20201014, 12:53  #34  
Feb 2017
Nowhere
4,517 Posts 
Quote:
N = p*q, p < q prime with b^{2k} < N < b^{2k+1} then, in order that p and q have the same number of baseb digits, it is necessary that p >= b^{k}. (The only possibility of equality is with k = 1 when the base b is prime.) The only question that arises is whether the indicated number p_{1}p_{2} is less than b^{2k+1}. I believe this will be true for sufficiently large k. I see no analogous argument for least brilliant greater than an odd power of the base, or for greatest brilliant number less than any power of the base. 

20201014, 13:59  #35 
May 2013
Germany
83 Posts 
Dr Sardonicus,
thank you for your explanation. 
20201101, 15:46  #36 
Oct 2018
2^{3}·3 Posts 
Continuing this work for 10^169c, I have found that,
10^16914319 = 2093963760229909907466815025292144577767961972509185032132596865267781491968551925027 * 4775631837535734107517020048684519409802862518997809812035307071144108182496827007803 I also attach proof files for some of the work I have done, for 10^n+c for n = 167 and 169. Every number that has a factor larger than 1000 has that factor listed in the files. I'll post the files for n=165 shortly, I seem to have lost some ECM work that I'll redo first. I intend to continue with n=171, but now I'm starting to get into the territory where the SNFS polynomials are getting rather large coefficients for the batch factorization approach and the relations I already have saved. I'm not sure if it would be quicker to sieve again for a new shared rational side, or if using the bad polynomials with the already existing relations is the least work, but for now I'm using what I have. 
20201101, 18:42  #37 
May 2013
Germany
83 Posts 
I think it is a good idea to share these informations.
Thank you. So anyone who is interested in can doublecheck the correctness of the statements easily. Last fiddled with by Alfred on 20201101 at 18:45 
20201102, 19:50  #38 
Oct 2018
2^{3}×3 Posts 
And finally here are the proof files for 10^165+c.

20201110, 04:04  #39 
Jan 2012
Toronto, Canada
3B_{16} Posts 
I am reserving 10^199+c to find the smallest 200digit number which splits into p100*p100. Likely to take at least a few months with an expected 160+ SNFS factorizations, thought I'd at least post here to prevent any potential duplicated efforts. If anyone is interested in crunching a few of these let me know, I can coordinate sieving efforts on another thread.

20201113, 20:52  #40 
Oct 2018
2^{3}·3 Posts 
The next one was quicker and only required 27 SNFS factorizations.
10^171+7467 = 15982339170654488061693029140006521400812407348641102533477071444640746972955602480993 * 62569063847432371483112919249240694575724386807642240564815844097979821472243476190219 Continuing with 10^171c. 
20201210, 09:06  #41 
Oct 2018
2^{3}·3 Posts 
Another 60 SNFS factorizations revealed that
10^17116569 = 10026073074372053022855343749617316836566548448825765741868691467529514155418582501607 * 99739947293634841017115301301296053264053136150611286806831026018776115664482913987233 Proof file is attached. The batch SNFS approach was still faster than regular SNFS but its getting close, I'll probably continue with 10^173 + c 
Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Could a Distributed Computing approach help find the smallest Brier number?  jasong  Math  5  20070529 13:30 
10^119+x brilliant number  Citrix  Prime Sierpinski Project  12  20060519 22:21 
smallest number used in a mathematical proof?  ixfd64  Lounge  22  20060201 17:06 
Can you find the smallest number?  Fusion_power  Puzzles  8  20031118 19:36 
Smallest untested number?  wirthi  Math  10  20031005 13:02 