OFFSET
1,2
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.
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.
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...
EXAMPLE
64 is a term since A000005(64) = 7 and gcd(7, 30) = 1.
MATHEMATICA
Select[Range[10^4]^2, CoprimeQ[DivisorSigma[0, #], 30] &]
PROG
(PARI) isok(k) = gcd(numdiv(k), 30) == 1;
for(k=1, 10650, if(isok(k^2), print1(k^2, ", ")))
CROSSREFS
KEYWORD
nonn
AUTHOR
Amiram Eldar, May 18 2022
STATUS
approved