20060213, 17:18  #34  
"Robert Gerbicz"
Oct 2005
Hungary
10110111010_{2} Posts 
Quote:


20060213, 21:41  #35 
Jun 2003
1,579 Posts 
I don't know about snfs, never used it.
I have a related question. Can 2^nx (for x small, say x<100) be n smooth. especially when x=1? Any solutions for this one? (Other than the first few trivial cases) Citrix 
20060213, 21:59  #36 
Jan 2005
Transdniestr
503 Posts 
I doubt if there's any solutions to it. There's no low solutions right? Unless you count n=1

20060213, 22:23  #37 
Jan 2005
Transdniestr
503 Posts 
Sorry, I was looking only at x=1

20060213, 22:26  #38 
Jun 2005
Near Beetlegeuse
2^{2}·97 Posts 
If is the first term in the sequence, then = integer when .
But I only checked for n is prime. Last fiddled with by Numbers on 20060213 at 22:31 
20060213, 22:30  #39 
Jan 2005
Transdniestr
111110111_{2} Posts 
Hi Numbers, would do you mean exactly by "a" being the first term in the sequence?

20060213, 22:38  #40 
Jun 2005
Near Beetlegeuse
110000100_{2} Posts 
Let , then integer when etc.
So 2 is the first term in the sequence and each subsequent term is 2+(7t) where t =(0, 10). 
20060213, 22:42  #41 
Jan 2005
Transdniestr
767_{8} Posts 
Gotcha, thanks

20060213, 22:48  #42  
Jun 2003
1,579 Posts 
Quote:
Numbers, n smooth does not mean it has to be divisible by n. (even if it does I meant, that all the prime factors are less than or equal to n) Citrix edit: you can also rule out all prime n since 2^p1 has factors of the forum 2*k*p+1 which is >p. For that matter all 2^n1 can also be ruled out, even 2^n2^k such that 2^k<100. Then the lowest case under consideration is 2^n3. Example 2^23 is 2 smooth, 2^73 is 5 smooth and so on. I would also like the question for the 2^n+x side? (2^3+1 is 3 smooth) Last fiddled with by Citrix on 20060213 at 23:00 

20060213, 22:54  #43  
Jun 2005
Near Beetlegeuse
2^{2}×97 Posts 
Quote:
My sequence identifies those cases (or some of them, at least) where there might be a solution. It was never intended to be a complete solution on its own. The next step is to determine those numbers in my sequence that are divisible by prime < n. Edit. In fact, on thinking about it, I realise that I have not excluded the possibility that these numbers are not divisible by some p > n, so it is in fact completely useless. Why is it that whenever I try to help you, all I do is confuse the issue :) Last fiddled with by Numbers on 20060213 at 23:07 

20060213, 23:40  #44  
"William"
May 2003
New Haven
3×787 Posts 
Quote:


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