Factors of $n^2$ and $n$

Let $n=p_1^{q_1}p_2^{q_2}$ , which implies $n^2=p_1^{2q_1}p_2^{2q_2}$ for some prime $p_1,p_2$ . The number of divisors of $n$ is $(q_1+1)(q_2+1)$ and $n^2$ is $(2q_1+1)(2q_2+1)$ .

Consider the factors of $n^2$ other than $n.$ Note that the number of such factors greater than $n$ are equal to the factors less than $n$ (If we group all of these factors (excluding $n$ ) into pairs that multiply to $n^2$ , then one factor per pair is less than $n$ ). Therefore the number of factors of $n^2$ less than $n$ is $\frac{(2q_1+1)(2q_2+1)-1}{2}= 2q_1q_2+q_1+q_2$ . Whereas the number of factors of $n$ other than itself is $(q_1+1)(q_2+1)-1=q_1q_2+q_1+q_2$

Thus the number of factors less than $n$ but do not divide $n$ are $(2q_1q_2+q_1+q_2)-(q_1q_2+q_1+q_2)=q_1q_2$ . In this case $q_1q_2=23 \times 17=\boxed{391}.$