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Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

22×2,281 Posts Quote:
 Originally Posted by sweety439 Base 11: d=1: conjectured k = 3 (3), it is proven. d=2: conjectured k = ? (?) d=3: conjectured k = ? (?) d=4: conjectured k = ? (?) d=5: conjectured k = 2 (2), it is proven. d=6: conjectured k = ? (?) d=7: conjectured k = 3 (3), it is proven. d=8: conjectured k = ? (?) d=9: conjectured k = ? (?) d=X: conjectured k = 861 (713), it is proven. Now, all bases 2<=b<=12 are researched.
Is cognitive dissonance your goal wherever you go?
Attached Thumbnails   2016-12-18, 08:09   #24
gd_barnes

May 2007
Kansas; USA

23×3×52×17 Posts Quote:
 Originally Posted by sweety439 Base 5: d=1: conjectured k = 3 (3), it is proven. d=2: conjectured k = 191115 (22103430), 258 k's remain. d=3: conjectured k = 585655 (122220110), 2274 k's remain. d=4: conjectured k = 346801 (42044201), 126 k's remain. Base 7: d=1: conjectured k = 76 (136), it is proven. d=2: conjectured k = 15979 (64405), 17 k's remain. d=3: conjectured k = 5629 (22261), 19 k's remain. d=4: conjectured k = 20277 (113055), 16 k's remain. d=5: conjectured k = 43 (61), it is proven. d=6: conjectured k = 408034255081 (41323316641135). Base 11: d=1: conjectured k = 3 (3), it is proven. d=2: conjectured k = ? (?) d=3: conjectured k = ? (?) d=4: conjectured k = ? (?) d=5: conjectured k = 2 (2), it is proven. d=6: conjectured k = ? (?) d=7: conjectured k = 3 (3), it is proven. d=8: conjectured k = ? (?) d=9: conjectured k = ? (?) d=X: conjectured k = 861 (713), it is proven. Now, all bases 2<=b<=12 are researched.
Your "research" mostly just involves copying others work with some additional trivial effort of your own. Did you see the work that I did on base 12 digit=5 a few posts back? The conjecture is k=33441, there are 42 k's remaining at n=10K, and I am currently searching them all to n=25K.

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 2016-12-18 at 08:43   2016-12-18, 12:24 #25 sweety439   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^n-1)/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 2016-12-18 at 12:26   2016-12-18, 12:28 #26 sweety439   Nov 2016 8B716 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^n-1)/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.   2016-12-18, 12:31 #27 sweety439   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^n-1)/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.   2016-12-18, 12:32 #28 sweety439   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^n-1)/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.   2016-12-18, 12:35 #29 sweety439   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^n-1)/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.   2016-12-18, 12:38 #30 sweety439   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^n-1)/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.   2016-12-18, 12:44 #31 sweety439   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 2016-12-18 at 12:51   2016-12-18, 16:47 #32 sweety439   Nov 2016 23·97 Posts base 11, d=6: Tested to 150000, I still found no such k. Continuing...   2016-12-18, 17:02 #33 sweety439   Nov 2016 23×97 Posts base 11, d=6: Tested to 200000, I still found no such k. Continuing...   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post a1call Miscellaneous Math 179 2015-11-12 14:59 sinide Factoring 12 2010-11-09 01:05 aaa120 Factoring 19 2010-09-04 09:16 storm5510 Other Mathematical Topics 14 2010-08-31 01:16 Unregistered Math 11 2004-11-30 22:53

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