$(C_1)\qquad\dim_k(k[x,y]/I)= \sum_{j=1}^d \binom{l_j+1}{2}=n.$$
Question 1: Given an integer $1 \leq d \leq n$ we seek a combinatorial formula for the number $D(l_1,..,l_d,n)$ of unordered tuples of integers $(l_1,...,l_d)$ with $l_i\geq 1$ and $\sum_j \binom{l_j+1}{2}=n$:
Let $S(l_1,..,l_d,n)$ be the following set:
$$S(l_1,..,l_d,n):=\{ (l_1,..,l_d)\text{ an unordered set of integers $l_i$}. l_i \geq 1, \sum_j \binom{l_j+1}{2}=n \}$$
By definition: $D(l_1,..,l_d,n)$ is the number of elements in $S(l_1,..,l_d,n)$.
Do you know such a formula or a reference to where this type of formula is studied? I ask for an explicit reference to a study of this problem and such formulas in the litterature. If you have seen these numbers appearing in the study of the $n!$-conjecture I ask for a reference.
Note 1: I'm "imprecise" when writing $D(l_1,..,l_d,n)$ - this reflects that the numbers arise when studying the ideals
$$\mathfrak{p}_1^{l_1}\cdots \mathfrak{p}_d^{l_d} \subseteq k[x,y]$$
with $\mathfrak{p}_i:=(x-a_i,y-b_i)$.
Note 2: We may generalize these numbers as follows: If $A:=k[x_1,..,x_n]$ is a polynomial ring in $n$ variables over $k$, we may want to "parametrize" the set of ideals $I \subseteq A$ with $dim_k(A/I)=k$ for some integer $k\geq 1$. Let $\mathfrak{p}_i:=(x_1-a(i)_1,\ldots ,x_n-a(i)_n)$ with $a(i)_j \in k$ It follows similarly (if we choose $d$ coprime maximal ideals $\mathfrak{p}_1,..,\mathfrak{p}_d$) there is an equality
$$\dim_k(A/\mathfrak{p}_i^{l_i+1})=\binom{l_i+n}{n},$$
and if $I:=\mathfrak{p}_1^{l_1}\cdots \mathfrak{p}_d^{l_d+1}$
it follows
$$\dim_k(A/I)= \sum_j \binom{l_j+n}{n}.$$
Let $S(l_1,..,l_d,n,d,k)$ denote the set of unordered tuples $(l_1,..,l_d)$ with $l_i \geq 1$ and with
$$ \sum_j \binom{l_j+n}{n}=k.$$
Let $D(l_1,..,l_d,n,d,k)$ denote the set of elements in $S(l_1,..,l_d,n,d,k)$. I'm asking a similar question for the numbers $D(l_1,..,l_d,n,d,k)$.
Again I'm using an "imprecise" notation to indicate that these numbers arise when parametrizing ideals.
Question 2: Given an arbitrary field $k$ and an arbitrary finitely generated $k$-algebra $A$ with a confinite ideal $I \subseteq A$ (this means $I$ is an ideal with $\dim_k(A/I)< \infty$). Are you able to give an "elementary" parametrization of all such ideals $I$ using methods similar to the ones introduced above?
