Let $n=a\_ka\_{k-1}\ldots a\_1a\_0$ be a $(k+1)$ digit number.


**Step 1:** Find $a\_k \times a\_{k-1} \times \ldots \times a\_1 \times a\_0= c\_tc\_{t-1}\ldots c\_1c\_0$.


**Step 2:** Find $a\_k + a\_{k-1} + \ldots + a\_1 + a\_0= d\_ld\_{l-1}\ldots d\_1d\_0$.


Then form $n\_1=c\_tc\_{t-1}\ldots c\_1c\_0d\_ld\_{l-1}\ldots d\_1d\_0$.


Repeat steps 1 and 2 for $n\_1$ if $n\_1$ is not a single digit number. Else, stop.


Examples: $1 \to 1; \ldots ; 9 \to 9; 10\to 01=1; 11\to 12 \to 23\to 65\to 3011\to 05=5 ; \ldots$


**Note:**
 All the permutations of the digits of a number yield the same result, i.e. for instance, $127,172,217,271,712,721 \to 1410\to 06$. Hence, once all numbers from 1 to 50 have been checked, we need only check 55, 66, 77, 88 and 99 to extend our result for the range 1-100.


**Facts:**
All numbers except $38$(or $83$), $66$, and $88$ converge. $38$ and $88$ fall into the 5-cycle: $$88\to 8416 \to 14417\to 11217\to 1412 \to 88$$ while $66$ falls into the interesting 1-cycle: $$3612\to 3612$$
Also $6417$ falls into the 5-cycle: $$16818\to 38424\to 76821\to 67224\to 67221\to 16818$$


**Question:** Is the Collatz phenomenon happening here?(i.e. Do all numbers yield a sequence that converges to a single digit number or a sequence that falls into a cycle?)


**My question has been answered**. But I had some thoughts going on based on my experiment: Is the number of cycles finite?


Thanks.(MO is cool! I can't disagree.)



In the sense of Collatz
Yes, the only way that you can get a sequence which doesn't converge to a repeating cycle is if you find a sequence with unbounded elements, but it is easy to show that this is not the case. If you start with a number $n$, with $k$ digits, then the number of digits in $f(n)$ (after you perform both steps) is at most $2+\log\_{10}(9^k)+\log\_{10}(9k)$ which is less than $k$ for large enough $k$.


