 mersenneforum.org Add repeated digits after a number until it is prime
 Register FAQ Search Today's Posts Mark Forums Read  2016-12-15, 11:27   #12
gd_barnes

May 2007
Kansas; USA

27D816 Posts Quote:
 Originally Posted by sweety439 The polynomial of 1{5}1 in base 12 is (16*12^n-49)/11. The polynomial of 23{1} in base 12 is (298*12^n-1)/11, in this case I found a proven prime (298*12^1676-1)/11. For other polynomials for the near-repdigits and quasi-repdigits in base 12 (use X and E for 10 and 11):
Are you sure about that for 23{1} base 12? Can you explain how the polynomial of 23{1} in base 12 = (298*12^n-1)/11 ?

Here is the way that I figure it:
23{1} base 12
23*12^n+(12^n-1)/11
(253*12^n+12^n-1)/11
(254*12^n-1)/11
Therefore 23{1} base 12 = (254*12^n-1)/11

And further:
27{1} base 12
27*12^n+(12^n-1)/11
(297*12^n+12^n-1)/11
(298*12^n-1)/11
Therefore 27{1} base 12 = (298*12^n-1)/11

Am I not understanding something?

One other thing: When representing numbers for bases > 10, the following symbols are used for digits > 10:
10 = A
11 = B
12 = C
(etc.)

Your representation of 10 = X and 11 = E looks strange and is confusing. I would suggest that you change that for future posts.

Last fiddled with by gd_barnes on 2016-12-15 at 11:36   2016-12-15, 12:58 #13 fivemack (loop (#_fork))   Feb 2006 Cambridge, England 5·31·41 Posts (16*12^n-49)/11 is 15{n}1 in base 12; and it's a difference of two squares divided by eleven (ie it's (4*12^{n/2}-7) * (4*12^{n/2}+7) The right-hand side is always divisible by eleven. So it's never prime, it factors as 4{n}5 * 3B{n}5 and then those probably factor further.   2016-12-15, 13:01 #14 fivemack (loop (#_fork))   Feb 2006 Cambridge, England 143238 Posts [QUOTE=gd_barnes;449201]Are you sure about that for 23{1} base 12? Can you explain how the polynomial of 23{1} in base 12 = (298*12^n-1)/11 ? Here is the way that I figure it: 23{1} base 12 23*12^n+(12^n-1)/11 ^ that is your mistake, '23' is in base 12 so you should be working with 27_10   2016-12-15, 13:53   #15
sweety439

Nov 2016

223010 Posts Quote:
 Originally Posted by gd_barnes Are you sure about that for 23{1} base 12? Can you explain how the polynomial of 23{1} in base 12 = (298*12^n-1)/11 ? Here is the way that I figure it: 23{1} base 12 23*12^n+(12^n-1)/11 (253*12^n+12^n-1)/11 (254*12^n-1)/11 Therefore 23{1} base 12 = (254*12^n-1)/11 And further: 27{1} base 12 27*12^n+(12^n-1)/11 (297*12^n+12^n-1)/11 (298*12^n-1)/11 Therefore 27{1} base 12 = (298*12^n-1)/11 Am I not understanding something? One other thing: When representing numbers for bases > 10, the following symbols are used for digits > 10: 10 = A 11 = B 12 = C (etc.) Your representation of 10 = X and 11 = E looks strange and is confusing. I would suggest that you change that for future posts.
23{1} base 12 is (all numbers below are written in base 12):

23*10^n+(10^n-1)/E
=(209*10^n+10^n-1)/E
=(20X*10^n-1)/E

converted to base 10, it is (298*12^n-1)/11.

10=A, 11=B, 12=C, ... is for base 16. In http://oeis.org/A057472 (the sequence for Riesel problem base 6 for k=2), 10 is written as X and 11 is written as E.

Last fiddled with by sweety439 on 2017-02-07 at 15:19   2016-12-15, 14:00   #16
sweety439

Nov 2016

2×5×223 Posts Quote:
 Originally Posted by fivemack (16*12^n-49)/11 is 15{n}1 in base 12; and it's a difference of two squares divided by eleven (ie it's (4*12^{n/2}-7) * (4*12^{n/2}+7) The right-hand side is always divisible by eleven. So it's never prime, it factors as 4{n}5 * 3B{n}5 and then those probably factor further.
You wrote "n/2", this n may be odd! n does not necessary to be even.

Besides, 16*12^n-49 = (4*12^{n/2}-7) * (4*12^{n/2}+7) is divisible by 11 for even n, but (16*12^n-49)/11 does not necessary divisible by 11.

Last fiddled with by sweety439 on 2016-12-15 at 14:00   2016-12-15, 14:02 #17 sweety439   Nov 2016 223010 Posts (16*12^n-49)/11 is divisible by 13 for odd n and has algebra factors for even n, like 25*12^n-1, it is also divisible by 13 for odd n and has algebra factors for even n. Thus, both of them are never prime. All numbers of the form 1{5}1 or 20{E} in base 12 are composite. Last fiddled with by sweety439 on 2017-02-07 at 15:17   2016-12-15, 19:54   #18
gd_barnes

May 2007
Kansas; USA

100111110110002 Posts Quote:
 Originally Posted by sweety439 23{1} base 12 is (all numbers below are written in base 12): 23*10^n+(10^n-1)/E =(209*10^n+10^n-1)/E =(20X*10^n-1)/E converted to base 10, it is (298*12^n-1)/11. 10=A, 11=B, 12=C, ... is for base 16. In http://oeis.org/A057472 (the sequence for Riesel problem base 16 for k=2), 10 is written as X and 11 is written as E.
OK I get it for 23{1} base 12.

