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Old 2019-08-26, 16:57   #298
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This is the minimal set of non-single digit primes in base b (this set would contain all two-digit 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 2019-08-26 at 17:13
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Old 2019-08-26, 17:12   #299
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some sets are not complete
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File Type: txt minimal set of primes 1 mod 4.txt (31.1 KB, 22 views)
File Type: txt minimal set of primes 3 mod 4.txt (19.9 KB, 17 views)
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Old 2019-08-26, 17:28   #300
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some sets are not complete
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File Type: txt minimal set of primes 1 mod 3.txt (37.3 KB, 17 views)
File Type: txt minimal set of primes 2 mod 3.txt (19.0 KB, 18 views)
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Old 2019-08-26, 17:48   #301
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Quote:
Originally Posted by sweety439 View Post
........
This file is wrong (for n = 4, 6, 8, 9, X), update the correct file.
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File Type: txt smallest prime with digit sum n.txt (807 Bytes, 22 views)
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Old 2019-08-26, 18:08   #302
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update the program (for minimal primes) that can show any base (not only the bases <= 72)
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File Type: txt program of minimal prime base b.txt (522 Bytes, 16 views)

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Old 2019-08-26, 18:14   #303
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update the program (for permutable primes) that can show any base (not only the bases <= 72)
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File Type: txt program of permutable prime base b.txt (1.4 KB, 17 views)

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Old 2019-08-26, 21:11   #304
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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
Attached Files
File Type: txt minimal51.txt (753.1 KB, 323 views)
File Type: txt minimal60.txt (49.2 KB, 18 views)
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Old 2019-08-26, 21:17   #305
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you can run any bases >=2, not only the bases <=72
Attached Files
File Type: txt program of minimal prime base b.txt (567 Bytes, 19 views)
File Type: txt program of permutable prime base b.txt (1.4 KB, 16 views)
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Old 2019-08-29, 01:53   #306
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Quote:
Originally Posted by sweety439 View Post
This is the minimal set of non-single digit primes in base b (this set would contain all two-digit 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.
Not including the single-digit primes, proof of that these sets are complete:

b=2: we obtain the 2-digit 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 2-kernel {10, 11} is complete.

b=3: we obtain the 2-digit 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 3-kernel {10, 12, 21, 111}.

b=4: we obtain the 2-digit 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 4-kernel {11, 13, 23, 31, 221}.

b=5: we obtain the 2-digit 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 2019-08-29 at 04:24
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Old 2019-09-02, 09:29   #307
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For odd primes p:

* p divides a(p) if and only if p is Bernoulli-irregular
* p divides a(2*p) if and only if p is Euler-irregular
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Old 2019-09-02, 09:29   #308
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Quote:
Originally Posted by sweety439 View Post
For odd primes p:

* p divides a(p) if and only if p is Bernoulli-irregular
* p divides a(2*p) if and only if p is Euler-irregular
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
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