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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.)

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closed as too localized by Andres Caicedo, Dmitri Pavlov, Pete L. Clark, Zev Chonoles, S. Carnahan Feb 28 '11 at 7:21

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3  
The Collatz process is more interesting than yours because it does not depend on the decimal representation of numbers. –  lhf Feb 27 '11 at 18:16
    
No oeis.org entry. –  Chulumba Feb 27 '11 at 18:16
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@lhf: but the Collatz problem depends on the number "3". –  Igor Rivin Feb 28 '11 at 2:13

1 Answer 1

up vote 5 down vote accepted

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 $k$.

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To be more precise, Mathematica tells me that $f(n)<n$ for all $n$ with at least 110 digits. –  Gjergji Zaimi Feb 27 '11 at 19:17
    
@Gjergji, thanks. I don't see why your answer is definitive. Here is why: Denote the number of digits ($1+\lfloor \log_{10}f(n)\rfloor $) by $N(f(n))$. Start with $n$ having number of digits $k$ large enough. I am okay with the proposition that $N(f(n))<k$ for $k$ large enough. Continuing the process, we may reach a number whose number of digits is not large enough. This number then may explode and surpass our original $n$. This means we now have a number whose number of digits is large enough...then back to the 'small number of digits' case...then back to the 'large number of digits' case... –  Chulumba Feb 27 '11 at 19:52
    
@Gjergji Zaimi. cont'd $\ldots$This shows why the "number of digits" argument might not be so definitive. However, your Mathematica computation seems to be in favor. Do you agree? –  Chulumba Feb 27 '11 at 19:56
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I don't understand your comment. I showed that there is a number $N$ so that $f(n)<n$ for all $n>N$. This means that no matter which number you start with, at some point you will end up in the range $[1,N]$, after which there are only finitely many possibilities for the sequence to continue. Yes, after it reaches this interval it may again jump outside the interval but it will decrease again until it becomes $<N$ etc. So essentially you can see that every sequence is eventually bounded (by $\max (f(n))$ for all $n\le N$) and therefore eventually periodic. –  Gjergji Zaimi Feb 27 '11 at 20:06
    
@Gjergji Zaimi. I loved your answer. I can see the reason now. On further experiment, may be there are a finite number of cycles into which a number may fall(just a thought). –  Chulumba Feb 27 '11 at 20:35

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