For any positive integer $n$, let $f(n)$ be the number of subsets of $\{1, 2, . . . , n\}$ whose sum is equal to $n$. Does there exist infinitely many positive integers $m$ such that $f(m) = f(m + 1)$?
(Note that each element in a subset must be distinct.)
Answer: No, only $m = 1,3$ satisfy this.
Suppose $m \ge 4$. Consider all such subsets which add up to $m$. Add 1 to their maximum element to obtain subsets for $m+1$. Now, if $m$ is odd, say $2k+1 (k > 1)$, then consider $\{1,k,k+1\}$ as a subset that adds up to $m+1$ but has not been listed already (since $1,k,k$ isn't valid for $m$). If $m$ is even, say $2k (k \ge 1)$, then consider $\{k, k+1 \}$ as a subset that adds up to $m+1$ but has not been listed already. Thus, $f(m+1) \ge f(m) + 1$ for all $m \ge 4$.
An interesting question would be to find $m$ for which equality holds.
Any way $f(n)$ could be computed exactly? It is the coefficient of $x^n$ in $(1+x)(1+x^2)(1+x^3) \ldots (1+x^n)$ however this doesn't seem to do much to help. Differentiating $n$ times and putting $0$ doesn't give any nontrivial results either.