Extracting sequence from generating function

Question

I was provided the following generating function, and was unsure how to use it. I have never seen an example where the function “involved” itself. The generating function is $F(z)^8$ Where $$F(z)=z+z^6 F(z)^5+z^{11} F(z)^{10}+z^{16} F(z)^{15}+z^{21} F(z)^{20}$$

Any help appreciated

Answer 1

Use Lagrange inversion. From the functional equation that $F(z)$ satisfies, we get, by multiplying through by $z$, $$ zF(z)=z^2\left(1+(zF(z))^5+(zF(z))^{10}+(zF(z))^{15}+(zF(z))^{20}\right). $$ Thus, $$ \frac{z}{1+z^5+z^{10}+z^{15}+z^{20}}\circ(zF(z))=z^2, $$ so $$ zF(z)=\left(\frac{z}{1+z^5+z^{10}+z^{15}+z^{20}}\right)^{\langle-1\rangle}\circ z^2, $$ where "$\langle-1\rangle$" denotes the compositional inverse. Thus, $zF(z)$ is an even function, i.e. $zF(z)=H(z^2)$ for the invertible power series $H(z)=\left(\frac{z}{1+z^5+z^{10}+z^{15}+z^{20}}\right)^{\langle-1\rangle}$. This means that $H=H(z)$ satisfies the functional equation $H=z\varphi(H)$ for $\varphi(z)=1+z^5+z^{10}+z^{15}+z^{20}$. Note that $\varphi(0)\ne 0$. We need to find the coefficients of $$ F(z)^8=\frac{H(z^2)^8}{z^8}=\frac{H(z)^8}{z^4}\circ z^2. $$ By Lagrange inversion, $$ [z^n]f(H(z))=\frac{1}{n}[z^{n-1}]\!\left(\varphi(z)^n f'(z)\right), $$ where $[z^n]$ means the coefficient at $z^n$. Here, $f(z)=z^8$, so $f'(z)=8z^7$, and we need essentially \begin{multline*} [z^n]\frac{H(z)^8}{z^4}=[z^{n+4}]H(z)^8=\frac{1}{n+4}[z^{n+3}]\!\left(8z^7\varphi(z)^{n+4}\right)=\\=\frac{8}{n+4}[z^{n-4}] (1+z^5+z^{10}+z^{15}+z^{20})^{n+4}. \end{multline*} And since $1+z^5+z^{10}+z^{15}+z^{20}=(1+z+z^2+z^3+z^4)\circ z^5$, only the terms where $n\equiv 4\pmod 5$ would be nonzero in the $[z^{n-4}] (1+z^5+z^{10}+z^{15}+z^{20})^{n+4}$. And since we are also composing with $z^2$ at the end, $F(z)^8$ will only have terms with powers congruent to $8\!\!\!\mod\!\!10$.

Finally, here is the pentanomial coefficient triangle. Scroll down to example, it's easier to see there.

Answer 2

A method that will take a while but gives the first few terms for the equation $$ F(x) = x + x^6 F(x)^5 + x^{11} F(x)^{10} + x^{16} F(x)^{15} + x^{21} F(x)^{20} $$ is to let $F(x) = a_{0} + a_{1} \, x + a_{2} \, x^2 + \cdots$.

First notice that the right-hand side does not have a constant term. This leads to $F(x)$ being $F(x) = x + a_{2} x^2 + \cdots$. Also notice that $a_2 = a_{3} = \cdots = a_{10} = 0$ and further reduces $F(x)$ to $F(x) = x + x^{11} + \cdots$

This leads to \begin{align} x + x^{11} + a_{21} \, x^{21} + \cdots &= x + x^{11} (1 + 5 \, x^{10} + \cdots) + x^{21} (1 + 10 x^{10} + \cdots) + \cdots \\ &= x + x^{11} + (5 + 1) \, x^{21} + \cdots. \end{align} Continuing gives the result mentioned by Gary in the comments.