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Fix a positive integer $d \geq 2$, and let $n,k$ be natural numbers with $k \leq n$.

Let $b(n,k)$ denote the number of monomials of degree $kd-(n+1)$ in $n+1$ variables $x_0,\ldots,x_n$ with all partial degrees $\leq d-2$.

Then I have observed that:

$b(n,k) = {d-1 \choose n}+ \sum_{k'=1}^{n-k} \sum_{n'=k'}^{k'+k-2} b(n',k') {d \choose n-n'}$

(where binomial coefficients are zero if they do not make sense). I was wondering if anyone can see some easy combinatorial argument for why this is true.

To provide some context, these numbers are closely related to the Hodge numbers of smooth projective hypersurfaces of degree $d$, but I am looking for an elementary argument.

An example, as requested:

To give an example, take $d=4$. Then $b(3,2)$ is the number of monomials in $x_0,\ldots,x_3$ of degree $4$ with all partial degrees at most $2$. There are $19$ such monomials:

One monomial $x_0 x_1 x_2 x_3$, ${4 \choose 2}=6$ monomials like $x_0^2 x_1^2$, and $4{3 \choose 2}=12$ monomials like $x_0^2 x_1 x_2$, so $1+6+12=19$.

The formula in this case says that:

$b(3,2)={3 \choose 3} + b(1,1){4 \choose 2}$, which is correct since one easily checks that $b(1,1)=3$.

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  • $\begingroup$ Could you give a small example of $b(n,k)$ and the associated monomials in some small case? $\endgroup$ May 22, 2013 at 15:03
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    $\begingroup$ Notice that $b(n,k)$ is the coefficient of $t^{kd-(n+1)}$ in $(1+t+t^2+\dots+t^{d-2})^{n+2}.$ (This is a standard combinatorial method). This might help you find a combinatorial proof, or even a closed formula. $\endgroup$ May 22, 2013 at 20:00

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