On the second part, so what! That's just bad representation and is inconsistent regardless of what OEIS does. So what is the representation for base 13. Is it:
10 = X
11 = E
12 = T or XII

What is the representation for base 14?

10 = X
11 = E
12 = T or W or XII
13 = T or H or XIII?

Why does base 16 only get the A/B/C/D/E/F representation?

Why not just stick with the same representation for all bases > 10.

I want to see you do this for base 14 or base 18. What will you do?

Last fiddled with by gd_barnes on 2016-12-15 at 19:55   2016-12-16, 08:24 #19 gd_barnes   May 2007 Kansas; USA 100111110110002 Posts base 12 d=5 I did some work on base 12 d=5. The conjecture is k=33441. The covering set is [7 13 19 157] All k's have been searched to n=10K. There are 42 k's remaining. Primes for n>3000: Code:  k n 1331 6190 2098 6542 2197 5457 2852 5908 2891 4196 5863 3073 6539 3005 7506 9492 10041 9840 10964 6510 16957 8290 17459 8249 19279 3667 23184 4298 23431 3320 24302 5430 26377 6319 26629 6932 27508 3117 30394 4471 33239 3857 42 k's remaining at n=10K: Code: 4446 4927 6123 6591 6656 7761 9313 10543 12043 12272 12498 13572 13811 15137 15691 15756 18504 19427 19609 20532 21086 21286 22417 23946 24492 24822 25251 25784 26187 26278 27001 27166 27383 28561 28826 29822 29874 30308 30862 31283 31499 32661 Note: All k's where k==(5 mod 12) are effectively the same search as (k-5)/12 -if- (k-5)/12 does not have a prime at n=1 and so are not considered. At CRUS we call these MOB (multiples of the base). There was only one k remaining at n=10K that fit this criteria: k=15137. k=(15137-5)/12 == k=1261 has a prime at n=1 therefore k=15137 must continue to be searched. Sieving is complete and I just started testing them all for n=10K-25K. It should take ~4 days on one slow core. Last fiddled with by gd_barnes on 2016-12-16 at 09:13   2016-12-16, 09:12 #20 gd_barnes   May 2007 Kansas; USA 100111110110002 Posts base 3 d=1 I did some doublecheck work on base 3 d=1. All k's have been searched to n=50K. There are 15 k's remaining. No problems were found with the nature and quantity of k's remaining but other problems were found as follows: First, there were significant inconsistencies in the display of k's remaining and primes. These differences were not consistent with the method by which k's were shown for other bases on the page. For example for base 6 d=1, no k's are shown where k==(1 mod 6) unless (k-1)/6 has a prime at n=1. In other words the most "reduced" k is shown, which is the correct method. That is not the case for base 3. Several primes and k's remaining are k==(1 mod 3) and can be reduced. Second, k=3689 was initially shown as remaining with a search depth of n=19200. This is incorrect. I found a prime for it at n=16856. Then in the later table it is shown as magically not remaining even though no prime was ever shown for it. Primes for n>5000: Code:  k n 59 8972 156 24761 498 20847 1668 12083 2055 12978 2718 9567 3059 28580 3158 15331 3368 17455 3515 5898 3689 16856 3755 26022 3788 5031 4029 47256 4376 16533 4589 21404 4655 11134 5625 24314 5759 11140 5876 36665 15 k's remaining at n=50K: Code: 806 915 968 1565 1794 2877 3393 3738 3813 3969 4356 4388 4905 5325 5798 Notes on base 6 d=1: I am now doing doublecheck work on this base/digit. I will post my findings when I reach n=10K. It will be at most a few day trivial effort for me to search it to n=25K so I will likely continue to that depth. I noticed that he only searched this base to n=13K or 15K. He shows two different search depths at different places so no one really knows...more inconsistency. Last fiddled with by gd_barnes on 2016-12-16 at 09:57   2016-12-17, 19:47 #21 sweety439   Nov 2016 2·5·223 Posts 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. Last fiddled with by sweety439 on 2016-12-17 at 19:49   2016-12-17, 19:53   #22
sweety439

Nov 2016

2×5×223 Posts Quote:
 Originally Posted by gd_barnes OK I get it for 23{1} base 12. On the second part, so what! That's just bad representation and is inconsistent regardless of what OEIS does. So what is the representation for base 13. Is it: 10 = X 11 = E 12 = T or XII What is the representation for base 14? 10 = X 11 = E 12 = T or W or XII 13 = T or H or XIII? What about base 15? What about bases 17, 18, or higher? Why does base 16 only get the A/B/C/D/E/F representation? Why not just stick with the same representation for all bases > 10. I want to see you do this for base 14 or base 18. What will you do?
At present, in this project, I only research for bases b<=12. For bases 12<b<=36, I uses A, B, C, D, ... for 10, 11, 12, 13, ..., but for base 11 and base 12, I uses X for 10 and E for 11.

See http://mersenneforum.org/showthread.php?t=21819, the generalized minimal primes project, in this project, I use A, B, C, D, ... for 10, 11, 12, 13, ... in all bases, because in that project I research all bases b<=36, but in this project (at present), I only research all bases b<=12. I will research this project for 12<b<=36 in the future, but not now. At that time, I will use A, B, C, D, ... for 10, 11, 12, 13, ... in all bases.

In fact, I will research both this project and the generalized minimal primes project to all bases b<=144 in the future. For bases b>36, I use ":" to separate the digits, for example, in base 40, I write 27:10:39 for the number 27*40^2+10*40+39, and in base 85, I write 68:19:47:35 for the number 68*85^3+19*85^2+47*85+35.

Last fiddled with by sweety439 on 2016-12-17 at 20:03   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

All times are UTC. The time now is 20:12.

Fri Sep 18 20:12:14 UTC 2020 up 8 days, 17:23, 1 user, load averages: 2.13, 2.14, 2.00