Growth series for the lamplighter group

Question

This paper gives the growth series of the lamplighter group as

$$ \frac{(1+x)(1-x^2)^2(1+x+x^2)}{(1-x^2+x^3)^2(1-x-x^2)} = 1 + 3x + 6x^2 + 8x^3 + 10x^4 + \ldots $$

The first 3 coefficients seem to be correct. However, I get 12 group elements with a word norm of 3. They are the following:

\begin{array}{l l} ttt & at^{-1}a \\ tta & at^{-1}t^{-1} \\ tat & t^{-1}at \\ tat^{-1} & t^{-1}at^{-1} \\ att & t^{-1}t^{-1}a \\ ata & t^{-1}t^{-1}t^{-1} \end{array}

What is the reason for this discrepancy?

Answer 1

You are right. On page 6 of that paper, equation $(3.5)$ is given as: $$f_{G\wr\mathbb{Z}}(x)=\frac{f_G(x)(1-x^2)^2(1+xf_G(x))}{(1-x^2f_G(x))^2(1-xf_G(x))}$$

On substituting $(1+x)$ for $f_G(x)$, I get:

$$f_{{\cal{L}}_2}(x)=\frac{(1+x)(1-x^2)^2(1+x+x^2)}{(1-x^2-x^3)^2(1-x-x^2)}$$

And when expanded, the result seems to agree with you:

$$1 + 3 x + 6 x^2 + 12 x^3 + 22 x^4 + 40 x^5 + 71 x^6 + 123 x^7 + 212 x^8 + 360 x^9 + 607 x^{10} +...$$

In the working at the bottom of the same page, a $^2$ goes missing, and then $(-x^2)(+x)$ becomes $+x^3$ in the denominator. It's just a mistake.