20200905, 05:13  #1  
Aug 2006
17·349 Posts 
18digit factorization challenge
In another thread, Alberico Lepore posted:
Quote:
390644893234047643 with the code Code:
rsp(59,2) As a sidenote, on my old machine, this takes 2.5 ms to factor in gp (using SQUFOF) and about 3 seconds to factor it by trial division. Factorizations of this size can be done by hand; see here for an account of Václav Šimerka's factorization of the 17digit repunit 11111111111111111 using a precursor of SQUFOF. 

20200905, 09:05  #2 
May 2017
ITALY
3×5×31 Posts 
N=390644893234047643 , (N1)/6=F
A=((4*N+1)/9+5)/8 ,((N1)/9)/2=B, C=((7*N1)/9+4)/14 F is integer A,B,C not integer > N=(6*a+5)*(6*b+5) to be optimized this method needs two factors 18 * t + 1 and 18 * s + 7 Last fiddled with by Alberico Lepore on 20200905 at 09:08 
20200905, 09:21  #3 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{3}·7·163 Posts 

20200905, 09:28  #4 
May 2017
ITALY
1D1_{16} Posts 

20200905, 09:41  #5  
May 2017
ITALY
3×5×31 Posts 
Quote:
Quote:
but don't cheat 

20200905, 16:52  #6  
"Dylan"
Mar 2017
521 Posts 
Quote:
F = 65107482205674607 A = 781289786468095309/36 B = 65107482205674607/3 C = 65107482205674608/3 so you are correct that F is an integer, and A, B and C are not. Then you claim N = (6*a+5)*(6*b+5) presuming a and b in this equation are the same as A and B above, we have 390644893234047643 = 101735621739893665192780582115190241/6 which is false. Also, what is t and what is s? You have not defined what these are. 

20200905, 17:13  #7 
Mar 2019
2^{3}×13 Posts 

20200905, 17:41  #8 
May 2017
ITALY
3×5×31 Posts 
Until Tuesday I will be on standby.
There is a serious possibility that I will not be able to continue my studies on factorization. I will read the posts but I will be unable to reply. We hope well. 
20200905, 17:57  #9 
"Curtis"
Feb 2005
Riverside, CA
2×47^{2} Posts 

20200906, 00:19  #10 
Aug 2006
13455_{8} Posts 
I hope you and your family are well. Take as long as you need.

20200907, 12:19  #11 
May 2017
ITALY
111010001_{2} Posts 

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