Proving number of partitions of $n$ to $3$ parts at most. I have an exercise, to prove that the number of partitions of $n$ to at most $3$ integers is $\frac{(n+3)^2}{12}$ rounded. I tried to prove by induction but I don't know how.

The number of partitions of $n$ into at most $k$ integers is equal to the number of partitions of $n$ into integers no bigger than $k$.  Let's call this number $p(n,k)$.  In general $$p(n,k)=p(n,k-1)+p(n-k,k)$$ starting with $p(0,0)=1$ and $p(n,0)=0$ for $n\not=0$ and zero if $n$ or $k$ are negative.
For given $k$, patterns are going to repeat every $k!$ terms, so we can find and show by induction that $p(m,1)=1$ and $p(2m,2)=p(2m+1,2)=m+1$ for $m \ge 0$ and thus

*

*$p(6m+0,3)= 3m^2+3m+1 = \frac{((6m+0)+3)^2}{12} + \frac1{4}$

*$p(6m+1,3)= 3m^2+4m+1 = \frac{((6m+1)+3)^2}{12} - \frac1{3}$

*$p(6m+2,3)= 3m^2+5m+2 = \frac{((6m+2)+3)^2}{12} - \frac1{12}$

*$p(6m+3,3)= 3m^2+6m+3 = \frac{((6m+3)+3)^2}{12} + 0$

*$p(6m+4,3)= 3m^2+7m+4 = \frac{((6m+4)+3)^2}{12} - \frac1{12}$

*$p(6m+5,3)= 3m^2+8m+5 = \frac{((6m+5)+3)^2}{12} - \frac1{3}$
Since all the fractions on the right hand side are smaller than $\frac12$, this  leads to the conclusion that you can round and say $$p(n,3)= \left[\frac{(n+3)^2}{12}\right].$$
