TITLE: In general what happens in Conway's Prime Game given $2^n$, with $n$ composite, as the initial value?
QUESTION [7 upvotes]: The fractions are $$\frac{17}{91}, \frac{78}{85}, \frac{19}{51}, \frac{23}{38}, \frac{29}{33}, \frac{77}{29}, \frac{95}{23}, \frac{77}{19}, \frac{1}{17}, \frac{11}{13}, \frac{13}{11}, \frac{15}{2}, \frac{1}{7}, \frac{55}{1}.$$ Obviously (at least to me after thinking about it for more than a few minutes) $2^n$ becomes $15^n$ as the evenness of the number is divided out and $3$s and $5$s are added to the factorization. The next step is then $55 \times 15^n$ followed by $65 \times 15^n$. After that it's less clear to generalize. At least for $n = 4, 6$ I have verified that the process gets back on track. I have started calculations for $n = 8$ (corresponding to $256$) but that's when I thought maybe I'm duplicating effort in some well-documented journal article.
Is it already known, or is it easy to prove, that starting with $2^n$ with $n$ composite always gets the process back on track to give prime numbers, or is it possible for the machine to get mired in an infinite loop?

REPLY [4 votes]: At least according to the Wikipedia entry on Conway's prime algorithm in FRACTRAN, the algorithm works when started with a number of the form $2^n 7^m$ where $0\le m < n$.  It then generates all numbers $2^{n'} 7^{k-1}$ (along with other numbers with prime factors besides $2$ and $7$) where $n' > n$ and $k$ is the largest proper divisor of $n'$.  In particular, it generates exactly those powers of $2$ for which the exponent is prime and larger than $n$.
In other words, what you're observing is exactly what the algorithm is supposed to do.  You can start it with any $2^n$, regardless of whether $n$ is prime or composite, or even with $2^n 7^m$ as long as $m < n$.