Binary Linear Codes of Minimum Distance 3 Let $B_n$ denote the maximum size of a binary linear code (a binary code that is closed under addition) whose codewords have length $n$ and whose minimum distance is $3$. I have been searching for the best-known lower bounds for $A(n,3)$, but I can't seem to find anything. My question is simply: What are some lower bounds for the numbers $B_n$? 
 A: Because you specify the minimum distance to $d=3$ and constrain yourself to binary linear codes, the question can be answered explicitly.
Let the code $C$ be given by a check matrix
$$
H=(H_1\ |\ H_2\ |\ H_3\ |\ \cdots\ H_n),
$$
where the columns $H_i, 1\le i\le n$, are binary vectors of with $m$ components. For the code to have minimum distance $\ge3$ it is necessary and sufficient that all the columns $H_i$ are non-zero and distinct, i.e. $H_i\neq H_j$ whenever $i\neq j$. This is easy to understand. If $H_i=0$ for some $i$, then the word of weight one with its unique non-zero component will be a codeword. Similarly the existence of a word of weight two in $C$ is equivalent to finding two linearly dependent columns from $H$ (the non-zero components are located at the positions of the dependent columns). Over the binary field a set of two vectors is linearly dependent, iff the two vectors are equal.
Because the columns have exactly $m$ bits, the maximum number of non-zero distinct column vectors is $2^m-1$. This leads to the following answer:

Given the code length $n$, find the unique positive integer $m$ such that $$2^{m-1}\le n\le 2^m-1.$$ The maximum dimension of a binary linear code of length $n$ and $d_{min}=3$ is then $k=n-m.$ The above recipe allows us to easily also construct a check matrix $H$ with required properties, and hence also a code. The number of words in such a linear code is 
  $$B_n=2^k=2^{n-\lfloor \log_2(n+1)\rfloor}.$$

A: 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).
$$
\begin{array}{cl} 
n & A(n,d) \\
5 & 4\\ 6& 8 \\ 7 & 16\\ 8 & 20\\ 9 & 40 \\ 10 & 72-79 \\ \cdots & \cdots \\ 16 & 2560-3276 \end{array}$$
A: Let $F$ be the finite field of size $2$ and let $K = F^n$. Of those $K$ subcodes of distance $3$ fix a $K$ subcode $C$ of maximum dimension. The unit $K$ codewords together generate $K$ while the non-trivial quotient code $K/C$ has a Hamel basis $Z$ made of unit $K/C$ codewords; i.e. made of $C$ cosets that contain a unit $K$ codeword.  
$Z$ has at most size $2$ or else $K/C$ would have a codeword of weight $3$ which is false. $C$ is not a hyperplane or else for distinct unit $K$ codewords $u,v$ we'd have $u+v$ in $C$ implying there exists a $C$ codeword of weight $2$ which is false.
Therefore, $Z = n - dim(C) = 2$ and $dim(C) = n - 2$ and $B_n = 2^{n-2}$.
