Two binary codewords of weight $4$ at distance $6$ from each other must
have a common $1$ in exactly one position, right? So for starters, here
are three such codewords
$$1111000000\\1000111000\\1000000111$$
Can you find two more? You cannot. So let us back up a little bit
and think some more since three codewords sharing a common $1$ in the
same position does not work. What if we chose the five putative codewords
so that for any given codeword, the other four had a common $1$ in
four different positions? So, now we have
$$1111000000\\1000111000$$
as before, but now codeword #3 shares a $1$ in position $2$ with codeword #1 and a $1$ in position $5$ with codeword #2, so that the first three codewords are
$$1111000000\\1000111000\\0100100110$$
The fourth codeword shares a $1$ in positions $3,6,8$ with codewords #1, #2, #3
respectively giving
$$1111000000\\1000111000\\0100100110\\0010010101$$
Can you now figure out the fifth codeword? Which positions have not been used
thus far for sharing?
Note that there are $\binom{5}{2} = 10$ pairs of codewords and each pair
shares a $1$ in a different position.
Thanks to Dilip in the comments, I was able to come up with a solution. I figured I would post a solution for anyone else interested in the problem.
As mentioned above, consider the first-order binary Reed-Muller code, $\mathcal{R}(1, 4)$. This is a $[2^4, 4+1, 2^{4-1}] = [16, 5, 8]$ code. Then, shorten this code by deleting the over-all parity check bit and taking only the remaining codewords of even weight. By doing this, we we are then left with a $[15, 4, 8]$ (which can be verified by looking at the generator matrix for $\mathcal{R}(1, 4)$).
Now, further shorten this code by deleting a nonzero coordinate. We will be left with a $[14, 4, 7]$ code. By shortening the $[14, 4, 7]$ code by deleting another nonzero coordinate, we will get a $[13, 4, 6]$ code.
Best Answer
The binary Hamming code with parameters $[n,k,d]=[2^m-1,2^m-m-1,3]$ is perfect (no larger code exists) and for the values of $n=2^m-1,$ $m\geq 3$ provides the lower bound $$A_{2^m-1}\geq 2^{(2^m-m-1)}$$.
So $A_7\geq 16,$ etc.
The Gilbert-Varshamov bound states that for alphabet size $q$ a prime power, there exists a linear $[n,k]$ code with minimum distance at least $d$ provided $$q^{n-k}\geq \sum_{i=0}^{d-2} {n-1 \choose i} (q-1)^i.$$ So you can take $q=2$ and find the largest $k$ satisfying this inequality.
There is a table in Roman's Coding Theory book, which has the following information (but for all, not necessarily linear codes). $$ \begin{array}{cl} n & A(n,d) \\ 5 & 4\\ 6& 8 \\ 7 & 16\\ 8 & 20\\ 9 & 40 \\ 10 & 72-79 \\ \cdots & \cdots \\ 16 & 2560-3276 \end{array}$$