20200209, 19:54  #1 
(loop (#_fork))
Feb 2006
Cambridge, England
3^{3}·239 Posts 
An unreasonable operation
Consider the operation where you double a number and then cross out all the zeroes in its base10 representation
eg 34 > 68 > 136 > 272 > 544 > 188 > 376 > 752 > 154 > 38 > 76 > 152 > 34 It's heuristically obvious that every number ends up in a cycle. Running on starts up to 10^6 and 3141592653589 .. 3141592753589 gives me a reasonably strong belief that the shortest cycle length is 12 and the longest 432 (starting 118), which is more structure than I'd have expected. You get the same sort of variouslimitcycles behaviour for multiplying by K rather than doubling; K=6 has fixed points (eg 18). K=7 has a much larger set of limit cycles, ranging from a fixed point at 15 to a cycle of length 6600 starting with 1157313. k=8 also gives quite a wilderness of limit cycles. By k=11 there is a cycle of length 30960 (hit by 925, where the first repeated value is 786699925). I don't have a clue how to prove that there aren't other cycles. 
20200209, 20:20  #2 
Feb 2020
11 Posts 
The likelihood of a random number containing a 0 increases as the size of the integer increases, so there will be a certain range of values at which numbers no longer increase on average. This is the range where the loops will be. I think you can conclude above a certain value that such things are unlikely, but attempting to prove that no more cycles exist seems similar to trying to prove the collatz conjecture. I happen to have posted a sequence in the "math" forum recently that is a similar kind of operation, though it doesn't have a dependence on its size (meaning the numbers can get extremely large if a sequence meanders in that direction).

20200209, 20:21  #3 
(loop (#_fork))
Feb 2006
Cambridge, England
1100100110101_{2} Posts 
An analogous question:
Prove there is no integer for which n, 2n, 4n, ..., (2^20)n all have no 0 in their decimal expansion Find an integer for which n, 2n, 4n, ..., (2^19)n all have no 0 in their decimal expansion Much harder if you look at powers of 3, 7 or 9 Last fiddled with by fivemack on 20200209 at 20:22 
20200210, 03:00  #4  
Aug 2006
3·1,993 Posts 
Quote:
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Code:
has(n)=vecmin(digits(n))>0 is(n)=for(k=1,19,if(!has(n<<k),return(0)));1 for(N=1,9^8, n=fromdigits(digits(N,9)); k=n%10; if((k==2k==7) && is(10*n+9), return(10*n+9))) 

20200210, 14:41  #5  
Feb 2017
Nowhere
2^{2}×3×13×37 Posts 
Quote:
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20200210, 15:41  #6  
"Robert Gerbicz"
Oct 2005
Hungary
1,567 Posts 
Quote:
It is easy to see that the length should be divisible by four, and n mod 10 is in {2,4,6,8}. And if we'd delete 1 in the numbers (instead of zero), then for every starting number it goes to a cycle, because the iterated numbers can't be longer, otherwise (2*n) would be start by one, but we delete that. 

20200210, 20:53  #7 
Feb 2017
Nowhere
5772_{10} Posts 

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