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A354799 Integers m in A001694 such that 3 | d(m^2), where d(n) = A000005(n).

%I A354799 #37 Jun 29 2022 01:50:08
%S A354799 16,81,128,144,324,400,432,625,648,784,1024,1152,1296,1936,2000,2025,
%T A354799 2187,2401,2500,2592,2704,3200,3456,3600,3888,3969,4624,5000,5184,
%U A354799 5488,5625,5776,6272,7056,8100,8192,8464,8748,9216,9604,9801,10000,10125,10368,10800
%N A354799 Integers m in A001694 such that 3 | d(m^2), where d(n) = A000005(n).
%H A354799 Michael S. Branicky, <a href="/A354799/b354799.txt">Table of n, a(n) for n = 1..10000</a>
%F A354799 Equals { A001694 \ A350014 }.
%F A354799 Equals { m in A001694 : d(m^2) mod 3 = 0 }.
%F A354799 Sum_{n>=1} 1/a(n) = zeta(2)*zeta(3)/zeta(6) - 5*zeta(3)/(2*zeta(2)) = 0.1166890133... . - _Amiram Eldar_, Jun 28 2022
%e A354799 A001694(5) = 16 is a term since d(16^2) = d(256) = 9, and 9 is a multiple of 3.
%e A354799 A001694(13) = 81 is a term since d(81^2) = d(6561) = 9, and 9 is a multiple of 3.
%e A354799 A001694(3) = 8 is not a term since d(8^2) = d(64) = 7, which is not divisible by 3.
%t A354799 With[{nn = 10800}, Select[Union@ Flatten@ Table[a^2*b^3, {b, nn^(1/3)}, {a, Sqrt[nn/b^3]}], Mod[DivisorSigma[0, #^2], 3] == 0 &]]
%o A354799 (PARI) isok(m) = ispowerful(m) && !(numdiv(m^2) % 3); \\ _Michel Marcus_, Jun 27 2022
%o A354799 (Python)
%o A354799 from sympy import divisor_count as d, factorint as f
%o A354799 def ok(k): return k > 1 and min(f(k).values()) > 1 and d(k*k)%3 == 0
%o A354799 print([k for k in range(11000) if ok(k)]) # _Michael S. Branicky_, Jun 28 2022
%Y A354799 Cf. A000005, A001694, A008585, A076400, A350014.
%K A354799 nonn,easy
%O A354799 1,1
%A A354799 _Michael De Vlieger_, Jun 21 2022