A354178 Numbers whose number of divisors is coprime to 30.

1, 64, 729, 1024, 4096, 15625, 46656, 59049, 65536, 117649, 262144, 531441, 746496, 1000000, 1771561, 2985984, 3779136, 4194304, 4826809, 7529536, 9765625, 11390625, 16000000, 24137569, 34012224, 43046721, 47045881, 47775744, 60466176, 64000000, 85766121, 113379904

Offset: 1

Author

Amiram Eldar, May 18 2022

Keywords

• nonn

Comments

Numbers k such that gcd(d(k), 30) = 1, where d(k) is the number of divisors of k (A000005).

All the terms are squares since their number of divisors is odd.

Examples

			64 is a term since A000005(64) = 7 and gcd(7, 30) = 1.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000 (first 1709 terms from Amiram Eldar)
• Titus Hilberdink, How often is d(n) a power of a given integer?, Journal of Number Theory, Vol. 236 (2022), pp. 261-279.
• Index entries for sequences computed from exponents in factorization of n.

Crossrefs

Subsequence of other sequences of numbers k such that gcd(d(k), m) = 1: A000290 (m=2), A336590 (m=3), A352475 (m=6).

Cf. A000005, A007775, A354179.

Programs

Mathematica

Select[Range[10^4]^2, CoprimeQ[DivisorSigma[0, #], 30] &]

PARI

isok(k) = gcd(numdiv(k), 30) == 1;
for(k=1, 10650, if(isok(k^2), print1(k^2,", ")))

Formula

a(n) = A354179(n)^2.

The number of terms <= x is (zeta(5)*zeta(5/3))/(zeta(4)*zeta(10/3))*x^(1/6) + (zeta(3)*zeta(3/5))/(zeta(2)*zeta(12/5))*x^(1/10) + O(x^(1/20 + eps)) for all eps > 0 (Hilberdink, 2022).

Sum_{n>=1} 1/a(n) = Product_{p prime} (p^2 + p^8 + p^12 + p^14 + p^18 + p^20 + p^24 + p^30)/(p^30 - 1) = 1.0183538548...