Why are numbers sum(2^(k*c)) smooth?
 

I made this experimental observation, that numbers of the form:

2^(0*c) + 2^(1*c) + 2^(2c) + ... + 2^(k*c) are smooth. (by this I mean that the largest factor in their factorization is very small; they also have many factors).

Somebody can explain me why?

[QUOTE=preda;494905]I made this experimental observation, that numbers of the form:

2^(0*c) + 2^(1*c) + 2^(2c) + ... + 2^(k*c) are smooth. (by this I mean that the largest factor in their factorization is very small; they also have many factors).

Somebody can explain me why?[/QUOTE]
My 2 cents:
In its simplest form for c=1, every consecutive-consecutive double pair are coprime.
Summing up bunch of coprimes will result in a number of small factors, but could also include some very large prime factors.
For c>1, you end up with multiples of the simplest form c= 1.

ETA One important variable which makes a great difference is if the total number of addend terms is odd or even.
See below for a parallel.
[url]http://www.mersenneforum.org/showpost.php?p=453119&postcount=9[/url]

[QUOTE=a1call;494906]My 2 cents:
In its simplest form for c=1, every consecutive-consecutive double pair are coprime.
Summing up bunch of coprimes will result in a number of small factors, but could also include some very large prime factors.
For c>1, you end up with multiples of the simplest form c= 1.

ETA One important variable which makes a great difference is if the total number of addend terms is odd or even.
See below for a parallel.
[url]http://www.mersenneforum.org/showpost.php?p=453119&postcount=9[/url][/QUOTE]

I don't really understand your explanation.. maybe a bit slower for me.

But I did test the odd/even hypothesis; in pari-gp (which I barely know):
f(c,n)=for(m=1,n,print(#factor(sum(k=0,m,2^(k*c)))[,1]))

f(10, 16)
2
4
4
4
9
6
6
9
9
9
13
6
12
11
9
4

? f(15, 16)
3
3
8
4
8
8
12
7
10
7
18
5
18
11
18
8

? f(20, 16)
2
7
4
8
11
10
7
17
14
13
17
13
14
22
11
12

Thus I don't see much support for odd/even.

Try it with s primorial initial valve and add your terms to it.
This will exaggerate the effect.
I predict you should see a frequency of 4 cycles.

[QUOTE=preda;494905]2^(0*c) + 2^(1*c) + 2^(2c) + ... + 2^(k*c) are ...


Somebody can explain me why?[/QUOTE]
Why is [$]{2^{c(k+1)} - 1} \over {2^c - 1}[/$] smooth?
Is it because [$]{2^{c(k+1)} - 1}[/$] has tons of algebraic factors, maybe? Someone played paper darts too much in the 8th grade math class?

[QUOTE=Batalov;494913]Why is [$]{2^{c(k+1)} - 1} \over {2^c - 1}[/$] smooth?
Is it because [$]{2^{c(k+1)} - 1}[/$] has tons of algebraic factors, maybe? Someone played paper darts too much in the 8th grade math class?[/QUOTE]

Now I see the error of my ways :)

But what happens when 2^c - 1 does not divide [$]2^{c(k+1)} - 1[/$], how are the factors affected?

In which situation is the number of factors maximized -- e.g. is it good for k+1 to be a power of 2? is it good or bad for c to be even? or c to be a power of 2?

Thanks!

[QUOTE=preda;494915]But what happens when 2^c - 1 does not divide [$]2^{c(k+1)} - 1[/$], ...
[/QUOTE]
That's not possible. This is like asking "at what length of a string of 9s, the number 999999999...99999999 is not divisible by 9?"

[QUOTE=preda;494915]... how are the factors affected?
[/QUOTE]
Well the factors of 2^c-1 are taken away. There are still tons left.

[QUOTE=a1call;494912]Try it with s primorial initial valve and add your terms to it.
This will exaggerate the effect.
I predict you should see a frequency of 4 cycles.[/QUOTE]

Are you saying that adding a primorial to the sum() would increase the number of factors? -- why?

[QUOTE=Batalov;494916]That's not possible. This is like asking "at what length of a string of 9s, the number 999999999...99999999 is not divisible by 9?"
[/QUOTE]
sorry, that was a silly question.

[QUOTE]Well the factors of 2^c-1 are taken away. There are still tons left.[/QUOTE]
Do you know which conditions (on k and c) would tend to maximize the number of factors left?

[QUOTE=preda;494918]Do you know which conditions (on k and c) would tend to maximize the number of factors left?[/QUOTE]

I'll attempt a rough analysis of the number of factors of

N(c, k) = sum(i=0, k, 2^(c*i)) == (2^(c*(k+1)) - 1) / (2^c - 1)

When k == 2^p - 1, then:

N(c, k) == prod(i=0, p-1, 2^(c * 2^i) + 1)

Let's take an example,
N(c, 7) = (2^(8c) - 1)/(2^c - 1) = (2^c + 1) * (2^(2c) + 1) * (2^(4c) + 1)

So the number of factors of N(c, k) is the sum of the nb. of factors of 2^(c*2^i) + 1, with i from 0 to log2(k). This may turn out to be O(log(k)^2), and seems to be larger for highly composite c.

Also, I confirm that indeed the number of factors of N(c, k) seems to be larger for odd k than for even k.

[URL="http://myfactors.mooo.com/#AlgebraicFactorsQuery"]This[/URL] might be useful for playing around with.