20030428, 19:51  #1  
Dec 2002
Frederick County, MD
562_{8} Posts 
Probability of finding a factor in TF
Alrighty,
At http://www.mersenne.org/math.htm, we are told: Quote:
Tim 

20030429, 15:07  #2 
Dec 2002
Frederick County, MD
370_{10} Posts 
In a posibly related note, is the following infinite series equal to 1?
[code:1]inf Σ 1/(n(n+1)) = 1 ? n=1[/code:1] 
20030429, 19:47  #3 
Jan 2003
far from M40
5^{3} Posts 
Yes, it is.
1/(n*(n+1)) = (n+1n)/(n*(n+1)) = (n + 1)/(n*(n+1)) n/(n*(n+1)) = 1/n  1/(n+1) So, the kth partial  sum has the value 1  1/(k+1) which's limit for k to infinity is 1 q.e.d. Sums like these, where a summand is (partly) nihilized by its successor are called telescope sums. Benjamin 
20030509, 02:53  #4  
"Richard B. Woods"
Aug 2002
Wisconsin USA
1110000010100_{2} Posts 
Re: Probability of finding a factor in TF
Quote:
Quote:
That is, it refers to the probability of the existence of a factor (known or unknown) of 2^p1 within the specified range, not necessarily the probability of discovering a previouslyunknown factor there after some factors of 2^p1 are already known. 

20030607, 05:56  #5 
"Richard B. Woods"
Aug 2002
Wisconsin USA
16024_{8} Posts 
For a thorough earlier discussion, see the thread "Does the LL test:s factorization save or waste CPU time?" in The Software forum at http://www.mersenneforum.org/viewtopic.php?t=78, especially svempasnake's table comparing the 1/n prediction to the actual number of factors found, on page 3 at Mon Sep 16, 2002 9:32 pm (http://www.mersenneforum.org/viewtop...highlight=#895).

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