Partition of number vs dividing into blocks Let $P(n)$ be the number of partitions of integer $n$ and let $A(n,k)$ be the number of ways to put $n$ indistinguishable toys into $k$ distinguishable boxes but with the following restriction: the number of toys in $i$-th box must be divisible by $i$.
For which $(n,k)$ we have $A(n,k) < P(n), \ A(n,k) = P(n), \ A(n,k) > P(n)$ ?
My guess is $k < n, \ n \ge k$ and third inequality is never satisfied. Am i right?
Why?
I suppose there is a bijection between $P$ and $A$. Let take random partition of $n$. Let's group up identical components and write them like $component \cdot number$. For those who appear only once we write $component \cdot 1$. It gives us the way of putting toys into boxes. For example, let's take:
$10 = 3 + 2 + 2 + 1 + 1 + 1$. We get $10 = 3 \cdot 1 + 2 \cdot 2 + 1 \cdot 3$. And it is exactly the way we can form it into boxes. We put one block of $3$ toys into box three (so $3$ toys), two blocks of $2$ toys into box number two (so $4$ toys) and three blocks of $1$ toys into box number one (so $3$ toys).
On the other hand from the box setup we can get the original partition. For example we have $3$ toys in box one, and $3$ toys in box three. So we have 3 blocks of $1$ toy in box one and 1 block of $3$ toys in box three, so we get $3 \cdot 1 + 1 \cdot 3 = 3 + 1 + 1 + 1$.
So having bijection we can say that if $k<n$ we won't have big enough box to get the partition of $n$ containing itself. If $n=k$ we are good, because we can get every partition and every box setup has its own partition as we have shown before. If $k>n$ nothing changes, because we we can't put anything else than $0$ in boxes number $n+1,...,k$
Is that OK?

Your reasoning looks fine. Nice bijection.
