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A342037 Numbers k such that A307437(k) is divisible by 3.

%I A342037 #19 Jul 20 2022 05:15:13
%S A342037 1,27,702,1107,1431,2187,3375,3456,4266,5157,5805,6561,6831,7668,8073,
%T A342037 11313,11961,12771,12825,13149,13176,13257,14526,14715,14796,15039,
%U A342037 16011,16227,16497,17388,17496,17631,19251,19332,19413,20223,20277,20871,20952,21654
%N A342037 Numbers k such that A307437(k) is divisible by 3.
%C A342037 Indices of terms of A307437 that is divisible by 3.
%C A342037 For e > 0, 3^e is a term if and only if 2*3^e+1 is composite. Hence this sequence is infinite.
%C A342037 All terms > 1 are divisible by 27. Proof: Write a term k = 3^a*r > 1, 3 does not divide r. Suppose A307437(k) = 3^e*s, 3 does not divide s, e >= 1.
%C A342037 i) a = 0. If s = 1, then 2k = 2r divides psi(3^e) = 2*3^(e-1) => k = r = 1, a contradiction. Hence s > 1, then 2k = 2r divides psi(s).
%C A342037 ii) a = 1. If s has a prime factor congruent to 1 modulo 3, then r | psi(s) => 2k = 6r divides psi(s). Otherwise, we must have 3 | psi(3^e) => e >= 2, then 2k = 6r divides psi(7*s), a contradiction.
%C A342037 iii) a = 2. If s has a prime factor congruent to 1 modulo 9, then r | psi(s) => 2k = 18r divides psi(s). Otherwise, we must have 9 | psi(3^e) => e >= 3, then 2k = 18r divides psi(19*s), a contradiction.
%H A342037 Chai Wah Wu, <a href="/A342037/b342037.txt">Table of n, a(n) for n = 1..1000</a>
%e A342037 The smallest k such that 2*702 | psi(k) is k = 4293 = 3^4 * 53, hence 702 is a term.
%e A342037 The smallest k such that 2*3375 | psi(k) is k = 20331 = 3^4 * 251, hence 3375 is a term.
%e A342037 The smallest k such that 2*3456 | psi(k) is k = 20817 = 3^4 * 257, hence 3456 is a term.
%o A342037 (PARI) print1("1, "); forstep(n=27, 10000, 27, if(A307437(n)%3==0, print1(n, ", "))) \\ see A307437 for its program
%Y A342037 Cf. A307437, A341887, A003306.
%K A342037 nonn
%O A342037 1,2
%A A342037 _Jianing Song_, Feb 26 2021
%E A342037 More terms from _Chai Wah Wu_, Feb 27 2021