20161217, 19:55  #23  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}×2,281 Posts 
Quote:


20161218, 08:09  #24  
May 2007
Kansas; USA
2^{3}×3×5^{2}×17 Posts 
Quote:
These are interesting conjectures, which is why I have done some work on them. You should consider doing the same instead of creating more conjectures. Batalov was even kind enough to give me some good information on sieving them so that I could search them to a reasonable depth. It is possible to sieve these sequences with srsieve/sr1sieve/sr2sieve. By the way your research is a bit off. For base=5 digit=4 that is the same as the Riesel base 5 conjecture, which is currently being worked on by PrimeGrid. We track it on the CRUS pages at http://www.noprimeleftbehind.net/cru...onjectures.htm. The very latest stats are at PrimeGrid at http://primegrid.com/stats_sr5_llr.php . There are currently 72 k's remaining at n=~2.28M. (not 126 k's remaining) I have a challenge for you: Determine 2 or more of these base 11 conjectures that currently have a "?" after them. It's pretty easy with PFGW if the conjecture is k<1e5. It is how I found the b=12 d=5 conjecture. Last fiddled with by gd_barnes on 20161218 at 08:43 

20161218, 12:24  #25 
Nov 2016
23·97 Posts 
base 11, d=3: conjectured k = 1520 (1162), cover set = {2, 7, 19, 37}, period=6
For the number 1520*11^n+3*(11^n1)/10: If n=0 (mod 2), then this number is divisible by 2. If n=1 (mod 6), then this number is divisible by 7. If n=3 (mod 6), then this number is divisible by 19. If n=5 (mod 6), then this number is divisible by 37. Last fiddled with by sweety439 on 20161218 at 12:26 
20161218, 12:28  #26 
Nov 2016
8B7_{16} Posts 
base 11, d=9: conjectured k = 716 (5X1), cover set = {2, 7, 19, 37}, period=6
For the number 716*11^n+9*(11^n1)/10: If n=0 (mod 2), then this number is divisible by 2. If n=1 (mod 6), then this number is divisible by 19. If n=3 (mod 6), then this number is divisible by 37. If n=5 (mod 6), then this number is divisible by 7. 
20161218, 12:31  #27 
Nov 2016
23×97 Posts 
base 11, d=2: conjectured k = 2627 (1X79), cover set = {3, 7, 19, 37}, period=6
For the number 2627*11^n+2*(11^n1)/10: If n=1 (mod 2), then this number is divisible by 3. If n=0 (mod 6), then this number is divisible by 37. If n=2 (mod 6), then this number is divisible by 7. If n=4 (mod 6), then this number is divisible by 19. 
20161218, 12:32  #28 
Nov 2016
23·97 Posts 
base 11, d=4: conjectured k = 973 (805), cover set = {3, 7, 19, 37}, period=6
For the number 973*11^n+4*(11^n1)/10: If n=1 (mod 2), then this number is divisible by 3. If n=0 (mod 6), then this number is divisible by 7. If n=2 (mod 6), then this number is divisible by 19. If n=4 (mod 6), then this number is divisible by 37. 
20161218, 12:35  #29 
Nov 2016
23·97 Posts 
base 11, d=8: conjectured k = 4625 (3525), cover set = {3, 7, 19, 37}, period=6
For the number 4625*11^n+8*(11^n1)/10: If n=1 (mod 2), then this number is divisible by 3. If n=0 (mod 6), then this number is divisible by 37. If n=2 (mod 6), then this number is divisible by 19. If n=4 (mod 6), then this number is divisible by 7. 
20161218, 12:38  #30 
Nov 2016
23·97 Posts 
base 11, d=X: conjectured k = 861 (713), cover set = {3, 7, 19, 37}, period=6
For the number 861*11^n+10*(11^n1)/10: If n=0 (mod 2), then this number is divisible by 3. If n=1 (mod 6), then this number is divisible by 19. If n=3 (mod 6), then this number is divisible by 7. If n=5 (mod 6), then this number is divisible by 37. 
20161218, 12:44  #31 
Nov 2016
23×97 Posts 
base 11, d=6: I found no such k<=40000.
Now, I have founded the conjectured k's for base 11 and all d except d=6. Last fiddled with by sweety439 on 20161218 at 12:51 
20161218, 16:47  #32 
Nov 2016
23·97 Posts 
base 11, d=6: Tested to 150000, I still found no such k.
Continuing... 
20161218, 17:02  #33 
Nov 2016
23×97 Posts 
base 11, d=6: Tested to 200000, I still found no such k.
Continuing... 
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