A352475 Numbers m such that gcd(d(m),6) = 1.

1, 16, 64, 81, 625, 729, 1024, 1296, 2401, 4096, 5184, 10000, 11664, 14641, 15625, 28561, 38416, 40000, 46656, 50625, 59049, 65536, 82944, 83521, 117649, 130321, 153664, 194481, 234256, 250000, 262144, 279841, 331776, 455625, 456976, 531441, 640000, 707281, 746496

Offset: 1

Author

Michael De Vlieger, Mar 26 2022

Keywords

• nonn
• easy

Comments

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

The square roots of terms are in A001694.

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

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Titus Hilberdink, How often is d(n) a power of a given integer?, Journal of Number Theory, Vol. 236 (2022), pp. 261-279.

Crossrefs

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

Programs

Mathematica

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

PARI

isok(m) = gcd(numdiv(m), 6) == 1; \\ Michel Marcus, Mar 29 2022

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

Formula

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

Sum_{n>=1} 1/a(n) = Pi^2/9 (A100044). - Amiram Eldar, Mar 31 2022

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