A318761 Composite k that divides 2^(k-2) + 3^(k-2) + 6^(k-2) - 1.

4, 6, 8, 12, 24, 25, 125, 174, 228, 276, 325, 348, 451, 1032, 1105, 1128, 1729, 2388, 2465, 2701, 2821, 3721, 5272, 5365, 6601, 8911, 10585, 12025, 12673, 15841, 18721, 22681, 23585, 23725, 29341, 31621, 32376, 35016, 35425, 41041, 41125, 46632, 46657, 47125

Offset: 1

Author

Jianing Song, Sep 02 2018

Keywords

• nonn

Comments

Note that for primes p >= 5, p always divides 2^(p-2) + 3^(p-2) + 6^(p-2) - 1 (see A318760).

It's interesting to study the squares of primes in this sequence. For primes p >= 5, x^(p^2-2) == x^(p-2) (mod p^2) for any integer x, so p^2 is a term if and only if p^2 divides 2^(p-2) + 3^(p-2) + 6^(p-2) - 1. It's easy to see that for any prime p, p^2 is a term of this sequence if and only if p is in A238201 and p != 3 (p = 2, 5, 61, 1680023, 7308036881, there are no others up to 7*10^10). - Jianing Song, Dec 25 2018

Examples

			(2^10 + 3^10 + 6^10 - 1)/12 = 5403854 which is an integer, so 12 is a term.
(2^22 + 3^22 + 6^22 - 1)/24 = 5484238967813377 which is also an integer, so 24 is a term.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..4127 (terms below 10^10)

Crossrefs

Cf. A238201, A318760.

A052155 is a proper subsequence.

Programs

Mathematica

Select[Range[48000], CompositeQ[#] && Mod[Sum[PowerMod[k, #-2, #],{k, {2, 3, 6}}], #] == 1 &] (* Amiram Eldar, Jul 17 2024 *)

PARI

b(n) = lift(Mod(2, n)^(n-2) + Mod(3, n)^(n-2) + Mod(6, n)^(n-2));
for(n=2, 30000, if(isprime(n)==0&&b(n)==1, print1(n, ", ")))