Sufficient but not necessary condition for primality?

[a1call]

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.

No soup for you, NEXT.

[rudy235]

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
Aⁿ-1/(A-1) 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

[slandrum]

Originally Posted by rudy235

Aⁿ-1/(10-1) is prime.

Shouldn't that be (Aⁿ-1)/(A-1)

[Andrew Usher]

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? Well-founded conjecture says there are, and infinitely many of them.

[rudy235]

[C]

Originally Posted by slandrum

Shouldn't that be (Aⁿ-1)/(A-1)

Yeah! Sorry, when I edited the post, I changed the part that had ‘10’ to A but then left a ‘10’ without changing it.