20070509, 16:24  #1 
Oct 2006
2^{2}·5·13 Posts 
Probability of ndigit factor?
I am interested in what the probability of a ndigit factor is in a random xdigit composite.
The reason I ask is, I am trying to factor a C120 and a C160. (For both, I am using Alpertrons ECM applet.) For the C160, I don't have a single factor, and have sieved to (so far) curve 437 (greater than 25 digits, 55% likely greater than 30 digits). So far, the largest factor (not the result, but smaller prime factor) is a P43. P20's, P25's, and P30's are pretty common. If it matters, it is a home prime sequence, but not one anyone is working on (a 18digit starting number); and this is just a small hobby, nothing of mathematical importance Thanks for anyones help! Roger 
20070509, 16:45  #2  
Nov 2003
2^{6}×113 Posts 
Quote:
A Practical Analysis of the Elliptic Curve Factoring Algorthm. As n >oo, the probability that n has a factor between a and a^(1+e) is e/(e+1). This applies uniformly for a^(1+e) < sqrt(n). If the factor is greater than sqrt(n), then it must be the largest factor and its distribution is given by Dickman's rho function. See Knuth, vol. II. 

20070509, 17:25  #3 
"Nancy"
Aug 2002
Alexandria
4640_{8} Posts 
Lets assume your "random" number N is chosen uniformly from large enough interval [1, B] so that we can estimate the probability that a prime p divides it as 1/p. Now you want the probability that there is at least one p in [10^{[I]n[/I]1}, 10^{[I]n[/I]}1] that divides N. This is messy to compute, lets look at the expected number of such primes instead, . If this expected value is much smaller than 1, it is quite close to the probability. Now for some constant M, so we can take the sum over the interval as roughly like log log (10^{[I]n[/I]})  log log (10^{[I]n[/I]1}) = log(n / (n1)). If n/(n1) is close to 1, you can approximate log(n / (n1)) by 1/n, or better 1/(n0.5).
However, this assumes that N was taken from all the integers not exceeding B. If you have done a PRP test already and it came out negative, you know that N is not any of the primes below B. A prime p divides a prime q iff p=q, so we should remove all the primes from the interval [1, B] except the ones in [10^{[I]n[/I]1}, 10^^{[I]n[/I]}1]. Since I assume 10^{[I]n[/I]} << B, we simply remove all of the primes below B instead. So the set of numbers you chose your random number from has cardinality about BB/log(B), the set of nonprime numbers not exceeding B that have a prime factor in [10^{[I]n[/I]1}, 10^{[I]n[/I]}1] has size about B*(log(n /(n1))) and if you number was chosen uniformly from all the nonprimes ≤ B, the probability becomes log(n / (n1)) / (1  1/log(B)) Everything changes again if you did trial division/ECM/whatever on your number. You can use a Bayesian model to account for unsuccessful factoring attempts. Silverman and Wagstaff "Practical Analysis of the Elliptic Curve Factoring Algorithm" treats this in detail. Alex Last fiddled with by akruppa on 20070509 at 23:09 Reason: small clarification 
20070509, 22:51  #4 
Oct 2006
2^{2}×5×13 Posts 
Wow, information dump!
Thanks for the explanation, Akruppa; both of your posts pointed to Wagstaff/Silverman's paper, so I'll have a look. Roger 
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