Conway's arithmetical theorem on look and say sequences I have trouble understanding the proof of the arithmetical theorem (see attachment), and especially this part : " On the other hand, our Chemical theorem shows that the descendants oh each of He, Li, ..., U increase as fast as any of them, and that this is at some rate > 1. ".
I don't understand why's that. The chemical theorem tells us that all sufficiently late descendants of all the common elements other than Hydrogen involve all of the 92 elements simultaneously. But it doesn't tell us anything about rates, does it ?
Thank you all !
Original Conway's article on the look and say sequences

 A: The way I would look at this is similar to a finite irreducible positive-recurrent aperiodic Markov Chain with a transition matrix where outcome elements (other than $H$) add up to $1$ for each element.
If you exclude the element $H (22)$, then you can consider the proportional presence of the other elements.  These proportions must tend to positive constants as the number of steps increase because:

*

*each element can be reached from every other (so must be recurrent and irreducible); I once drew a diagram of some of the paths to count the number of complete orderings of the elements



*there are a finite number of elements (so must be positive recurrent)


*the lengths of some loops are coprime to the lengths of other loops (so must be aperiodic): for example $Li \to He \to Li$ and $Ho \to Dy \to Tb \to Ho$
This implies that there is a single stationary distribution and this is the limiting distribution of proportions of the non-$H$ elements as the number of steps increases.
The stationary distribution is an eigenvector of the transition matrix with corresponding eigenvalue $1$.
Considering growth of numbers, the growth rates for each element must converge to the same limit so as to maintain the proportions over time.
You could say that, instead of a transition matrix where outcome elements add up to $1$ for each element,  you use a transition matrix where every outcome element is $1$ (or $0$), then it will also have the stationary distribution as an eigenvector and the growth rate $\lambda$ as the corresponding eigenvalue.
