<s>[INST] Partition with minimum size of parts In how many ways can we partition a number $N$? The answer is given by the partition function $p(N)$. So for instance, the partitions of the number $4$ are:
\begin{equation}
4; ~~~ 3+1;~~~ 2+2;~~~ 2+1+1;~~~1+1+1+1
\end{equation}
and hence $p(4) = 5$. 
How is this modified if I require each part to be at least of some size $k$? So for instance, if I required that each part must be at least of size $k=2$, the valid partitions of $4$ would be reduced to 
\begin{equation}
4; ~~~ 2+2. 
\end{equation}
How is this formalized? And is there a systematic way of enumerating all possibilities? 
 [/INST] Another  technique is based upon  generating functions.  We  encode   the  summands $j$ as powers  of $x$ and  the  coefficient of $x^j$ provides the  number of contributions of the  summand  $j$.
In the   example   with   $N=4$  the number  of partitions $p(4)$  is

\begin{align*}
p(4)&=[x^4](x^0+x^1+x^2+x^3+x^4)(x^0+x^2+x^4)(x^0+x^3)(x^0+x^4)\\
&=[x^4](1+x+x^2+x^3+x^4)(1+x^2+x^4)(1+x^3)(1+x^4) \tag{1}
\end{align*}
  Here we use the coefficient of operator $[x^j]$ to   denote  the  coefficient  of $x^j$ in a series. There are four  factors in (1)  indicating  that the numbers  $1,2,3$  and $4$ are the building blocks of a partition of $4$. The  first factor in (1) tells  us    we  can use the number $1$ zero or once or twice or up to four times, whereas e.g. the number $3$ can only be used zero or one time.

Now we want to analyse restricted partitions having smallest summand $k=2$. This means we are looking for

\begin{align*}
[x^4]&(1+x^2+x^4)(1+x^3)(1+x^4)\tag{2}\\
&\qquad=[x^4](1+x^2+x^4)(1+x^3+x^4)\tag{3}\\
&\qquad=[x^4](1+x^2+x^3+2x^4)\tag{4}\\
&\qquad=2
\end{align*}
  and we are finished.

Comment:


*

*In   (2)   we  take   contributions  only from   $k=2,3$    and      $4$.

*In    (3)   we        multiply   the   two    right-most   factors     out and skip thereby all summands with powers greater then $4$ since they do nothing contribute to $[x^4]$.

*In (4) we multiply out and again skip summands with powers greater than four. We observe, the coefficient of $x^4$ is $2$ and we are done.
</s>