Let $f:\mathbb N^0 \times \mathbb N \rightarrow \mathbb N^{0}$ be a piece-wise function defined recursively as follows $$f(x,y) = \begin{cases} 0 & (\text{if } x< y \text{ and (if } y \not =1)) \\ 1 & (\text{$y$ = 1}) \text{ or } y = x \\ \sum_{i=1}^yf(x-y,i) & x> y \end{cases}$$
Is there a way to explicitly define this function? I'm having trouble and for some reason I believe it may be in the form of series like this $$f(x,y) = \sum_{i=2}^y \lfloor \frac{x}{i} \rfloor$$ But i got this only by shear intuition and am not sure how i got it but i have a funny feeling it'll be similar to this.
For $x>0$, $f(x,y)$ is the number of partitions of $x$ into $y$ parts, or equivalently into parts with maximum equal to $y$ (equivalence is due to taking the conjugate partition). This can be proved by showing these satisfy the same recurrence. For our "base" cases we have:
$x < y \implies f(x,y) = 0$: There are no ways to partition $x<y$ with some term of magnitude $y$.
$y \in \{x, 1\} \implies f(x,y) = 1$: There is exactly one way to partition $x$ into one part, namely $x$. Likewise, there is only one way to partition $x$ into $x$ terms, $1+1+\cdots +1$.
$x > y \implies f(x,y) = \sum_i f(x-y,i)$: Finally, the partitions of $x$ with maximum $y$ is the number of partitions of $x-y$ with maximum less than or equal to $y$(as we know all such partitions start $y + $ some partition of $x-y$ using smaller terms). But this is just the sum over second biggest values for partitions of $x-y$, so: $\sum_i^y f(x-y,i)$
A quick reading suggests we can compute a closed form for any fixed $y$, but the general closed form appears quite messy, especially since the generating function for the total number of partitions is messy. See here for why we don't expect a nice closed form, as these give partitions of particular numbers. Likewise, see here for a summary of what we do know about the partition function.