6. Given that $m, n, k$ are positive integers, if there exists a pair of positive integers $(a, b)$ satisfying
$$
(1+a) n^{2}-4(m+a) n+4 m^{2}+4 a+b(k-1)^{2}<3 \text {, }
$$

then the number of possible values of $m+n+k$ is $\qquad$ .
$$
\begin{array}{l}
(1+a) n^{2}-4(m+a) n+4 m^{2}+4 a+b(k-1)^{2} \\
=n^{2}-4 m n+4 m^{2}+a\left(n^{2}-4 n+4\right)+b(k-1)^{2}=(n-2 m)^{2}+a(n-2)^{2}+b(k-1)^{2}<3,
\end{array}
$$

Since $m=1,2 ; n=1,2,3 ; k=1,2$, thus $(m, n, k)=(1,1,1),(1,2,1),(1,2,2),(1,3,1),(2,3,1)$. Therefore, $m+n+k=3,4,5,6$ for a total of 4 values.