Not sure how to apply generating functions

Question

I am trying to find the generating function $A(x)$

$[x^n]A(x) = \sum_{a+b+c=n \\ a,b,c \geq 0 \\ a,b,c \in \mathbb{Z}}\frac{a^2b}{c}$

My initial thoughts are that the 3 variables can be their own series and can have their own generating functions, but I am struggling to see how the constraints on the variables would apply to their generating functions.

Answer 1

Here's how I would approach the problem. We have three variables $a, b, c$ which are all conveniently non-negative and sum to $n$, so it is clear that we want something of the form

$$ G(x) = (a_0 + a_1x + a_2x^2 + \cdots)(b_0 + b_1x + b_2x^2 + \cdots)(c_0 + c_1x + c_2x^2 + \cdots) $$

This way, we have $[x^n]G(x) = \sum_{i + j + k = n \\ i, j, k \in \mathbb{Z}_{\geq 0}} a_ib_jc_k$.

Hopefully now it becomes simple. If you still don't see it, note that $a_i$ just means a function in $i$, so you can write $a(i)$ if it's clearer. Then look at your summand $\frac{a^2b}{c}$ and see how it corresponds to $a_ib_jc_k$.

Also, if you don't know how to find the coefficient from this power series, I have written an answer on solving generating functions literally a week ago. You can find it here.

Hope this helps!

Answer 2

Welcome to MSE. Observe that \begin{align}\left(\sum_{\alpha\ge0}x_\alpha\,t^\alpha\right)\left(\sum_{\beta\ge0}y_\beta\,t^\beta\right)\left(\sum_{\gamma\ge0}z_\gamma\,t^\gamma\right) &=\sum_{\alpha,\beta,\gamma\ge0}x_\alpha\,y_\beta\,z_\gamma\,t^{\alpha+\beta+\gamma}\\[.4em] &=\sum_{n=0}^\infty\left(\sum_{\substack{\alpha,\beta,\gamma\ge0\\\alpha+\beta+\gamma=n}}x_\alpha\,y_\beta\,z_\gamma\right)t^n. \end{align}

Answer 3

After correction (see last comment above), $$A(x)=\sum_{a,b\ge0, c>0}\frac{a^2b}cx^{a+b+c}=F(x)G(x)H(x)$$ where $$F(x)=\sum_{k=0}^\infty k^2x^k,\quad G(x)=\sum_{k=0}^\infty kx^k,\quad H(x)=\sum_{k=1}^\infty\frac{x^k}k.$$ There are myriads of posts on this site to calculate formal power series of the form $\sum k^px^k$ where $p\in\Bbb Z,$ so let me recall only those useful here.

• The particular case $p=0$ corresponds to the geometric power series: $\sum_{k=0}^\infty x^k=\frac1{1-x}.$
• Differentiating it twice, we derive $\sum_{k=0}^\infty kx^{k-1}=\frac1{(1-x)^2}$ and $\sum_{k=0}^\infty k(k-1)x^{k-2}=\frac2{(1-x)^3},$ hence $$G(x)=\frac x{(1-x)^2}\quad\text{and}\quad F(x)=\frac{2x^2}{(1-x)^3}+G(x)=\frac{x^2+x}{(1-x)^3}.$$
• Integrating it once, we derive $$H(x)=-\ln(1-x).$$

Therefore $$A(x)=-\frac{x^2(1+x)}{(1-x)^5}\ln(1-x).$$