A note on the size of N-free families
  The $\mathcal{N}$ poset consists of four distinct sets $W,X,Y,Z$ such that $W\subset X$, $Y\subset X$, and $Y\subset Z$ where $W$ is not necessarily a subset of $Z$. A family $\mathcal{F}$ as a subposet of the $n$-dimensional Boolean lattice, $\mathcal{B}_n$, is $\mathcal{N}$-free if it does not contain $\mathcal{N}$ as a subposet. Let $\text{La}(n, \mathcal{N})$ be the size of a largest $\mathcal{N}$-free family in $\mathcal{B}_n$. Katona and Tarj\'{a}n proved that $\text{La}(n,\mathcal{N})\geq {n \choose k}+A(n,4,k+1)$, where $k=\lfloor n/2\rfloor$ and $A(n, 4, k+1)$ is the size of a single-error-correcting code with constant weight $k+1$. In this note, we prove for $n$ even and $k=n/2$, $\text{La}(n, \mathcal{N}) \geq {n\choose k}+A(n, 4, k)$, which improves the bound on $\text{La}(n, \mathcal{N})$ in the second order term for some values of $n$ and should be an improvement for an infinite family of values of $n$, depending on the behavior of the function $A(n,4,\cdot)$. 
Ryan R. Martin and Shanise Walker
math.CO