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November 4th, 2010, 09:13 PM  #1 
Joined: Nov 2010 Posts: 7 Thanks: 0  Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 ...?
where the next number has n+1 divisors of the previous number in the sequence, and is the lowest number with that many divisors Code: number divisors 1 1 2 1,2 6 1,2,3 12 1,2,3,4 24 1,2,3,4,6 48 1,2,3,4,6,8 .... 
November 4th, 2010, 09:58 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 5,874 Thanks: 154  Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 .
Your example has omissions: Code: number divisors 1 1 2 1,2 6 1,2,3 12 1,2,3,4,6 24 1,2,3,4,6,12 48 1,2,3,4,6,8,12,24 
November 4th, 2010, 10:06 PM  #3 
Joined: Nov 2010 Posts: 7 Thanks: 0  Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 .
Thanks...that leads me directly to another question How is it determined that there isn't a number, less than 48, with 7 divisors? 
November 4th, 2010, 10:12 PM  #4  
Joined: Jun 2009 Posts: 13 Thanks: 0  Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 . Quote:
 
November 4th, 2010, 10:30 PM  #5  
Joined: Nov 2010 Posts: 7 Thanks: 0  Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 . Quote:
 
November 4th, 2010, 10:39 PM  #6  
Joined: Oct 2009 Posts: 105 Thanks: 0  Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 . Quote:
Every integer has an even number of divisors, correct? Unless it is a perfect square. Therefore, it is trivial to check 9, 16, 25, 36, etc...  
November 4th, 2010, 10:58 PM  #7  
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 5,874 Thanks: 154  Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 . Quote:
Code: 1 1 2 1, 2 6 1, 2, 3 12 1, 2, 3, 4, 6 24 1, 2, 3, 4, 6, 8, 12 48 1, 2, 3, 4, 6, 8, 12, 16, 24 Quote:
24 has 1, 2, 3, 4, 5, 8, 12 as divisors (not including 24). The pattern does not hold (per example); for example 6 has 3 divisors (not including 6) and 12 has 5 divisors (not including 12) and there is no number between 6 and 12 that has 4 divisors if that number is not included as a divisor.  
November 4th, 2010, 11:06 PM  #8 
Joined: Nov 2010 Posts: 7 Thanks: 0  Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 .
Doh...of course...I was so intrigued by the mathematics of it, I didn't go back and check whether the statement or example were sound before posting it up! Is it OK to evolve the thought? As you're pointed out, n+1 divisors cannot hold...in that case, how would the sequence be if the statement was edited as follows? "where the next number has more divisors than the previous number in the sequence, and is the lowest number with that many divisors" 
November 5th, 2010, 12:18 AM  #9  
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 5,874 Thanks: 154  Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 . Quote:
Maybe the n+1 idea holds (with a different table). Here's a start: Code: 1 1 2 1, 2 6 1, 2, 3 16 1, 2, 4, 8 18 1, 2, 3, 6, 9 30 1, 2, 3, 5, 10, 15 42 1, 2, 3, 6, 7, 14, 21 256 1, 2, 4, 8, 16, 32, 54, 128 Perhaps of interest: Code: 2^1 1, 2 2^2 1, 2, 4 2^3 1, 2, 4, 8 2^4 1, 2, 4, 8, 16  
November 5th, 2010, 12:27 AM  #10 
Joined: Nov 2010 Posts: 7 Thanks: 0  Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 .
Actually, if it's legal to include 1, then should it be permitted to use the number itself? How does the sequence look then? As soon as you get over half a dozen divisors, it becomes really laborious to check candidates. Currently also mulling the significance of the binary sequence...v.interesting 

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