A Counting Parametric Problem

Clearly, if $n$ is odd, $n^2+1$ is even. In addition $\dfrac{n^2+1}{2}$ is odd because $n^2+1\equiv 2\pmod{4}$ . Therefore, if $n$ is odd, then $\dfrac{n^2+1}{2}\le n$ , which gives $n\le 1$ .

Now, suppose we have an even integer $n$ . We can divide this by two repeatedly until we get an odd integer. Since we know that the only odd integer that reduces to $1$ after $f(n)$ is applied some number of times, the rest of the solutions must be a power of two. There are $6$ even powers of two less than $100$ , and including the $1$ , we have a total of $\boxed{7}$ numbers.

1) we will only get 1 if n = 2 or 1 finally...notice ff(1) = f(2) = 1

2)now if n = 2^any number we will finally end up with f(2) which we saw was 1

3) also we can see for any odd n except 1 the value will keep increasing..therefore we can reject all odd values of n except 1.

4) for any even value of n other than (2^any number), we will finally end up with a odd number after a few applications, which as we saw in step 3 cant end up in 1 as it keeps increasing.

5) therefore only possible values of n will be 1,2,4,8,16,32 and 64