mersenneforum.org > Math Odds a largish number has N divisors?
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2006-02-13, 17:18   #34
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

101101110102 Posts

Quote:
 Originally Posted by Citrix cool. May be this will solve the problem for 88.
Yes, maybe. Do you know that these numbers are Snfs targets? However it's Snfs difficulty is 1024 bits, larger than the current world record...

 2006-02-13, 21:41 #35 Citrix     Jun 2003 1,579 Posts I don't know about snfs, never used it. I have a related question. Can 2^n-x (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
 2006-02-13, 21:59 #36 grandpascorpion     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
 2006-02-13, 22:23 #37 grandpascorpion     Jan 2005 Transdniestr 503 Posts Sorry, I was looking only at x=1
 2006-02-13, 22:26 #38 Numbers     Jun 2005 Near Beetlegeuse 22·97 Posts If $a$ is the first term in the sequence, then $\large \frac{2^{n}-x}{n}$= integer when $\large x\,=\,a+nt,\,t=(0,\,10),\,x<100$. But I only checked for n is prime. Last fiddled with by Numbers on 2006-02-13 at 22:31
 2006-02-13, 22:30 #39 grandpascorpion     Jan 2005 Transdniestr 1111101112 Posts Hi Numbers, would do you mean exactly by "a" being the first term in the sequence?
 2006-02-13, 22:38 #40 Numbers     Jun 2005 Near Beetlegeuse 1100001002 Posts Let $\large n\,=\,7$, then $\large \frac{2^{7}-x}{7}=$ integer when $\large x\,=\,2,\,9,\,16$ etc. So 2 is the first term in the sequence and each subsequent term is 2+(7t) where t =(0, 10).
 2006-02-13, 22:42 #41 grandpascorpion     Jan 2005 Transdniestr 7678 Posts Gotcha, thanks
2006-02-13, 22:48   #42
Citrix

Jun 2003

1,579 Posts

Quote:
 Originally Posted by Numbers Let $\large n\,=\,7$, then $\large \frac{2^{7}-x}{7}=$ integer when $\large x\,=\,2,\,9,\,16$ etc. So 2 is the first term in the sequence and each subsequent term is 2+(7t) where t =(0, 10).

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^p-1 has factors of the forum 2*k*p+1 which is >p. For that matter all 2^n-1 can also be ruled out, even 2^n-2^k such that 2^k<100.

Then the lowest case under consideration is 2^n-3. Example 2^2-3 is 2 smooth, 2^7-3 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 2006-02-13 at 23:00

2006-02-13, 22:54   #43
Numbers

Jun 2005
Near Beetlegeuse

22×97 Posts

Quote:
 Originally Posted by Citrix I meant that all the prime factors are less than or equal to n.
Actually, I should have said that t = (0, 1, 2, ...), then it would be a proper sequence.

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 2006-02-13 at 23:07

2006-02-13, 23:40   #44
wblipp

"William"
May 2003
New Haven

3×787 Posts

Quote:
 Originally Posted by Citrix Can 2^n-x (for x small, say x<100) be n smooth.
28-81 is 8 smooth

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