&0  &0  &0  &a_5&b_1\\
0  &0  &0  &0  &b_1&a_6
\end{array}
\right),  \label{rhoT1}$$ which is still of the block-diagonal form, and computation of its eigenvalues is straightforward. There are only two eigenvalues which are possibly negative. The negativity is thus given by $$\begin{aligned}
{\cal N}(\rho_{12})=&\frac{1}2\max \big[0,\sqrt{(a_1-a_2)^2+4b_2^2}-a_1-a_2)\big]\nonumber\\
+&\frac{1}2\max\big[0,\sqrt{(a_5-a_6)^2+4b_1^2}-a_5-a_6\big].\label{N}\end{aligned}$$

Substituting Eqs. (\[para1\])–(\[para2\]) leads to the analytical result of negativity $$\begin{aligned}
{\cal N}(\rho)=&\max
\big[0,\sqrt{(a_1-a_2)^2+4b_2^2}-a_1-a_2)\big]\nonumber\\
=&2\max[0,a_2-2a_1]\nonumber\\
=&\frac{1}{3}\max\Big[0,\frac{e^\beta-4e^{-\frac{1}{2}\beta}}
{e^\beta+2e^{-\frac{1}{2}\beta}}\Big],\label{N1}\end{aligned}$$ where the second equality follows from Eq. (\[para2\]).

We can see that the negativity is a function of the single parameter $\beta$. In the limit of $T\rightarrow 0$, the negativity becomes 1/3 and the ground state is entangled. From Eq. (\[N1\]), it is direct to check that the negativity is a monotonically decreasing function when temperature increases. After a certain threshold value of the temperature, the entanglement disappears. This threshold value $T_\text{th}$ can be obtained as $$T_\text{th}=3/(4\ln2)\approx 1.0820.$$ For a ring of spin-half particles interacting via the Hisenberg Hamiltonian, it was shown that the pairwise thermal entanglement is determined by the internal energy [@WangPaolo]. It is natural to ask if similar relations exist in the present mixed-spin system. The internal energy can be obtained from the partition function as $$U=-\frac{1}{Z}\frac{\partial Z}{\partial\beta}.$$ Substituting Eq. (\[para1\]) into the above 