Let $S_n$ be the collection of subsets of $\{1,3,5,...\}$ whose element sum is $n$. We’ll build a function $f:S_n\to S_{n+1}$. Let $S\in S_n$. Case 1: $1\notin S$. Define $f(S)=S\cup\{1\}$. Case 2: $1\in S$. Let $m$ be the maximal element of $S$. Since $n>1$, $m>1$. Define $f(S)=(S\setminus\{1\})\cup\{m+2\}$. Clearly $f$ is injective. Hence $A_n=|S_n|\leq |S_{n+1}|=A_{n+1}$.

math90 wrote: Let $S_n$ be the collection of subsets of $\{1,3,5,...\}$ whose sum is $n$. We’ll build a function $f:S_n\to S_{n+1}$. Let $S\in S_n$. Case 1: $1\notin S$. Define $f(S)=S\cup\{1\}$. Case 2: $1\in S$. Let $m$ be the maximal element of $S$. Since $n>1$, $m>1$. Define $f(S)=(S\setminus\{1\})\cup\{m+2\}$. Clearly $f$ is injective. Hence $A_n=|S_n|\leq |S_{n+1}|=A_{n+1}$. Nice! Removing $1$ And adding $m+2$ guarantees that the function is injective. However, why is it not sufficient to just create a function that increases the maximum element by $1$, regardless of whether the representation has a $1$ present or not?

Because all elements must be odd.

math90 wrote: Because all elements must be odd. Oh. Ofc. Duh.