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%I #25 Dec 26 2018 13:33:46
%S 4,6,8,12,24,25,125,174,228,276,325,348,451,1032,1105,1128,1729,2388,
%T 2465,2701,2821,3721,5272,5365,6601,8911,10585,12025,12673,15841,
%U 18721,22681,23585,23725,29341,31621,32376,35016,35425,41041,41125,46632,46657,47125
%N Composite k that divides 2^(k-2) + 3^(k-2) + 6^(k-2) - 1.
%C Note that for primes p >= 5, p always divides 2^(p-2) + 3^(p-2) + 6^(p-2) - 1 (see A318760).
%C 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
%e (2^10 + 3^10 + 6^10 - 1)/12 = 5403854 which is an integer, so 12 is a term.
%e (2^22 + 3^22 + 6^22 - 1)/24 = 5484238967813377 which is also an integer, so 24 is a term.
%o (PARI) b(n) = lift(Mod(2, n)^(n-2) + Mod(3, n)^(n-2) + Mod(6, n)^(n-2));
%o for(n=2, 30000, if(isprime(n)==0&&b(n)==1, print1(n, ", ")))
%Y Cf. A238201, A318760.
%Y A052155 is a proper subsequence.
%K nonn
%O 1,1
%A _Jianing Song_, Sep 02 2018
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