A341670 -id:A341670 - cp's OEIS frontend

A309906 a(n) is the smallest number of divisors of p^n - 1 that may possibly occur for arbitrarily large primes p.

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

Jon E. Schoenfield, Aug 21 2019

nonn

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)

			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.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A000005, A005385, A006863, A079612, A161460, A341658, A341662, A341670.

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

a(n) = A000005(A079612(n))*2^A000005(n).

a(n) = 2^(A000005(n)+1) for odd n. - Jianing Song, Dec 05 2021

Name edited by Jon E. Schoenfield, Mar 04 2021

A342065 Primes p such that p^9 - 1 has 16 divisors.

Original entry on oeis.org

383, 12227, 44519, 44687, 56003, 97523, 130259, 148727, 160739, 169007, 208799, 258887, 270563, 281783, 331883, 336143, 353099, 364979, 498119, 501707, 550679, 573107, 577667, 716747, 753023, 775367, 781007, 784727, 861299, 887543, 1084247, 1085159, 1099139

Offset: 1

Jon E. Schoenfield, Feb 27 2021

nonn

Conjecture: sequence is infinite.
The only primes p such that p^9 - 1 has fewer than A309906(9)=16 divisors are p=2 (2^9 - 1 = 511 = 7*73 has 4 divisors) and p=3 (3^9 - 1 = 19682 = 2*13*757 has 8 divisors).
For every term p, p^9 - 1 is of the form 2*q*r*s, where q = (p-1)/2, r = (p^2 + p + 1), and s = (p^6 + p^3 + 1) are primes (see Example section).
The Generalized Dickson's Conjecture implies there are infinitely many p such that p, (p-1)/2, p^2+p+1 and p^6+p^3+1 are prime. - Robert Israel, Feb 28 2021

			                       factorization of p^9 - 1
    p =   ===================================================
n   a(n)  2 * (p-1)/2 * (p^2+p+1) *      (p^6 + p^3 + 1)
-  -----  ---------------------------------------------------
1    383  2 *   191   *    147073 *          3156404483062657
2  12227  2 *  6113   * 149511757 * 3341330794198073514753973

Robert Israel, Table of n, a(n) for n = 1..1000

Cf. A000005, A000040, A309906, A341670.

R:= NULL: count:= 0: q:= 1:
while count < 100 do
  q:= nextprime(q);
  p:= 2*q+1;
  if isprime(p) and isprime(p^2+p+1) and isprime(p^6+p^3+1) then
    count:= count+1; R:= R, p;
  fi
od:
R; # Robert Israel, Feb 28 2021

A342066 Primes p such that p^10 - 1 has 256 divisors.

Original entry on oeis.org

1187, 4723, 33037, 66973, 72797, 87973, 100523, 197123, 219683, 229693, 276293, 278827, 440653, 448997, 482837, 562963, 601333, 621443, 670493, 742723, 846877, 892357, 1033427, 1149307, 1166027, 1245067, 1256747, 1614413, 1679773, 1865693, 1950323, 1970467

Offset: 1

Jon E. Schoenfield, Feb 27 2021

nonn

Conjecture: sequence is infinite.
The only primes p such that p^10 - 1 has fewer than A309906(10)=256 divisors are 2, 3, 5, 7, 11, 13, and 43.
p^10 - 1 = (p-1)*(p+1)*(p^4 - p^3 + p^2 - p + 1)*(p^4 + p^3 + p^2 + p + 1). For every p > 11, one of these five factors is divisible by 11; one of p-1 and p+1 is divisible by 3; and p-1 and p+1 are consecutive even numbers, so one of them is divisible by 4 and their product is divisible by 8; thus, p^10 - 1 is divisible by 2^3 * 3 * 11.
For every term p with the exception of a(1)=1187, p^10 - 1 is of the form 2^3 * 3 * 11 * q * r * s * t, where q, r, s, and t are distinct primes > 11.

			For p = a(1) = 1187, p^10 - 1 = 2^3 * 3^3 * 11 * 593 * 1983522604541 * 1986867499321;
for p = a(2) = 4723, p^10 - 1 = 2^3 * 3 * 11 * 787 * 1181 * 45245048697451 * 497484826300381.

Cf. A000005, A000040, A309906, A341670.

A342067 Primes p such that p^11 - 1 has 8 divisors.

Original entry on oeis.org

3, 467, 2039, 4679, 5399, 5939, 6899, 8783, 12347, 16487, 18443, 23879, 25583, 33647, 35879, 36299, 44819, 47207, 53147, 57119, 67499, 74507, 90239, 93287, 96059, 119759, 125003, 133499, 135119, 136223, 157019, 159539, 164999, 165059, 168887, 178799, 188159

Offset: 1

Jon E. Schoenfield, Feb 28 2021

nonn

Conjecture: sequence is infinite.
The only primes p such that p^11 - 1 has fewer than A309906(11)=8 divisors are 2 and 5.
p^11 - 1 = (p-1)*(p^10 + p^9 + p^8 + p^7 + p^6 + p^5 + p^4 + p^3 + p^2 + p + 1).
For every term p, p^11 - 1 is of the form 2*q*r, where q and r are distinct odd primes. With the exception of p=a(1)=3, each term p is a number such that (p-1)/2 and (p^10 + p^9 + p^8 + ... + p^2 + p + 1) are primes.

			   p =
n  a(n)             factorization of p^11 - 1
-  ----  ------------------------------------------------
1     3  2 *   23 *                                  3851
2   467  2 *  233 *           494424256962371823779424877
3  2039  2 * 1019 *    1242754384106847037173120489949801
4  4679  2 * 2339 * 5030640462820574591105701447273296601

Cf. A000005, A000040, A309906, A341670.

isok(p) = isprime(p) && (numdiv(p^11-1) == 8); \\ Michel Marcus, Feb 28 2021

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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.

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A342065 Primes p such that p^9 - 1 has 16 divisors.

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A342066 Primes p such that p^10 - 1 has 256 divisors.

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A342067 Primes p such that p^11 - 1 has 8 divisors.

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