Let $r(m)=\sum _{d|m} 1$ which is number of positive dividers $m$ and $P(a)$ which is number of possible divisions of $a$.
Prove that the total number of ingredients in all $n$ divisions is equal $\sum _{k=0}^{n-1} P(k)r(n-k)$.
My try:
Let's see what happens for $5$:
$$5=5$$
$$5=4+1$$
$$5=3+2$$
$$5=3+1+1$$
$$5=2+2+1$$
$$5=2+1+1+1$$
$$5=1+1+1+1+1$$
So $P(5)=7$.
Hence we have that the total number of ingredients in all $5$ divisions is equal: $$5+4+3+3+2+2+1=20$$
Now we will use the given formula: $$\sum _{k=0}^{4} P(k)r(n-k)=P(0)r(5)+P(1)r(4)+P(2)r(3)+P(3)r(2)+P(4)r(1)=1 \cdot 2 +1 \cdot 3 +2 \cdot 2 + 3 \cdot 2 + 5 \cdot 1 =20$$
I think that exists bijection between my way of counting and given pattern but I can't find it.
Can you give me some tips?