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Old 2016-12-08, 19:35   #1
sweety439
 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

47·79 Posts
Default generalized minimal (probable) primes

There are researches for minimal primes in base b: https://cs.uwaterloo.ca/~cbright/reports/mepn.pdf, and data for minimal primes and remaining families in bases 2 to 30: https://github.com/curtisbright/mepn...ee/master/data, data for minimal primes and remaining families in bases 28 to 50: https://github.com/RaymondDevillers/primes.

This is a text file for minimal primes in bases 2 to 16.
Attached Files
File Type: txt minimal primes in base 2 to 16.txt (12.1 KB, 543 views)
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Old 2016-12-16, 18:15   #2
sweety439
 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

47·79 Posts
Default

These are unsolved families in base 2 to base 36, given by the links. (see the links for the top (probable) primes)
Attached Files
File Type: txt unsolved families for base 2 to 36.txt (1.9 KB, 603 views)

Last fiddled with by sweety439 on 2016-12-16 at 18:30
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Old 2016-12-21, 19:59   #3
sweety439
 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

47×79 Posts
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Is anyone reserving these families?

Bases <= 36 with only few families remaining:

Base 17:

F1{9}: (4105*17^n-9)/16

Base 19:

EE1{6}: (15964*19^n-1)/3

Base 21:

G{0}FK: 7056*21^n+335

Base 26:

{A}6F: (1352*26^n-497)/5
{I}GL: (12168*26^n-1243)/25

Base 28:

O{A}F: (18424*28^n+125)/27

Base 36:

O{L}Z: (4428*36^n+67)/5
{P}SZ: (6480*36^n+821)/7

Last fiddled with by sweety439 on 2019-11-27 at 23:36 Reason: this n should the number of digits in {}
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Old 2016-12-21, 20:00   #4
sweety439
 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

47·79 Posts
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The letters A, B, C, D, ... are the digits: A=10, B=11, C=12, D=13, ...
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Old 2017-01-16, 18:24   #5
sweety439
 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

371310 Posts
Default

The status:

Base 17:

F1{9}: (4105*17^n-9)/16: at n=1M, no (probable) prime found.

Base 19:

EE1{6}: (15964*19^n-1)/3: at n=707K, no (probable) prime found.

Base 21:

G{0}FK: 7056*21^n+335: at n=506K, no (probable) prime found.

Base 25:

EF{O}: 366*25^n-1: at n=660K, no (probable) prime found.
OL{8}: (4975*25^n-111)/8: at n=303K, no (probable) prime found.
CM{1}: (7729*25^n-1)/24: at n=303K, no (probable) prime found.
E{1}E: (8425*25^n+311)/24: at n=303K, no (probable) prime found.
...
(all other unsolved families in base 25 may be tested to n=303K)

Base 26:

{A}6F: (1352*26^n-497)/5: at n=486K, no (probable) prime found.
{I}GL: (12168*26^n-1243)/25: at n=497K, no (probable) prime found.

Base 27:

8{0}9A: 5832*27^n+253: at n=368K, no (probable) prime found.
C{L}E: (8991*27^n-203)/26: at n=368K, no (probable) prime found.
999{G}: (88577*27^n-8)/13: at n=368K, no (probable) prime found.
E{I}F8: (139239*27^n-1192)/13: at n=368K, no (probable) prime found.
{F}9FM: (295245*27^n-113557)/26: at n=368K, no (probable) prime found.

Base 28:

O{A}F: (18424*28^n+125)/27: at n=543K, no (probable) prime found.

Base 29:

All the unsolved families may be tested to n=242K.

Since the page https://github.com/curtisbright/mepn...ee/master/data only solve the minimal prime problem to bases b<=30, for 31<=b<=36, these bases are reserved by me (these bases have already reserve to n=10K). Now, I am reserving bases 31, 35 and 36, use factordb.

