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I am interested in computing the number of natural solutions $(x_0,y_0,z_0)$ of the equation $w_0x+w_1y+w_2z=d$, with $w_0,w_1,w_2$ and $d$ natural numbers and $\gcd (w_i,w_j)=1$ . Or equivalently, the number of monomials in three variables of quasihomogeneus degree d and weights $(w_0,w_1,w_2)$. Does anybody now any formula in terms of $w_0,w_1,w_2$ and $d$ or any reference in the literature?

With the previous notations I really need to compute $a_{\alpha+w_0w_1w_2}-a_{\alpha}$ being $\alpha< w_0w_1w_2$. I do not know if there are some kind of relations among the terms of this series...I want to look at the properties of these series because I need an explicit expresion in terms of the weights and $\alpha$.

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  • $\begingroup$ This is a well-known, classic problem in partition theory (for any number of weights), studied by Issai Schur. The g.f. for the numbers is easily derived, and from it one obtains a linear recurrence and an asymptotic formula for the number of ways for $d$. An exact closed formula is also available by decomposing the gf in partial fractions and expanding; not a very nice formula though. See en.wikipedia.org/wiki/Schur%27s_theorem#Combinatorics $\endgroup$ Oct 26, 2012 at 22:32

2 Answers 2

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Your problem is equivalent to the computation of $\dim_k H^0(\mathcal{O}_{\mathbb{P}(w_0, w_1, w_2)}(d))$ over any field $k$, where $\mathbb{P}(w_0, w_1, w_2)$ denotes the weighted projective plane with weights $w_0$, $w_1$, $w_2$.

I do not know any closed formula, but it is possible to give an answer in terms of generating functions. Let me state it for general weighted projective spaces.

Set $\boldsymbol{w}=(w_0, \ldots, w_r)$ and $$P_{\boldsymbol{w}}(t)=\prod_{i=0}^r(1-t^{w_i})^{-1}=\sum_{i=0}^{+\infty}a_it^i.$$

Then one has $$\dim_k H^0(\mathcal{O}_{\mathbb{P}(\boldsymbol{w})}(d))=a_d. \quad (*)$$

The reason is that $P_{\boldsymbol{w}}(t)$ is the Poincaré series of the graded polynomial algebra $S_{\boldsymbol{w}}=k[x_0, \ldots, x_r]$, where $x_i$ has degree $w_i$. Hence $a_d=\dim_k (S_{\boldsymbol{w}})_d$ and, since $\mathbb{P}(\boldsymbol{w})=\textrm{Proj}(S_{\boldsymbol{w}})$, we obtain $(*)$.

In your case one has $r=2$, so the number you are looking for is given by the term $a_d$ in the expansion $$P_{\boldsymbol{w}}(t)= \frac{1}{(1-t^{w_0})(1-t^{w_1})(1-t^{w_2})}=\sum_{i=0}^{+\infty} a_it^i.$$

You can find more details looking at Section 1 of Dolgachev's paper Weighted projective varieties, in “Group Actions and Vector Fields”, Lect. Notes in Math. 956, Springer-Verlag 1982, pp. 34-72.

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    $\begingroup$ Surely, that generating function identity can be explained without the appeal to weighted projective spaces. :-) $\endgroup$ Oct 15, 2012 at 1:40
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    $\begingroup$ Maybe you are right and there exists a purely combinatorial proof. But saying that the generating function is the Poincaré series of a graded polynomial algebra is - in my opinion - a more conceptual argument. Of course, you can avoid talking about Weighted Projective Spaces and just consider the graded algebra $S_w$; in fact, the WPS is just $\textrm{Proj}(S_w)$. But my way of thinking is more geometrical then algebraic, so I like to focus on geometrical objects, when it is possible :-) $\endgroup$ Oct 15, 2012 at 5:37
  • $\begingroup$ Just for the sake of clarity, I added in the answer the relationship between $\mathbb{P}(w)$ and $S_{w}$. $\endgroup$ Oct 15, 2012 at 5:54
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There is an asymptotic formula: If $w_0,...,w_r$ are nonnegative integers such that $gcd(w_0,...,w_r)=1$, then for large $n$:

$$a_n := |\lbrace (k_0,...,k_r) \in \mathbb{N}_0^r \mid \sum_i k_iw_i = n \rbrace| = \frac{n^r}{r!\;w_0 \cdots w_r} + O(n^{r-1})$$

For, as in Francesco's answer we have $$\sum_{n \ge 0} a_n t^n = \frac{1}{(1-t^{w_0})\ldots (1-t^{w_r})}$$ (let me know if you need more details for this step). Now the result follows from Robert Israel's answer of the following question:

Asymptotics for the coefficients of a rational function


Moreover, it's known that for large $n$, $a_n$ is a "polynomial with periodic coefficients" in $n$, i.e. there are polynomials $f_i$ over $\mathbb{Q}$ of degree $r$ such that $a_{nw+i} = f_i(n)$ for $0 \le i < w$ and all large enough $n$, where $w := lcm(w_0,...,w_r)$.

This is just Excercise 10.12 in [Eisenbud: Commutative Algebra].

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