jmc

If $n$ is prime, then $f(n)=n+1$. If $n+1$ is prime, then $n$ must be even. Therefore, the only prime value of $n$ for which $n+1$ is prime is $n=2$. If $n=p^a$ for some prime $p$ and an integer $a>1$, then $f(n)=\frac{p^{a+1}-1}{p-1}$. This value is not guaranteed to be composite, so we must check all powers of primes. Checking powers of $2$ first, $f(4)=7$, $f(8)=15$, and $f(16)=31$. Two of these powers of 2 work. Checking powers of $3$, $f(9)=13$ and $f(27)$ is beyond our boundary for $n$, so one power of $3$ works. Finally, $f(25)=31$, which gives one more value of $n$ that works. Finally, if $n$ is any other composite integer, it can be written as the product of two distinct primes $p$ and $q$. Since $n\le 25$, $n$ cannot be the product of three distinct primes, so $n=p^a{q}^b$ for positive integers $a$ and $b$. As a result, $f(n)=\left(\frac{p^{a+1}-1}{p-1}\right)\left(\frac{q^{b+1}-1}{q-1}\right)$, but then $f(n)$ is the product of two integers which are greater than $1$, so $f(n)$ is composite. Therefore, there are $2+1+1+1=\boxed{5}$ values of $n$ for which $f(n)$ is prime.