In fact, I decide to solve the minimal prime problem to all bases b<=64 in the future. However, at present, I only solve this problem to all bases b<=36. I will reserve bases 37<=b<=64 if all the bases b<=36 have been tested to at least n=1M.

Last fiddled with by sweety439 on 2017-01-16 at 18:30
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Old 2017-03-06, 12:10   #6
sweety439
 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

47×79 Posts
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For the two unsolved families in base 36:

O{L}Z: (30996*36^n+469)/35: tested up to n=15815, no (probable) prime found.
{P}SZ: (6480*36^n+821)/7: tested up to n=15815, no (probable) prime found.

Last fiddled with by sweety439 on 2017-06-01 at 14:11 Reason: this n should the number of digits in {}
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Old 2017-04-13, 18:41   #7
sweety439
 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

47×79 Posts
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Base 31:

E8{U}P: 13733*31^n-6: at n=15K, no (probable) prime found.
{P}I: (155*31^n-47)/6: at n=15K, no (probable) prime found.
{R}1: (279*31^n-269)/10: at n=15K, no (probable) prime found.
{U}P8K: 29791*31^n-5498: at n=15K, no (probable) prime found.

Last fiddled with by sweety439 on 2017-06-01 at 14:13 Reason: the same as base 36 reason
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Old 2017-05-02, 11:14   #8
sweety439
 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

47·79 Posts
Default

These problems are to find a prime of the form (k*b^n+c)/gcd(k+c,b-1) with integer n>=1 for fixed integers k, b and c, k>=1, b>=2, gcd(k,c)=1 and gcd(b,c)=1.

For some (k,b,c), there cannot be any prime because of covering set (e.g. (k,b,c) = (78557,2,1), (334,10,-1) or (84687,6,-1)) or full algebra factors (e.g. (k,b,c) = (9,4,-1), (2500,16,1) or (9,4,-25) (the case (9,4,-25) can produce prime only for n=1)) or partial algebra factors (e.g. (k,b,c) = (144,28,-1), (25,17,-9) or (1369,30,-1)). It is conjectured that for every (k,b,c) which cannot be proven that they do not have any prime, there are infinitely primes of the form (k*b^n+c)/gcd(k+c,b-1). (Notice the special case: (k,b,c) = (8,128,1), it cannot have any prime but have neither covering set nor algebra factors)

However, there are many such cases even not have a single known prime, like (21181,2,1), (2293,2,-1), (4,53,1), (1,185,-1), (1,38,1), (269,10,1), (197,7,-1), (4105,17,-9), (16,21,335), (5,36,821), but not all case will produce a minimal prime to base b, e.g. the form (197*7^n-1)/2 is the form 200{3} in base 7, but since 2 is already prime, the smallest prime of this form (if exists) will not be a minimal prime in base 7.

The c=1 and gcd(k+c,b-1)=1 case is the Sierpinski problem base b, and the c=-1 and gcd(k+c,b-1)=1 case is the Riesel problem base b.

Last fiddled with by sweety439 on 2017-05-02 at 11:27
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Old 2017-05-02, 12:18   #9
sweety439
 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

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Other special cases:

k=1, c=1, b is even: the generalized Fermat primes in base b.
k=1, c=1, b is odd: the generalized half Fermat primes in base b.
k=1, c=-1: the repunit primes in base b.
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Old 2017-05-02, 12:24   #10
sweety439
 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

47×79 Posts
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Also,

k=1, c>0: the dual Sierpinski problem base b.
k=1, c<0: the dual Riesel problem base b.
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Old 2017-05-02, 12:25   #11
sweety439
 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

371310 Posts
Default

(k*b^n+c)/gcd(k+c,b-1) has full algebra factors if and only if at least one of the following conditions holds:

* There is an integer r>1 such that k, b and -c are all perfect r-th powers.

or

* b and 4kc are both perfect 4th powers.

Last fiddled with by sweety439 on 2017-05-03 at 17:46
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