6. Given positive numbers $a, b, c$ satisfying $a+b+c=13$. Then the minimum value of $\frac{a^{2}+b^{3}+c^{4}+2019}{10 b+123 c+26}$ is $\qquad$ .
6.4.
$$
\begin{array}{l}
\text { Given } \frac{a^{2}+b^{3}+c^{4}+2019}{10 b+123 c+26} \\
=\frac{a^{2}+4^{2}+b^{3}+4^{3}+4^{3}+c^{4}+5^{4}+5^{4}+5^{4}}{2 a+12 b+125 c} \\
\geqslant \frac{8 a+48 b+500 c}{2 a+12 b+125 c}=4,
\end{array}
$$

Equality holds if and only if \(a=b=4, c=5\). Therefore, the minimum value is 4.