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. 
 A: As a matter of fact, OEIS has the corresponding integer sequence A053722, although shifted by 1, as it corresponds to the number of solutions to $X^2=1$ in $GL_n(2)$, i.e. it counts the identity along with the
involutions. I thought OEIS does this kind of elementary transformations of sequences automatically, but in fact it does not.
Incidentally, this (shifted) number can be obtained as the sum of dimensions of real representations of $GL_n(2)$, see e.g. paper by Fulman and Vinroot.
This still does not answer the original question, though.
A: 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.
