# Average number of full nodes in rooted m-ary tree

I am looking for a formula to express the average number of full nodes (i.e. nodes having exactly $$m$$ children) in a $$m$$-ary tree having $$n$$ nodes, i.e., $$\mu_{n}^{(m)} = \frac{\# \text{full nodes in all m-ary trees having n nodes} }{\# \text{nodes in all m-ary trees having n nodes}}$$.

As to the denominator, according to this answer, it sholud be given by $$n C_m^n = \binom{mn}{n} \frac{n}{(m-1)n + 1}$$ As to numerator, at the first glance, I tought that this answer would have provided a hint; but then I have realized that it is not the case because it gives how many trees have $$n$$ full nodes, but I need to know how many full nodes are in the set of all the $$m$$-ary trees having $$n$$ nodes.

Could someone help me to understand how to solve it and provide (if any) a usable formula for $$\mu_{n}^{(m)}$$?

Every $$m$$-ary tree is a node together with up to $$m$$ children, which are also $$m$$-ary trees. Let $$T(x,y)$$ be the generating function in which the coefficient of $$x^ny^k$$ is the number of $$m$$-ary trees with $$n$$ nodes and $$k$$ full nodes. Then $$T(x,y) = x(1 + T(x,y) + \cdots + T(x,y)^{m-1} + yT(x,y)^m).$$ The generating function of the total number of $$m$$-ary trees is $$A(x) = T(x,1)$$, and the generating function of the total number of full nodes in $$m$$-ary trees is $$B(x) = \frac{\partial}{\partial y} T(x,y)|_{y=1}$$.
The generating function $$A(x)$$ satisfies the equation $$A(x) = x(1 + A(x) + \cdots + A(x)^m).$$ The generating function $$B(x)$$ satisfies the equation $$B(x) = x(B(x) + 2A(x)B(x) + 3A(x)^2B(x) + \cdots + mA(x)^{m-1}B(x) + A(x)^m),$$ and so $$B(x) = \frac{xA(x)^m}{1-x(1+2A(x)+3A(x)^2+\cdots+mA(x)^{m-1})}.$$
For example, in the trivial case $$m=1$$, we have $$A(x) = x(1 + A(x))$$, and so $$A(x) = \frac{x}{1-x} = \sum_{n \geq 1} x^n.$$ It follows that $$B(x) = \frac{x\cdot \frac{x}{1-x}}{1-x} = \frac{x^2}{(1-x)^2} = \sum_{n \geq 1} (n-1)x^n.$$ Therefore the average number of full nodes in $$1$$-ary trees with $$n$$ nodes is $$\frac{n-1}{1} = n-1$$.