My Math Forum Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 ...?

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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 .... How would one express the sequence mathematically? Is the sequence interesting?
 November 4th, 2010, 09:58 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 5,542 Thanks: 21 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

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Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 .

Quote:
 Originally Posted by greg1313 Your example has omissions:
where's 36??

November 4th, 2010, 10:30 PM   #5

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Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 .

Quote:
Originally Posted by scoracle
Quote:
 Originally Posted by greg1313 Your example has omissions:
where's 36??
36 has divisors 1,2,3,4,6...the same as 24, therefore it is trumped by 24, as it is the lower of the two.

November 4th, 2010, 10:39 PM   #6

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Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 .

Quote:
 Originally Posted by Lupeto Thanks...that leads me directly to another question How is it determined that there isn't a number, less than 48, with 7 divisors?

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
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Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 .

Quote:
 Originally Posted by Lupeto 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 ....
The above table is incomplete -- here is a complete version:

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:
 Originally Posted by Lupeto 36 has divisors 1,2,3,4,6...the same as 24, therefore it is trumped by 24, as it is the lower of the two.
36 has 1, 2, 3, 4, 6, 9, 12, 18 as divisors (not including 36).
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
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Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 .

Quote:
 Originally Posted by Lupeto "where the next number has more divisors than the previous number in the sequence, and is the lowest number with that many divisors"
I'd imagine that's possible.

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
I don't know if there is a number between 42 and 256 with 8 divisors (not including the number itself).

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
A handy site: http://www.research.att.com/~njas/sequences/

 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 $2^{12}*3^{10}*5^6=3779136000000$ 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++)) Of course there are 2000 terms at the link, so you don't need to generate this yourself. (Faster code would, at a minimum, use the fact that the numbers will be products of primorials.)
 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

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Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 .

Quote:
Originally Posted by greg1313
Quote:
 Originally Posted by Lupeto "where the next number has more divisors than the previous number in the sequence, and is the lowest number with that many divisors"
I'd imagine that's possible.

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
30 is missing 6. This method will only work if every other number in the sequence is a perfect square.

November 5th, 2010, 11:16 AM   #15
Math Team

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From: Lexington, MA

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Thanks: 46

Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 .

Hello, six!

Quote:
 How is it determined that there isn't a number less than 48 with 7 divisors?

$\text{A number with 7 divisors has the form: }\,a^6$

$\text{Its divisors are: }\:1,\:a,\:a^2,\:a^3,\:a^4,\:a^5,\:a^6$

$\text{The least number occurs when }a=2:\;\;2^6\:=\:64$

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