20161208, 19:35  #1 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
47·79 Posts 
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. 
20161216, 18:15  #2 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
47·79 Posts 
These are unsolved families in base 2 to base 36, given by the links. (see the links for the top (probable) primes)
Last fiddled with by sweety439 on 20161216 at 18:30 
20161221, 19:59  #3 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
47×79 Posts 
Is anyone reserving these families?
Bases <= 36 with only few families remaining: Base 17: F1{9}: (4105*17^n9)/16 Base 19: EE1{6}: (15964*19^n1)/3 Base 21: G{0}FK: 7056*21^n+335 Base 26: {A}6F: (1352*26^n497)/5 {I}GL: (12168*26^n1243)/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 20191127 at 23:36 Reason: this n should the number of digits in {} 
20161221, 20:00  #4 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
47·79 Posts 
The letters A, B, C, D, ... are the digits: A=10, B=11, C=12, D=13, ...

20170116, 18:24  #5 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
3713_{10} Posts 
The status:
Base 17: F1{9}: (4105*17^n9)/16: at n=1M, no (probable) prime found. Base 19: EE1{6}: (15964*19^n1)/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^n1: at n=660K, no (probable) prime found. OL{8}: (4975*25^n111)/8: at n=303K, no (probable) prime found. CM{1}: (7729*25^n1)/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^n497)/5: at n=486K, no (probable) prime found. {I}GL: (12168*26^n1243)/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^n203)/26: at n=368K, no (probable) prime found. 999{G}: (88577*27^n8)/13: at n=368K, no (probable) prime found. E{I}F8: (139239*27^n1192)/13: at n=368K, no (probable) prime found. {F}9FM: (295245*27^n113557)/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 20170116 at 18:30 
20170306, 12:10  #6 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
47×79 Posts 
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 20170601 at 14:11 Reason: this n should the number of digits in {} 
20170413, 18:41  #7 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
47×79 Posts 
Base 31:
E8{U}P: 13733*31^n6: at n=15K, no (probable) prime found. {P}I: (155*31^n47)/6: at n=15K, no (probable) prime found. {R}1: (279*31^n269)/10: at n=15K, no (probable) prime found. {U}P8K: 29791*31^n5498: at n=15K, no (probable) prime found. Last fiddled with by sweety439 on 20170601 at 14:13 Reason: the same as base 36 reason 
20170502, 11:14  #8 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
47·79 Posts 
These problems are to find a prime of the form (k*b^n+c)/gcd(k+c,b1) 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,b1). (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^n1)/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,b1)=1 case is the Sierpinski problem base b, and the c=1 and gcd(k+c,b1)=1 case is the Riesel problem base b. Last fiddled with by sweety439 on 20170502 at 11:27 
20170502, 12:18  #9 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
47×79 Posts 
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. 
20170502, 12:24  #10 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
47×79 Posts 
Also,
k=1, c>0: the dual Sierpinski problem base b. k=1, c<0: the dual Riesel problem base b. 
20170502, 12:25  #11 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
3713_{10} Posts 
(k*b^n+c)/gcd(k+c,b1) 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 rth powers. or * b and 4kc are both perfect 4th powers. Last fiddled with by sweety439 on 20170503 at 17:46 
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