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A309906
a(n) is the smallest number of divisors of p^n - 1 that may possibly occur for arbitrarily large primes p.
27
4, 32, 8, 160, 8, 384, 8, 384, 16, 256, 8, 7680, 8, 128, 32, 1792, 8, 4096, 8, 3840, 32, 256, 8, 36864, 16, 128, 32, 2560, 8, 24576, 8, 4096, 32, 128, 32, 327680, 8, 128, 32, 36864, 8, 18432, 8, 2560, 128, 256, 8, 344064, 16, 1024, 32, 2560, 8, 20480, 32
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
FORMULA
a(n) = A000005(A079612(n))*2^A000005(n).
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
KEYWORD
nonn
AUTHOR
Jon E. Schoenfield, Aug 21 2019
EXTENSIONS
Name edited by Jon E. Schoenfield, Mar 04 2021
STATUS
approved