"Divides Phi" category

[sweety439]

Originally Posted by carpetpool

There is also this possible sequence I am approaching the T5K range for. It would be interesting to find a prime (k=6) making the "Divides Phi" achievable class (Probability 1/3). There is one here found about two decades ago.

I can attach the sieve file associated for this sequence if anyone's interested in testing further.

You can search 6*p^n+1 for all primes p, if p != 1 mod 7 and p != 34 mod 35, then there should be infinitely many primes of the form 6*p^n+1, but there is not always an easy prime, e.g. for p = 409, the first such prime is 6*409^369832+1, a 965900-digit prime!!

[carpetpool]

Originally Posted by sweety439

You can search 6*p^n+1 for all primes p, if p != 1 mod 7 and p != 34 mod 35, then there should be infinitely many primes of the form 6*p^n+1, but there is not always an easy prime, e.g. for p = 409, the first such prime is 6*409^369832+1, a 965900-digit prime!!

p = 1 mod 4 increases the odds of such a prime, as N = 6*p^n+1 is congruent 7 mod 8, but there is still 1/3 chance --- which at 965K digits, I'm not sure if anyone would be willing to test if N | Phi(p^k,2). If more primes were found, probably then it could be worth it. I was considering working on such a project if others were interested. AFAIK, Ryan is testing 6*5^n+1 and he's had quite extensive computation power lately so maybe he'll luck out soon...

6*13^n+1 is next on the list so, I might give that sequence a quick check...

[sweety439]

3*2^n+1 divides Phi(3*2^(n-1),2) for n = 5, 6, 8, 12, and the cofactor of them are all primes (this is the cofactor for n=12), are there any other such n?

[Batalov]

Originally Posted by sweety439

... are there any other such n?

Is that a riddle? We give up. What is the answer?