Combinatorial interpretation of $j$-canonical representation of naturals? In the paper

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*P. McMullen (1971). The number of faces of simplicial polytopes. Israel J. Math. 9: 559-570

P. McMullen defines the $j$-canonical representation of a natural number $n$ as a particular representation
$$ n = \binom{n_j}{j}+\binom{n_{j-1}}{j-1} + \dots + \binom{n_i}{i} $$
satisfying $n_j>n_{j-1}>\dots>n_i\geq i\geq 1$, and asserts that it is the unique such representation satisfying this. I convinced myself that this assertion is true, using the hockey-stick identity. But I would love a more combinatorial way to understand what's going on.
My question: Is there a combinatorial interpretation of the numbers $i$ and $n_j,\dots,n_i$ that make clear what they are revealing about the pair $(n,j)$? (If so, what is it?)
Further context: Given $n,j$, one defines a number $n^{\langle j\rangle}$ by taking the $j$-canonical representation of $n$ and then setting
$$n^{\langle j\rangle} := \binom{n_j+1}{j+1}+\binom{n_{j-1}+1}{j} + \dots + \binom{n_i+1}{i+1} $$
This number is related (see Theorem 2.2 here) to the the growth rate of Hilbert functions of standard-graded algebras over a field. I would like more intuition about what $n^{\langle j\rangle}$ is, and am hoping that an answer to the above question will help with this.
 A: This is inspired in the Kruskal-Katona theorem stated in the form of Uniform hypergraphs. Let $[n]=\{1,2,\cdots ,n\}$ and consider the set $\binom{[m]}{j}=\{A\subseteq [m]:|A|=j\}$ of subsets of $[m]$ with fixed length $j.$ Take $m$ big enough such that $n\leq \binom{m}{j}.$ Now, consider the reverse lexicographic ordering on $\binom{[n]}{j}$ that is $$[j]\geq [j-1]\cup {j+1}\geq \cdots \geq [m]\setminus [m-j].$$ For example, for $m=3,j=2$ the list would be

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*$\{2,3\}$

*$\{1,3\}$

*$\{1,2\}$
Suppose that $A_n$ is the set that is exactly in position $n$ of the list of all these sets using the lexicographic order and take $$A_n=\{1\leq n_1<n_2<\cdots <n_j\},$$ is clear then that if $n_j$ is the maximum is because we already went thru all $\binom{n_j-1}{j}$ possibilities. Now fix $n_j,$ then we have to went over all $\binom{n_{j-1}-1}{j-1}$ possibilities so at the end of the day the $n-$th set is in position $$1+\sum _{k=1}^j\binom{n_k-1}{k}.$$
So the $\{n_j\}$ are the elements(indexed from $0$) of the set that is exactly at position $n$ when you order lexicographically the elements of $\binom{[m]}{j}$ for $m$ such that $\binom{m}{j}\geq n.$
Note: this is not much different than any base decomposition. Numbers are the same but the order is given by the order of the digits in the base.
