A(k,4,r) is the independence number of the Johnson graph J(k,r).
What is the best known asymptotic lower bound on A(k,4,floor(k/2)) ?
I only obtained $\frac{{k\choose\lfloor{k/2}\rfloor}}{\lfloor{k^2/4}\rfloor}$.



lower bound on A(k,4,floor(k/2))
A better bound than that is known. Define a function $f:\{0,1\}^n\to \{0,1,\dots,n-1\}$ as follows:
$$ f(c\_0,c\_1,\dots,c\_{n-1})=\left(\sum\_{i=0}^{n-1} i\cdot c\_i\right)\bmod n.$$
Then define the following codes,
$$ C\_a=\{ c=(c\_0,c\_1,\dots,c\_{n-1})\in\{0,1\}^n ~|~ wt(c)=w, f(c\_0,c\_1,\dots,c\_{n-1})=a \},$$
where $wt(c)$ denotes the Hamming weight of $c$.


It is a simple exercise to see that each $C\_a$ is a constant-weight code with minimum Hamming distance 4, i.e., an independent set in $J(n,w)$. Since the $n$ codes form a partition of the vertices of $J(n,w)$, there is one of size at least $\frac{1}{n}\binom{n}{w}$.


In your case this translates to $A(k,4,\lfloor k/2\rfloor)\geq \frac{\binom{k}{\lfloor k/2\rfloor}}{k}$.


This taken from:
R. L. Graham and N. J. A. Sloane, "Lower bounds for constant weight codes," IEEE Transactions on Information Theory, vol. IT-26, pp. 37-43, 1980.