Answer: The following construction gives a relation with Question 2 and the Chinese Remainder Theorem (CRT) via Noether nomalization lemma (NNL). Let $k$ be any field and let $A$ be any finitely generated $k$-algebra and $I \subseteq A$ a cofinite ideal. By Atiyah-Macdonald Thm 8.7 (AM) and the NNL, since $B:=A/I$ is Artinian there is a decomposition
$$B \cong B_1\oplus \cdots \oplus B_d$$
with $(B_i, \mathfrak{m}_i)$ an Artinian local ring for all $i$. Since $dim_k(B_i)< \infty$ it follows there is an integer $l_i \geq 2$ with $\mathfrak{m}_i^{l_i}=0$. Let
$$\mathfrak{p}_i:=B_1\oplus \cdots \oplus \mathfrak{m}_i \oplus \cdots \oplus B_d \subseteq B,$$
and let
$J_i:=B_1 \oplus \cdots \oplus (0)\oplus \cdots \oplus B_d.$$
It follows $J_i \subseteq \mathfrak{p}_i$ and that $\mathfrak{p}_i$ is a maximal ideal. There is an euqlity $J_1\cdots J_d=(0)$. The ideals $\mathfrak{p}_i, \mathfrak{p}_j$ are coprime for $i \neq j$. Similar for $J_i,J_j$. Hence there is an equality
$$(0)=J_1\cdots J_d = J_1 \cap \cdots \cap J_d.$$
Let $p: A \rightarrow A/I$ and let $I_i:=p^{-1}(J_i)$. It follows
$$I:=p^{-1}((0))=p^{-1}(J_1\cdots J_d)=p^{-1}(J_1 \cap \cdots \cap J_d)=$$
$$p^{-1}(J_1) \cap \cdots \cap p^{-1}(J_d)=I_1\cap \cdots \cap I_d=I_1\cdots I_d.$$
This is because the ideals $I_i,I_j$ are coprime when $i\neq j$.
We may lift the maximal ideals $\mathfrak{p}_i$ to maximal ideals $\mathfrak{q}_i:=p^{-1}(\mathfrak{p}_i) \subseteq A$ and it follows the ideals $\mathfrak{q}_i, \mathfrak{q}_j$ are coprime when $i \neq j$.
Since there is ain inclusion $J_i \subseteq \mathfrak{p}_i$ it follow $I_i \subseteq \mathfrak{q}_i$. There is an integer $l_i$ with
$$J_i=\mathfrak{p}_{l_i+1} \subsetneq \mathfrak{p}_i^{l_i}$$
and it follows there are inclusions
$$\mathfrak{q}_i^{l_i+1} \subseteq I_i \subseteq \mathfrak{q}_i^{l_i}$$
Lemma. Given any cofinite ideal $I \subseteq A$, it follows there are maximal ideals $\mathfrak{q}_1,..,\mathfrak{q}_d$ and integers $l_1,..,l_d\geq 1$ and cofinite ideals $\mathfrak{q}_i^{l_i+1} \subseteq I_i \subseteq \mathfrak{q}_i^{l_i}$ with
$$I=I_1\cdots I_d.$$
Proof: The construction is given above QED.
Note: A product of powers of maximal ideals is cofinite, and by the Lemma we may study ideals "squeezed" between powers of maximal ideals
$$\mathfrak{m}^{l+1} \subseteq I \subseteq \mathfrak{m}^l$$
to obtain all cofininite ideals. We say such an ideal $I$ is an "$(\mathfrak{m},l)$-squeezed ideal". Note that if $A$ is a regular ring of dimension $d$ it follows $\mathfrak{m}^l/\mathfrak{m}^{l+1} \cong Sym^l(\mathfrak{m}/\mathfrak{m}^2)$ hence we have good control on the vector space $\mathfrak{m}^l/\mathfrak{m}^{l+1}$ when $A$ is regular.
Example: If $A:=k[x,y]$ with $k$ an algebraically closed field we get the following: Consider the set
$$\{(a_1,b_1,l_1),\ldots ,(a_d,b_d,l_d) \}\in Sym^d(k^2 \times \mathbb{Z})$$
with $(a_i,b_i,l_i)\in k^2 \times \mathbb{Z}, l_i\geq 1$ and $\sum_j \binom{l_j+1}{2}=n.$
Here $Sym^d(k^2 \times \mathbb{Z})$ is the "set theoretic symmetric product" of $k^2 \times \mathbb{Z}$. The symmetric group on d elements $S_d$ acts on $(k^2 \times \mathbb{Z})^d$ and $Sym^d(k^2\times \mathbb{Z}):=(k^2 \times \mathbb{Z})^d/S_d$ is the "quotient". Let
$$H^i:=Sym^i(k^2 \times \mathbb{Z})$$
and consider
$$H(n):=H^1\times H^2 \cdots \times H^n.$$
We get a construction of all cofinite ideals $I \subseteq A$ of length $n$ as a subset
$$"Hilb^n(k[x,y])" \subseteq H(n).$$
Note: $"Hilb^n(k[x,y])"$ is a set, not a scheme.