How many composition of $n$ are there in which each part is at least $c$ ?
I don't think my way of doing it is completely correct. Let $S=\bigcup_{k \geq 0} N_{\geq c}^k$ where $k=0$ is the empty composition of 0 . The generating function for a single part of the composition that is at least $c$ would be $x^c+x^{c+1}+x^{c+2}+\ldots$ This can be simplified using the formula for an infinite geometric series as $\frac{x^c}{1-x}$, since each term is $x$ times the previous term. To find the number of compositions of $n$, we don't limit the number of parts in the composition, so we consider any number of these parts. The compositions of $n$ are elements of $S$ of weight $n$, retaining our weight functions from before, and so the number is $$ \left[x^n\right] \Phi_S(x) =\left[x^n\right] \Phi_{\bigcup_{k \geq 0} N_{\geq c}^k}(x) $$
$$ =\left[x^n\right] \sum_{k \geq 0} \Phi_{N_{\geq c}^k}(x), \quad \text { by the Sum Lemma } $$
$$ =\left[x^n\right] \sum_{k \geq 0}\left(\Phi_{N_{\geq c}}(x)\right)^k, \quad \text { by the Product Lemma } $$
$$ =\left[x^n\right] \sum_{k \geq 0}\left(\frac{1-x}{1-x-x^c}\right)^k, \quad \text { by Geometric Series } $$
$$ =\left[x^n\right] \frac{1}{1-\frac{1-x}{1-x-x^c}}, \quad \text { by Geometric Series } $$
$$=\left[x^n\right] \frac{x^c-1+x}{x^c} $$
$$ =\left[x^n\right] 1+\frac{\left[x^n\right](-1+x)}{x^c} $$
$$ =\left[x^n\right] 1+\left[x^{n-c}\right](-1+x) $$ The term $\left[x^n\right] 1$ is straightforward; it signifies the coefficient of $x^n$ in the constant 1, which is 0 for all $n>0$ and 1 for $n=0$. For the term $\left[x^{n-c}\right](-1+$ $x$ ), there are two components to consider: -1 and $x$. The coefficient of $x^{n-c}$ in -1 is always 0 , since there's no $x$-term in -1 . For the term $x$, the coefficient of $x^{n-c}$ equals 1 if $n-c=1$, meaning $n=c+1$, and 0 for all other values of $n$. Combining these two parts, we get: $$ \left[x^n\right] 1+\left[x^{n-c}\right](-1+x)= \begin{cases}1, & n=0 \\ 1, & n=c+1 \\ 0, & \text { otherwise }\end{cases} $$
This implies there is one way to compose 0 (i.e., having no parts), and if $n=c+1$, there's also one way to compose $n$ from parts that are at least $c$ (namely, one part of size $c+1$ ). For all other values of $n$, no compositions are possible under these specific conditions.