## Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 ...?

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### 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: Select all
number  divisors1       12       1,26       1,2,312      1,2,3,424      1,2,3,4,648      1,2,3,4,6,8....

How would one express the sequence mathematically? Is the sequence interesting?
Lupeto
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### Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 .

Code: Select all
number  divisors1             12             1,26             1,2,312           1,2,3,4,624           1,2,3,4,6,1248           1,2,3,4,6,8,12,24
Numbers don't lie. They hide, but they don't lie.

greg1313
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### 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?
Lupeto
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### Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 .

where's 36??
scoracle
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### Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 .

scoracle wrote:

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

Lupeto wrote: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...
six
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### Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 .

Lupeto wrote: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: Select all
number  divisors1       12       1,26       1,2,312      1,2,3,424      1,2,3,4,648      1,2,3,4,6,8....

The above table is incomplete -- here is a complete version:

Code: Select all
1    12    1, 26    1, 2, 312  1, 2, 3, 4, 624  1, 2, 3, 4, 6, 8, 1248  1, 2, 3, 4, 6, 8, 12, 16, 24

Lupeto wrote: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.
Last edited by greg1313 on Fri Nov 05, 2010 2:35 am, edited 1 time in total.
Numbers don't lie. They hide, but they don't lie.

greg1313
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### 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"
Lupeto
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### Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 .

Lupeto wrote:"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: Select all
1     12     1, 26     1, 2, 316   1, 2, 4, 818   1, 2, 3, 6, 930   1, 2, 3, 5, 10, 1542   1, 2, 3, 6, 7, 14, 21256 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: Select all
2^1    1, 22^2    1, 2, 42^3    1, 2, 4, 82^4    1, 2, 4, 8, 16

A handy site: http://www.research.att.com/~njas/sequences/
Last edited by greg1313 on Fri Nov 05, 2010 3:30 am, edited 1 time in total.
Numbers don't lie. They hide, but they don't lie.

greg1313
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### 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
Lupeto
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### 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
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### Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 .

This is Sloane's A005179. Pari/GP code:
Code: Select all
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.)
Pari/GP: this is the program I probably mentioned in my post. Windows users can get it at http://pari.math.u-bordeaux.fr/pub/pari ... -2-6-2.exe

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

greg1313 wrote:
Lupeto wrote:"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: Select all
1     12     1, 26     1, 2, 316   1, 2, 4, 818   1, 2, 3, 6, 930   1, 2, 3, 5, 10, 1542   1, 2, 3, 6, 7, 14, 21256 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.
six
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### Re: Does anyone recognise the sequence 1, 2, 6, 12, 24, 48 .

Hello, six!

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

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

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

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

soroban
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