20230506, 23:50  #23  
"Rashid Naimi"
Oct 2015
Remote to Here/There
41×59 Posts 
The problem with Wilson’s theorem is that, while being deterministic, it also works for all primes and not a subset of them like the Sweety’s concept. As such it does not concur with the OP.
Quote:
Last fiddled with by a1call on 20230507 at 00:09 

20230518, 03:12  #24 
Jun 2015
Vallejo, CA/.
2^{3}×3×7^{2} Posts 
Although this is probably already known (by analogy) you can say that for n to be prime, a sufficient but not necessary condition is that
A^{n}1/(A1) is prime. Of course A can be any positive integer>1 _ as long it is not of the form j^{k} where k ≥ 2_ we have For A=2 .A000043 Mersenne Primes For A=3 .A028491 For A=5 .A004061 For A=6 .A004062 For A=7 .A004063 For A=10 A004023 Repunits For A=12 A004064 E&OE 
20230518, 10:49  #25 
Jan 2021
California
2×7×41 Posts 
Last fiddled with by slandrum on 20230518 at 10:50 
20230518, 12:11  #26 
Dec 2022
2×3×7×13 Posts 
Yes, I'm assuming that as the only sensible meaning.
That (A^n  1)/(A  1) is prime is certainly a sufficient condition (A, n integers > 1). Now does it become necessary when generalised? That is, is there always for n prime some base A that yields a prime? Wellfounded conjecture says there are, and infinitely many of them. 
20230518, 18:59  #27 
Jun 2015
Vallejo, CA/.
1176_{10} Posts 

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