asymptotic for the number of involutions in GL(n,2) Is it known how the number of involutions in $GL_n(2)$, the group of $n\times n$ matrices over $\mathbb{Z}/2\mathbb{Z}$, behaves as $n\to\infty$ ?
Equivalently, one may ask this for the number of $n\times n$ matrices $A$ over $\mathbb{Z}/2\mathbb{Z}$ satisfying $A^2=0$, as $(A+I)^2=I \mod 2$.
Needless to say, there are $\lfloor n/2\rfloor$ conjugacy classes of involutions in $GL_n(2)$, and one can write down formula for the centraliser order for each class, 
$$\frac{\left|GL_n(2)\right|}{2^{k^2+2k(n-2k)}\left|GL_k(2)\right|\left|GL_{n-2k}(2)\right|},\quad 1\leq k\leq \left\lfloor \frac{n}{2} \right\rfloor,$$
but it's quite a mess to just sum them up. 

We may write $|{\rm GL}(n,2)| = 2^{n^{2}} \prod_{j=1}^{n}( 1- \frac{1}{2^{j}}).$
As $n \to \infty$, the rightmost factor tends to $\left( \sum_{r=0
}^{\infty} \frac{p(r)}{2^{r}} \right)^{-1}$, where $p(r)$ is the number of partitions of $r$. Let us write $|{\rm GL}(n,2)| = 2^{n^{2}}f(n)$.
Then we see that the number of involutions of ${\rm GL}(n,2)$ is given by 
$\sum_{k = 1}^{\lfloor \frac{n}{2} \rfloor} 2^{2k(n-k)} \frac{f(n)}{f(k)f(n-2k)}$ using the formula given in the question ( if $k = \frac{n}{2}$, we should interpret $|{\rm GL}(n-2k,2)|$ as $1$).
Hence the number of involutions in ${\rm GL}(n,2)$ may be expressed as 
$2^{\frac{n^{2}}{2}} \left(\sum_{k=1}^{\lfloor \frac{n}{2} \rfloor} 2^{- 2\left(\frac{n}{2}-k\right)^{2}}\left( \frac{\prod_{j=k+1}^{n}( 1- \frac{1}{2^{j}})}{ \prod_{m=1}^{n-2k}( 1- \frac{1}{2^{m}})}\right)\right)$ if $n$ is odd,
and $2^{\frac{n^{2}}{2}} \left( \prod_{j=\frac{n}{2}+1}^{n}( 1- \frac{1}{2^{j}})+ \sum_{k=1}^{\lfloor \frac{n}{2} \rfloor-1} 2^{- 2\left(\frac{n}{2}-k\right)^{2}}\left( \frac{\prod_{j=k+1}^{n}( 1- \frac{1}{2^{j}})}{ \prod_{m=1}^{n-2k}( 1- \frac{1}{2^{m}})}\right)\right)$ if $n$ is even.
Hence when $n$ is even, the number of involutions is at most
$2^{\frac{n^{2}}{2}} 
\prod_{j=\frac{n}{2}+1}^{n}( 1- \frac{1}{2^{j}}) 
\left( 1 +  \frac{1}{\prod_{m=1}^{n-2}( 1- \frac{1}{2^{m}})} 
\left( \sum_{k=1}^{\frac{n}{2}} 2^{- 2(\frac{n}{2}-k)^{2}}\right)\right)$ and is at least $2^{\frac{n^{2}}{2}} \prod_{j=\frac{n}{2}+1}^{n}( 1- \frac{1}{2^{j}})$.
As (even) $n \to \infty$, the first product appearing approaches $1$ from below
and (the reciprocal of) the second product appearing tends to $\sum_{r=0}^{\infty} \frac{p(r)}{2^{r}}$, while the inner sum is at most $\sum_{j=0}^{\infty} \left(\frac{1}{4} \right)^{j^{2}}$. 
I omit the similar analysis when $n$ is odd.
