20180830, 04:50  #1 
"Mihai Preda"
Apr 2015
3^{2}·151 Posts 
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? 
20180830, 05:03  #2  
"Rashid Naimi"
Oct 2015
Remote to Here/There
2·1,009 Posts 
Quote:
In its simplest form for c=1, every consecutiveconsecutive 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. http://www.mersenneforum.org/showpos...19&postcount=9 Last fiddled with by a1call on 20180830 at 05:16 

20180830, 06:14  #3  
"Mihai Preda"
Apr 2015
2517_{8} Posts 
Quote:
But I did test the odd/even hypothesis; in parigp (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. 

20180830, 06:46  #4 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2×1,009 Posts 
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. Last fiddled with by a1call on 20180830 at 06:47 
20180830, 07:01  #5  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2·3·1,571 Posts 
Quote:
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? 

20180830, 07:27  #6  
"Mihai Preda"
Apr 2015
3^{2}×151 Posts 
Quote:
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! 

20180830, 07:35  #7  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2·3·1,571 Posts 
Quote:
Well the factors of 2^c1 are taken away. There are still tons left. 

20180830, 07:35  #8 
"Mihai Preda"
Apr 2015
3^{2}·151 Posts 

20180830, 08:16  #9  
"Mihai Preda"
Apr 2015
3^{2}·151 Posts 
Quote:
Quote:


20180830, 12:03  #10  
"Mihai Preda"
Apr 2015
3^{2}×151 Posts 
Quote:
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, p1, 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. 

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