
Number Theory Number Theory Math Forum 
 LinkBack  Thread Tools  Display Modes 
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: 5,543 Thanks: 22  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: 5,543 Thanks: 22  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: 5,543 Thanks: 22  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 
November 5th, 2010, 01:57 AM  #11 
Math Team Joined: Apr 2010 Posts: 1,763 Thanks: 19  Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 .
What also may be of help is primefactorisation. Consider these examples, shown by factorisation, excluding the number itself. 2=2^1. There are 2 options for the exponent. 0 or 1. only 2^0 is a factor of 2^1, and below 2. So 2 has 1 divisor but itself, being 1. 4=2^2. Now, we have 3 choises for the exponent, 0, 1 or 2. Exclude the number itself, and get as divisors: 2^0=1 and 2^1=2. 6=2^1*3^1. For all exponents, there are options, 0 or 1. Informal: 2 bases * 2 options is 4 numbers. Exclude the number itself (both exponents 1) : 3 factors, beiing 2^0*3^0=1, 2^1*3^0=2 and 2^0*3^1=3. The 3 divisors are 1, 2 and 3. Up until now, the method is a huge overkill, but I think, the efford pays of for larger numbers: 30=2^1*3^1*5^1. For all exponents, we have 2 choises, 0 or 1. 3 bases gives 2^3=8 divisors. Exclude the number itself and find the divisors beiing: 2^0*3^0*5^0=1 2^1*3^0*5^0=2 2^0*3^1*5^0=3 2^0*3^0*5^1=5 2^1*3^1*5^0=6 2^1*3^0*5^1=10 2^0*3^1*5^1=15 A number with 7 divisors. 7 is not factorable. The lowest prime is 2. So 2^7=128 is the first and only number with 7 divisors, excluding 128. Are you looking for the lowest number with 1000 divisors excluding the number itself: add 1 to 1000 and find 1001. factor 1002, gives 7^1*11^1*13^1 add the exponents, 1+1+1=3. So you will use 3 bases. The 3 lowest prime bases are 2, 3 and 5. So we find that substract 1 of all the primes is the lowest number with 1000 divisors excluding itself. Hoempa 
November 5th, 2010, 05:58 AM  #12 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 12,862 Thanks: 94  Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 .
This is Sloane's A005179. Pari/GP code: Code: find(k)=if(k%2,for(n=1,9e9,if(numdiv(n^2)==k,return(n^2))),for(n=1,9e9,if(numdiv(n)==k,return(n)))); n=1;k=1;while(1,print1(n", ");n=find(k++)) 
November 5th, 2010, 07:04 AM  #13 
Joined: Nov 2010 Posts: 7 Thanks: 0  Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 .
What an excellent forum! I'll digest the replies later, but thanks for your thoughts and for being forgiving of my errors...(I'm more used to forums where rudeness is the default position) Cheers :P 
November 5th, 2010, 09:10 AM  #14  
Joined: Oct 2009 Posts: 105 Thanks: 0  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  
November 5th, 2010, 11:16 AM  #15  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 2,598 Thanks: 46  Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 . Hello, six! Quote:
 

Tags 
recognise, sequence 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Gp sequence  taylor_1989_2012  Algebra  4  February 12th, 2013 05:58 PM 
A sequence  Albert.Teng  Algebra  2  October 5th, 2012 06:14 PM 
Sequence  15sliu  Elementary Math  0  November 14th, 2010 07:32 AM 
sequence of the average of a sequence  elim  Real Analysis  8  May 23rd, 2010 12:23 PM 
cauchy sequence that isn't a fast cauchy sequence  babyRudin  Real Analysis  6  October 10th, 2008 11:11 AM 