Let $p_{2}(n)$ denote the number of partitions of the positive integer n into at most 2 parts. Prove for each $n \ge 4$ that $p_{2}(n) + p_{2}(n-3) = n$
I understand that I need to split this into odd and even cases - I would just like my reasoning validating. Any comments/feedback on my proof or notation are very welcome.
Starting with the even numbers:
The number of partitions of size 2 of an even number $n$ is given by $\frac{n}{2} + 1$
The number of partitions of size 2 of an even number $n-3$ is given by $\frac{n-3}{2} + 1$ Since $n-3$ is odd, we only take the integer part of $\frac{n-3}{2}$ which is $\frac{n}{2} -2$. Adding on the 1 gives $\frac{n}{2} -1$.
Now for the even case, we do $p_{2}(n) + p_{2}(n-3) = \frac{n}{2} +1 + \frac{n}{2} -1 =n$
Odd Numbers
The number of partitions of size 2 of an odd number $n$ is given by $\frac{n}{2} + 1$. Since $n$ is odd, we only need the integer part of $\frac{n}{2}$ which is $\frac{n-1}{2}$. So we have $\frac{n-1}{2} +1$.
The number of partitions of size 2 of an odd number $n-3$ is given by $\frac{n-3}{2} + 1$. Since $n-3$ is even, we get $\frac{n-1}{2} -1 +1 = \frac{n-1}{2}$.
Now for the odd case, we do $p_{2}(n) + p_{2}(n-3) = \frac{n-1}{2} +1 + \frac{n-1}{2}=n$
It is simpler to have a "compact view" of the arguments. From the post, the formula $$p_2(n)=\left[\frac n2\right]+1$$ is known, and was shown. Now we can cover all cases in one breath: $$ \begin{aligned} p_2(n)+p_2(n-3) &= \left[\frac n2\right]+1 + \left[\frac {n-3}2\right]+1 \\ &= \left[\frac n2\right] + \left[\frac {n-3}2+1+1\right] \\ &= \left[\frac n2\right] + \left[\frac {n+1}2\right] \\ &= n\ . \end{aligned} $$ We have used some properties of the floor function $[x+1]=[x]+1$ and $[x] + \left[x+\frac 12\right]= [2x]$, where $x\in \Bbb R$, which are simpler to check in the case $x$ is half an integer.