How many compositions of $n$ are there where each part is at least $m$?
My attempt:
So far I have found the generating function and have that the desired number of compositions is $$ [x^{n}]\frac{1-x}{1-x-x^{m}}. $$
But I am unsure how to proceed, I have made a few attempts to transform this (for example using the geometric series or through rewriting), perhaps there is a simple way to do this and I am overlooking it. Does anybody have any suggestions. Thanks in advance.
Edit: I have summarized how I've obtained the previous expression, kindly mention if you notice any mistakes. Note that $\Phi_{N\geq m} = \left(\sum_{i \geq m} x^{i}\right) = \left(\sum_{i \geq 0} x^{i+m}\right)$
$$\Phi_{\cup_{k\geq 0}N_{\geq m}^{k}}(x) = \sum_{k \geq 0}(\Phi_{N\geq m}(x))^{k} = \sum_{k\geq 0}\left(\sum_{i\geq 0} x^{i+m}\right)^{k} = \sum_{k\geq 0}\left(x^{m}\sum_{i\geq 0} x^{i}\right)^{k}$$ $$ = \sum_{k\geq 0}(x^{m}(1-x)^{-1})^{k} = \frac{1}{1-x^{m}(1-x)^{-1}} = \frac{1-x}{1-x-x^{m}} $$
We obtain for $m,n\geq 2$ using the geometric series expansion \begin{align*} \color{blue}{[x^n]}&\color{blue}{\frac{1-x}{1-x-x^m}}=[x^n]\frac{1-x}{1-x\left(1+x^{m-1}\right)}\\ &=[x^n](1-x)\sum_{q=0}^{\infty}x^q\left(1+x^{m-1}\right)^q\\ &=\sum_{q=0}^n[x^{n-q}]\left(1+x^{m-1}\right)^q(1-x)\tag{1}\\ &=\sum_{q=0}^n[x^{n-q}]\sum_{j=0}^q\binom{q}{j}x^{(m-1)j}(1-x)\tag{2}\\ &=\sum_{q=0}^n[x^{q}]\sum_{j=0}^{n-q}\binom{n-q}{j}x^{(m-1)j}(1-x)\tag{3}\\ &=\sum_{q=0}^{\left\lfloor\frac{n}{m-1}\right\rfloor} [x^{(m-1)q}]\sum_{j=0}^{n-(m-1)q}\binom{n-(m-1)q}{j}x^{(m-1)j}(1-x)\tag{4}\\ &\,\,\color{blue}{=\sum_{q=0}^{\left\lfloor\frac{n}{m-1}\right\rfloor}\binom{n-(m-1)q}{q} -\sum_{q=0}^{\left\lfloor\frac{n-1}{m-1}\right\rfloor}\binom{n-1-(m-1)q}{q}}\tag{5} \end{align*}
Comment:
In (1) we apply $[x^{p-q}]A(x)=[x^p]x^qA(x)$. We also set the upper limit of the series to $n$ since other terms do not contribute to $[x^n]$.
In (2) we make a binomial expansion.
In (3) we change the order of summation $q\to n-q$.
In (4) we note the index $j$ needs to be a multiple of $m-1$ for coefficient selection.
In (5) we select the coefficient of $x^{(m-1)q}$. We take the factor $(1-x)$ into account by subtracting the sum from the same sum, but replacing $n$ with $n-1$.