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Old 2016-12-15, 11:27   #12
gd_barnes
 
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Quote:
Originally Posted by sweety439 View Post
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):
<snip>
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
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Old 2016-12-15, 12:58   #13
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(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.
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Old 2016-12-15, 13:01   #14
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[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
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Old 2016-12-15, 13:53   #15
sweety439
 
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Quote:
Originally Posted by gd_barnes View Post
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
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Old 2016-12-15, 14:00   #16
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Quote:
Originally Posted by fivemack View Post
(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
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Old 2016-12-15, 14:02   #17
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(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
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Old 2016-12-15, 19:54   #18
gd_barnes
 
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Quote:
Originally Posted by sweety439 View Post
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?

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?

Last fiddled with by gd_barnes on 2016-12-15 at 19:55
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Old 2016-12-16, 08:24   #19
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Default 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
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Old 2016-12-16, 09:12   #20
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Default 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.

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Old 2016-12-17, 19:47   #21
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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
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Old 2016-12-17, 19:53   #22
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Quote:
Originally Posted by gd_barnes View Post
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
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