The same multiple as before

$G(n+1) = \mbox{lcm}(n+1,G(n)) = \left( \frac{(n+1) \left(G(n)\right)}{\gcd\left(n+1,G(n)\right)}\right).$ Suppose the integer $m$ can be written as the product of two relatively prime factors $p,q.$ Then $p < m, q < m$ so $p\ \vert\ G(m-1), q\ \vert \ G(m-1)$ and so $m\ \vert \ G(m-1).$ If $m$ cannot be written as a product of two relatively prime factors, then $m$ must be a power of a prime. Thus, $G(k) = G(k+1)$ if and only if $k+1$ is not a power of a prime. There are 26 primes from $2$ to $101,$ including $2,3,5,7,$ which are less than 10. Powers of 2 less than 101 are $4,8,16,32,64$ , powers of 3 less than 101 are $9,27,81$ , powers of 5 less than 101 are $25$ and powers of 7 less than 101 are $49.$ This is an additional $5 + 3 + 1 + 1 = 10$ numbers. Thus, there are $26 + 10 = 36$ powers of primes from 2 to 101, so $100 - 36 = 64$ of the numbers from 1 to 100 have the property that $G(k) = G(k+1).$