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Primes which divide numbers of the form 3^k + 2 for k >= 1.
3

%I #33 May 04 2018 22:41:36

%S 5,7,11,17,19,29,31,43,53,59,79,83,89,97,101,107,113,127,131,137,139,

%T 149,163,173,179,197,199,211,223,227,233,241,251,257,269,281,283,293,

%U 317,331,337,347,353,379,389,401,409,419,439,443,449,457,461,463,467

%N Primes which divide numbers of the form 3^k + 2 for k >= 1.

%C The first odd prime not to appear in the sequence is 3 because 3^k + 2 == 2 mod 3 for k >= 1.

%C Primes p such that the order of -2 (mod p) divides the order of 3 (mod p). - _Joerg Arndt_, Mar 31 2018, corrected by _Robert Israel_, May 04 2018

%H Robert Israel, <a href="/A301913/b301913.txt">Table of n, a(n) for n = 1..10000</a>

%e 5 divides 245 which is 3^5+2 so 5 is in the sequence.

%e 7 divides 245 which is 3^5+2 so 7 is in the sequence.

%e The values of x = (3^k+2) mod 13 for k = 0, 1, 2, 3, ... are 3, 5, 11, 3, 5, 11, ...; 13 never divides any 3^k + 2, so 13 is not in the sequence.

%p select(t -> numtheory:-mlog(-2,3,t)<>FAIL, [seq(ithprime(i),i=3..100)]);

%t fQ[p_] := IntegerQ@ MultiplicativeOrder[3, p, -2]; Select[ Prime@ Range@ 100, fQ] (* _Robert G. Wilson v_, Apr 07 2018 *)

%o (PARI) is(n)=n>4 && isprime(n) && znorder(Mod(-2,n))%znorder(Mod(3,n))==0 \\ _Charles R Greathouse IV_, May 04 2018

%Y Cf. A168607.

%K nonn

%O 1,1

%A _Luke W. Richards_, Mar 28 2018