I have to find $n $ s.t. $A (n)=A (n+1) $. Let  $M=\lbrace 1, 3, 5, 7, 9,...\rbrace $.
For every $n\in \mathbb {N} $, $n\geq 2$, let $A (n)= $ the number of subsets of $M $ with the sum of the elements  $n$.
For example $A (9)=2$ because there are $\lbrace 9\rbrace$ and  $\lbrace 1, 3, 5\rbrace$ .
I proved that $A (n)\leq A (n+1)$. To every subset with the sum of the element $n $ I associated a subset with the sum of the elements $n+1$. Actually I constructed an injective function.
Now I have to find $n $ s.t. $A (n)=A (n+1) $.
I noticed that every $n\geq 16$ can be written as a sum of more than $3$ odd numbers.
If $n=x_1+x_2+...+x_k $ with $k\geq 3$ then I generate  two subsets $$\lbrace x_2,..., x_{k-1}, x_1+x_k -1\rbrace$$ and 
 $$\lbrace x_1,x_3,..., x_{k-1}, x_2+x_k -1\rbrace $$, where $x_1 <x_2 <... <x_k $.They are different.
So $A (n)> A (n-1) $ in this case. I have to check for $n <16$.
It's this a complete proof?
 A: First we claim that $A(n+1)\ge A(n)$. It's straightforward. Imagine $(x_1,x_2,...,x_m)$ is a set whose sum is $n$. If $x_1$ is equal to $1$ then $(x_2,...,x_m+2)$ is a set whose sum is $n+1$ ($x_m$ is replaced with $x_m+2$). Let us call this method of generating a set for $n+1$ A. 
If $x_1$ is not equal to $1$ then $(1,x_1,x_2,...,x_m)$ is a set whose sum is $n+1$. So, obviously  $A(n+1)\ge A(n)$. Again, let us call this method of generating a set for $n+1$ B.
Now, imagine $n+1$ is of the form of $4k$ numbers ($k\ge2)$. Notice that $(2k-1,2k+1)$ can be obtained neither from the method A nor from the method B. Therefore, in this case $A(n+1)$ is bigger than $A(n)$.
Now, imagine $n+1$ is of the form of $4k+3$ numbers ($k\ge3)$. Notice that $(3,2k-1,2k+1)$ can be obtained neither from the method A nor from the method B. Therefore, in this case $A(n+1)$ is bigger than $A(n)$.
Now, imagine $n+1$ is of the form of $4k+1$ numbers ($k\ge5)$. Notice that $(5,2k-3,2k-1)$ can be obtained neither from the method A nor from the method B. Therefore, in this case $A(n+1)$ is bigger than $A(n)$.
Now, imagine $n+1$ is of the form of $4k+2$ numbers ($k\ge7)$. Notice that $(3,7,2k-5,2k-3)$ can be obtained neither from the method A nor from the method B. Therefore, in this case $A(n+1)$ is bigger than $A(n)$.
Therefore we just need to check a few numbers less than $30$. 
A: As for the original question of finding an $n$ such that $A(n)=A(n+1)$, note that $A(3)=A(7)=1$. In light of the analysis that $A(n)\le A(n+1)$, what do we know about $A(4),A(5), A(6)$.
We also note that $A(8)=A(11)=2$. 
