20190826, 16:57  #298 
Nov 2016
2^{3}·317 Posts 
This is the minimal set of nonsingle digit primes in base b (this set would contain all twodigit primes), I do not know whether the sets are complete (however, of course, the b=13 case is not complete, since it must contain the largest PRP 13^32020*8+183)
If we also include the single digit primes (i.e. primes < b), please see https://github.com/curtisbright/mepn...ee/master/data and https://cs.uwaterloo.ca/~cbright/reports/mepn.pdf for more information. Last fiddled with by sweety439 on 20190826 at 17:13 
20190826, 17:12  #299 
Nov 2016
2^{3}·317 Posts 
some sets are not complete

20190826, 17:28  #300 
Nov 2016
2^{3}×317 Posts 
some sets are not complete

20190826, 17:48  #301 
Nov 2016
2^{3}·317 Posts 
This file is wrong (for n = 4, 6, 8, 9, X), update the correct file.

20190826, 18:08  #302 
Nov 2016
2^{3}×317 Posts 
update the program (for minimal primes) that can show any base (not only the bases <= 72)
Last fiddled with by sweety439 on 20190826 at 18:15 
20190826, 18:14  #303 
Nov 2016
2^{3}·317 Posts 
update the program (for permutable primes) that can show any base (not only the bases <= 72)
Last fiddled with by sweety439 on 20190826 at 18:15 
20190826, 21:11  #304 
Nov 2016
2^{3}×317 Posts 
these two files are the data for the first few minimal primes in bases 51 and 60 (see https://github.com/curtisbright/mepn...ee/master/data and https://github.com/RaymondDevillers/primes for bases <= 50)
the "minimal primes problem" is proven only for bases 2~16, 18, 20, 22~24, 30, 42, I think this problem for base 60 can also be proven, but this problem [was] only reserving for bases <= 50 
20190826, 21:17  #305 
Nov 2016
2^{3}×317 Posts 
you can run any bases >=2, not only the bases <=72

20190829, 01:53  #306  
Nov 2016
4750_{8} Posts 
Quote:
b=2: we obtain the 2digit primes 10 and 11, since for any prime p > 11, p must start and end with digit 1, and we have 11 <<< p, thus the 2kernel {10, 11} is complete. b=3: we obtain the 2digit primes 10, 12 and 21, for any prime p > 21, if p end with 2 and 12 !<<< p, then p must contain only 0 and 2, thus p is divisible by 2 and > 2, thus not prime, therefore, p must end with 1 (p cannot end with 0, or p is divisible by 10 and not prime), since 21 !<<< p, p must contain only 0 and 1, but since 10 !<<< p and p cannot have leading zeros, thus p can only have the digit 1, i.e. p is a repunit, and the smallest repunit prime is 111, thus completed the 3kernel {10, 12, 21, 111}. b=4: we obtain the 2digit primes 11, 13, 23 and 31, for any prime p > 31, if p end with 3 and 13 !<<< p and 23 !<<< p, then p must contain only 0 and 3, thus p is divisible by 3 and > 3, thus not prime, therefore, p must end with 1 (p cannot end with 0 or 2, or p is divisible by 2 and not prime), since 11 !<<< p and 31 !<<< p, thus p (before the final digit 1) must contain only 0 and 2, and we obtain the prime 221, since p cannot have leading zeros, the remain case is only 2{0}1, but all numbers of the form 2{0}1 are divisible by 3 and > 3, thus not prime, thus we completed the 4kernel {11, 13, 23, 31, 221}. b=5: we obtain the 2digit primes 10, 12, 21, 23, 32, 34 and 43, for any prime p>43: p end with 1 > before this 1, p cannot contain 2 > if p end with 11, then we find the prime 111, and all other primes contain at most two 1, ... p end with 2 > before this 2, p cannot contain 1 or 3 > p only contain 0, 2 and 4 > p is divisible by 2 and > 2 > p is not prime (thus, 12 and 32 are the only such primes end with 2) p end with 3 > before this 3, p cannot contain 2 or 4 > p only contain 0, 1 and 3 > we obtain the primes 133 and 313, thus other primes p cannot contain both 1 and 3 (before the final digit 3), and since p cannot have leading zeros, if p begin with 1, then p is of the form 1{0,1}3 > it must be of the form {1}3 (to avoid the prime 10) > but 113 is not prime and all primes except 111 contain at most two 1 > this way cannot find any primes, if p begin with 3, then p is of the form 3{0,3}3 > p is divisible by 3 and > 3 > p is not prime (thus, all such primes end with 3 are 23, 43, 133 and 313) p end with 4 > before this 4, p cannot contain 3 > since all primes > 2 are odd, p must contain at least one 1 > we obtain the prime 414 and we know that no 2 can before this 1 (to avoid the prime 21) and no 0 or 2 can after this 1 (to avoid the primes 10 and 12) > 1 must be the leading digit (since p cannot have leading zeros, and no 2, 3, 4 can before this 1 (to avoid the primes 21, 34 and 414, respectively) > p must be of the form 1{4} or 11{4} (since all primes except 111 contain at most two 1) > and we obtain the prime 14444 (thus, all such primes end with 4 are 34, 414 and 14444) Last fiddled with by sweety439 on 20190829 at 04:24 

20190902, 09:29  #307 
Nov 2016
2^{3}×317 Posts 
For odd primes p:
* p divides a(p) if and only if p is Bernoulliirregular * p divides a(2*p) if and only if p is Eulerirregular 
20190902, 09:29  #308 
Nov 2016
2536_{10} Posts 
Conjecture: a(n) = 1 only for n = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 24, 25, 27, 30, 33, 35, 42, 45
