OFFSET
1,2
COMMENTS
Numbers k such that gcd(d(k^2), 30) = 1, where d(k) is the number of divisors of k (A000005).
LINKS
Michael De Vlieger, Table of n, a(n) for n = 1..10000 (first 1709 terms from Amiram Eldar)
FORMULA
a(n) = sqrt(A354178(n)).
Sum_{n>=1} 1/a(n) = Product_{p prime} (p + p^4 + p^6 + p^7 + p^9 + p^10 + p^12 + p^15)/(p^15 - 1) = 1.2449394393...
EXAMPLE
8 is a term since A000005(8^2) = 7 and gcd(7, 30) = 1.
MATHEMATICA
Select[Range[25000], CoprimeQ[DivisorSigma[0, #^2], 30] &]
PROG
(PARI) isok(m) = gcd(numdiv(m^2), 30) == 1; \\ Michel Marcus, May 19 2022
CROSSREFS
KEYWORD
nonn
AUTHOR
Amiram Eldar, May 18 2022
STATUS
approved