%I #18 May 19 2022 03:44:28

%S 1,16,64,81,625,729,1024,1296,2401,4096,5184,10000,11664,14641,15625,

%T 28561,38416,40000,46656,50625,59049,65536,82944,83521,117649,130321,

%U 153664,194481,234256,250000,262144,279841,331776,455625,456976,531441,640000,707281,746496

%N Numbers m such that gcd(d(m),6) = 1.

%C All terms are square since numbers coprime to 6 are odd.

%C The square roots of terms are in A001694.

%C Intersection of A000290 and A336590, i.e., numbers whose prime factorization has only exponents that are congruent to {0, 4} mod 6 (A047233). - _Amiram Eldar_, Mar 31 2022

%H Amiram Eldar, <a href="/A352475/b352475.txt">Table of n, a(n) for n = 1..10000</a>

%H Titus Hilberdink, <a href="https://doi.org/10.1016/j.jnt.2021.07.020">How often is d(n) a power of a given integer?</a>, Journal of Number Theory, Vol. 236 (2022), pp. 261-279.

%F a(n) = A350014(n)^2.

%F Sum_{n>=1} 1/a(n) = Pi^2/9 (A100044). - _Amiram Eldar_, Mar 31 2022

%F The number of terms <= x is (zeta(3/2)/zeta(2))*x^(1/4) + (zeta(2/3)/zeta(4/3))*x^(1/6) + O(x^(1/8 + eps)), for all eps > 0 (Hilberdink, 2022). - _Amiram Eldar_, May 18 2022

%t Select[Range[864]^2, GCD[DivisorSigma[0, #], 6] == 1 &] (* or, more efficiently, *)

%t With[{nn = 864}, Select[Union[Flatten@ Table[a^2*b^3, {b, nn^(1/3)}, {a, Sqrt[nn/b^3]}]]^2, Mod[DivisorSigma[0, #], 3] > 0 &]]

%o (PARI) isok(m) = gcd(numdiv(m), 6) == 1; \\ _Michel Marcus_, Mar 29 2022

%o (PARI) m = 100000; seq = direuler(p=2, m, (1 - X^8)/(1 - X^4)/(1 - X^6)); for(n=1, m, if(seq[n] != 0, print1(n, ", "))) \\ _Vaclav Kotesovec_, May 19 2022

%Y Cf. A000005, A000290, A001694, A007310, A047233, A076400, A100044, A336590, A350014.

%K nonn,easy

%O 1,2

%A _Michael De Vlieger_, Mar 26 2022