20160919, 00:29  #1 
Apr 2014
200_{8} Posts 
Number of sequences that merge with any given sequence  infinite?
Are the number of sequences that merger with any given sequence infinite in number? Does changing the definition of 'any give sequence' change the answer?

20160919, 00:40  #2  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:


20160919, 01:09  #3  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9263_{10} Posts 
Quote:
Take sequence 2. Nothing merges into it and this is easy to prove. Then, consider sequence 28. Again, nothing (except 28 itself). Or sequence 3 (for which the single merging seq is alq(4)). Etc. __________________________ Forking this question now into a separate thread, because it has nothing to do with the subject of alq(4788). 

20160919, 14:03  #4  
"Andrew Booker"
Mar 2013
85_{10} Posts 
Quote:
For even numbers it's more complicated. For all of the ones that terminate, it's true by the above, since they will hit an odd prime different from 5. On the other hand, as Batalov pointed out, there are cycles consisting entirely of even numbers. Some of those also have infinitely many merging sequences (e.g. \(6=s(25)\) and \(284=s(80089)\)), but others provably do not (e.g. 28). 

20160919, 14:21  #5  
"Forget I exist"
Jul 2009
Dumbassville
20300_{8} Posts 
Quote:
edit:in fact given that argument you can say that until 2p for all prime p has at least 2 goldbach partitions you can't prove it. also if every odd number has to go back to a odd semiprime that implies either odd perfect numbers ( if they exist) must either be odd semiprimes ( I don't think their known form if they exist allows for that) or must have at least 2 sequences that merge with them directly. Last fiddled with by science_man_88 on 20160919 at 14:34 

20160919, 14:45  #6 
"Andrew Booker"
Mar 2013
5·17 Posts 
It's likely true that every even number at least 8 is the sum of two distinct primes (and this is what I meant by a strong form of Goldbach's conjecture). The claim is still true for 7 since \(7=s(8)=s(s(49))=\dots\).
Last fiddled with by arbooker on 20160919 at 14:56 
20160919, 14:47  #7 
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
now you have to guarantee it's not going to hit an untouchable number.

20160919, 15:03  #8 
"Andrew Booker"
Mar 2013
5·17 Posts 
Isn't that clear? Just keep going, e.g. \(49=s(215)=s(s(633))=\dots\). Each lift will be odd, so again we can find a product of two primes for the next one. Of course I can't prove that it will keep working since that would require proving the Goldbach conjecture, but that doesn't stop it being true.
Last fiddled with by arbooker on 20160919 at 15:06 Reason: changed to products of two primes 
20160920, 11:23  #9 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
5×19×61 Posts 
What are the largest untouchable numbers known?

20160920, 14:40  #10 
"Andrew Booker"
Mar 2013
85_{10} Posts 
It's known that a positive proportion of the natural numbers are untouchable, and in that sense there is no largest known. The proportion is conjectured to be about 17%, so they're fairly common.
I did a few experiments and was surprised to find that most aliquot cycles seem to be in the orbit of some odd number (and thus presumably infinitely many odd numbers). After 28, the next smallest counterexample is the amicable pair (356408,399592). 
20160920, 15:02  #11  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
Last fiddled with by science_man_88 on 20160920 at 15:06 

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