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Primes p such that p - 1 = (tau(p - 1) - 1)^k for some k >= 0, where tau(n) is the number of divisors of n (A000005).
3

%I #18 Apr 28 2024 16:24:37

%S 2,5,17,65537

%N Primes p such that p - 1 = (tau(p - 1) - 1)^k for some k >= 0, where tau(n) is the number of divisors of n (A000005).

%C Conjecture: the sequence is finite.

%C Corresponding values of numbers k: 0, 2, 2, 4, ...

%C A Fermat prime from A019434 of the form F(n) = 2^(2^n) + 1 is a term if k = 2^n * log(2) / log(2^n) is an integer.

%e 65537 (prime) is in the sequence because 65537 - 1 = (tau(65536) - 1)^4 = 16^4.

%o (Magma) [2] cat [n+1: n in [A219338(n)] | IsPrime(n+1)]

%o (Magma) Set(Sort([n: n in[1..1000000], k in [0..100] | IsPrime(n) and (n-1) eq (NumberOfDivisors(n-1) - 1)^k]))

%o (PARI) listp(nn) = {print1(p=2, ", "); forprime(p=5, nn, expo = valuation(x=(p-1), y=(numdiv(p-1)-1)); if (x == y^expo, print1(p, ", ")););} \\ _Michel Marcus_, Jun 04 2015

%Y Cf. A000005, A019434, A219338, A007516, A004249, A249759.

%K nonn

%O 1,1

%A _Jaroslav Krizek_, May 29 2015