OFFSET
1,1
COMMENTS
The existence of infinitely many primes p such that p^n - 1 has exactly a(n) divisors is conjectured. E.g., although it is known that p-1 has fewer than 4 divisors for only finitely many primes p (see Example section), it is not known whether there exist infinitely many primes p such that p-1 has exactly 4 divisors. (Thanks to Jianing Song, who pointed out the need for this clarification.) - Jon E. Schoenfield, Mar 04 2021
For each prime q, every number k that has exactly q divisors is a prime power k = p^(q-1) for some prime p. As a result, a(q-1) can be useful in identifying numbers of the form p^(q-1) - 1 that are terms of A161460 (see Example section).
From Bernard Schott, Aug 22 2019: (Start)
For n prime >= 3, a(n) = 8;
for n = q^2, q prime >= 3, a(n) = 16. (End)
LINKS
Antti Karttunen, Table of n, a(n) for n = 1..10000
FORMULA
a(n) = 2^(A000005(n)+1) for odd n. - Jianing Song, Dec 05 2021
EXAMPLE
a(1) = 4: The only primes p for which p-1 has fewer than 4 divisors are 2, 3, and 5; for all primes p > 5, p-1 has at least 4 divisors, and the terms in A005385 (Safe primes) except 5 are primes p such that p-1 has exactly 4 divisors.
a(2) = 32: p^2 - 1 = (p-1)*(p+1) has fewer than 32 divisors only for p = 2, 3, 5, 7, 11, 13, 17, 19, 23, 31, 37, 47, and 73; for all primes p such that the product of the 3-smooth parts of p-1 and p+1 is 24 and p-1 and p+1 each have one prime factor > 3, p^2 - 1 has exactly 32 divisors (see A341658).
a(4) = 160: primes p such that p^4 - 1 has exactly 160 divisors are plentiful (see A341662), but only p = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 59, 61, 71, 79, and 101 yield tau(p^4 - 1) < 160. Of these, p = 13, 29, 59, and 61 all give tau(p^4 - 1) = 80; 37 and 101 both give 120 divisors; and 41 and 71 both give 144. For each of the ten remaining primes (p = 2, 3, 5, 7, 11, 17, 19, 23, 31, 79), the value of tau(p^4 - 1) is unique, so each of those ten values of p^4 - 1 is a term in A161460.
PROG
(PARI) f(n) = {if (n%2, 2, res = 1; forprime(p=2, n+1, if (!(n % (p-1)), t = valuation(n, p); if (p==2, if (t, res *= p^(t+2)), res *= p^(t+1)); ); ); res; ); } \\ A079612
a(n) = numdiv(f(n))*2^numdiv(n); \\ Michel Marcus, Aug 22 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Jon E. Schoenfield, Aug 21 2019
EXTENSIONS
Name edited by Jon E. Schoenfield, Mar 04 2021
STATUS
approved