• **Entropy related to the** *ABS* **index of subdivision** *H***3***BO***<sup>3</sup>**

Let *S*(*H*3*BO*3) be a subdivision of *H*3*BO*3(*s*, *t*). Then by using Equation (2) and Table 1, the atom–bond sum connectivity is

$$\begin{aligned} \langle ABSS(H\_3BO\_3) \rangle &= \sum\_{\tilde{\xi}\_{(1 \sim 2)}} x^{\sqrt{\frac{1 + 2 - 2}{1 + 2}}} + \sum\_{\tilde{\xi}\_{(2 \sim 2)}} x^{\sqrt{\frac{2 + 2 - 2}{2 + 2}}} + \sum\_{\tilde{\xi}\_{(2 \sim 3)}} x^{\sqrt{\frac{2 + 3 - 2}{2 + 3}}} \\ &= 2(s + t + 1)x^{\sqrt{\frac{1}{3}}} + 12(st + s + t)x^{\sqrt{\frac{1}{2}}} \\ &+ 6(3s + 3t + 4st - 1)x^{\sqrt{\frac{3}{5}}} \end{aligned}$$

Taking the first derivative of Equation (14) at *x* = 1, we get the atom–bond sum connectivity index

$$\Delta ABC(S(H\_3BO\_5)) = 2(s+t+1)\sqrt{\frac{1}{3}} + 12(st+s+t)\sqrt{\frac{1}{2}} + 6(3s+3t+4st-1)\sqrt{\frac{3}{5}}.\tag{15}$$

Here, we determine the atom–bond sum connectivity entropy by using Table 1 and Equation (15) in Equation (6) according to the following:

$$\begin{split} \text{ENT}\_{ABS}(S(H\_5BO\_3)) &= \log\left(ABS) - \frac{1}{ABS} \log\left\{ \prod\_{\xi \in \mathbb{Z}\_+} \sqrt{\frac{(V\_{d\_i} + V\_{d\_j} - 2)}{(V\_{d\_i} + V\_{d\_j})}} \right\} \right\} \\ &\times \prod\_{\xi \in \mathbb{Z}\_+} \left[ \sqrt{\frac{(V\_{d\_i} + V\_{d\_j} - 2)}{(V\_{d\_i} + V\_{d\_j})}} \right] \frac{\sqrt{\frac{(V\_{d\_i} + V\_{d\_j} - 2)}{(V\_{d\_i} + V\_{d\_j})}}}{\sqrt{\frac{(V\_{d\_i} + V\_{d\_j} - 2)}{(V\_{d\_i} + V\_{d\_j})}}}} \times \prod\_{\xi \in \mathbb{Z}\_3} \left[ \sqrt{\frac{(V\_{d\_i} + V\_{d\_j} - 2)}{(V\_{d\_i} + V\_{d\_j})}} \right] \\ &= \log\left(ABS) - \frac{1}{ABS} \log\left\{ 2(s + t + 1) (\sqrt{\frac{1}{3}})^{\sqrt{\frac{3}{5}}} \times 12(st + s + t) (\sqrt{\frac{1}{2}})^{\sqrt{\frac{3}{5}}} \right\} \\ &\times \ 6(3s + 3t + 4st - 1) (\sqrt{\frac{3}{5}}) \sqrt{\frac{2}{3}} \end{split} \tag{16}$$
