This is easily equivalent to \(v_{p}(n) \not \equiv 1,5(\bmod 6)\) for all primes \(p\). We just count: \(p \geq 11 \Longrightarrow v_{p}(n)=1\) is clear, so we only look at the prime factorizations with primes from \(\{2,3,5,7\}\). This is easy to compute: we obtain 13.
Let a positive integer \(n\) be called a cubic square if there exist positive integers \(a, b\) with \(n=\operatorname{gcd}\left(a^{2}, b^{3}\right)\). Count the number of cubic squares between 1 and 100 inclusive.