Las fracciones son $$\frac{17}{91}, \frac{78}{85}, \frac{19}{51}, \frac{23}{38}, \frac{29}{33}, \frac{77}{29}, \frac{95}{23}, \frac{77}{19}, \frac{1}{17}, \frac{11}{13}, \frac{13}{11}, \frac{15}{2}, \frac{1}{7}, \frac{55}{1}.$$ 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 $55 \tiempos de 15^n$ 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 $256$) pero cuando pensé que tal vez me estoy duplicando esfuerzos en algunas bien documentado artículo de revista.
Es ya conocido, o es fácil demostrar, que a partir de $2^n$ $n$ compuesto siempre hace que el proceso vuelva a la pista para darle los números primos, o es posible que la máquina para conseguir sumido en un bucle infinito?