Let $n=d_m...d_2d_1$ where $d_i$ is digit of $n$ .
Define function $F(n)$, If there are any digits of $n$ being repeated consecutively, they convert those digits to that digit and generate new number.
Example: $F(10225)=1025$ because 2 repeat itself
$F(10000)=10$, $F(223335300)=23530$, $F(23)=23$.
Define recurrence relation $a_k= F(2\cdot a_{k-1})$ for $k\ge 1$ and $a_0=1$
$\{a_0,a_1,a_2,...\}=\{1 , 2 , 4 , 8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,6536,13072,2614,528,1056,212,424,848,1696,392,784,1568,3136,6272,1254,2508,5016,1032,2064,4128,8256,16512,3024,6048,12096,24192,48384,96768,19356,387072,...\}$
Problem: Does above sequence will ends with loop or is it approaches to infinity?
Generalization of this problem posted by Vepir : Can we reach infinity if we remove repeating digits from $a_k=m\cdot a_{k-1}$?
I haven't finished programming. It is very possible that you will get some loop. Thanks