We have the functions $f_1 , \dots , f_{30}$ :

$$\frac{n}{\lg n} , \ \ n^{\lg n} ,\ \ (\sqrt{2})^{\lg n}, \ \ n^2, \ \ n!, \ \ (\lg n)! ,\ \ \left( \frac{3}{2} \right)^n ,\ \ n^3 ,\ \ \lg^2 n ,\ \ \lg(n!) ,\ \ 2^{2^n}, \ \ n^{\frac{1}{\lg n}}, \ \ \ln{\ln n}, \ \ e^{\log_{10} n }, \ \ n \cdot 2^n, \ \ n^{\lg{\lg n}}, \ \ \ln n, \ \ 1 ,\ \ 2^{\lg n}, \ \ (\lg n)^{\lg n}, \ \ e^n ,\ \ 4^{\lg n}, \ \ (n+1)!, \ \ \sqrt{\lg n}, \ \ \lg{\lg{\lg n}}, \ \ 2^{\sqrt{2 \lg n}}, \ \ n, \ \ 2^n ,\ \ n \lg n, \ \ 2^{2^{n+1}}$$

We have the functions $f_1 , \dots , f_{30}$ :

$$\frac{n}{\lg n} , \ \ n^{\lg n} ,\ \ (\sqrt{2})^{\lg n}, \ \ n^2, \ \ n!, \ \ (\lg n)! ,\ \ \left( \frac{3}{2} \right)^n ,\ \ n^3 ,\ \ \lg^2 n ,\ \ \lg(n!) ,\ \ 2^{2^n}, \ \ n^{\frac{1}{\lg n}}, \ \ \ln{\ln n}, \ \ e^{\log_{10} n }, \ \ n \cdot 2^n, \ \ n^{\lg{\lg n}}, \ \ \ln n, \ \ 1 ,\ \ 2^{\lg n}, \ \ (\lg n)^{\lg n}, \ \ e^n ,\ \ 4^{\lg n}, \ \ (n+1)!, \ \ \sqrt{\lg n}, \ \ \lg{\lg{\lg n}}, \ \ 2^{\sqrt{2 \lg n}}, \ \ n, \ \ 2^n ,\ \ n \lg n, \ \ 2^{2^{n+1}}$$

@DanielFischer I have tried to prove that for the initial pseudocode that implements the Horner's method the following relation expresses an invariant for the while-loop at the lines 3-5.
At the beginning of each iteration of the while-loop at the lines 3-5
$$k=\sum_{d=0}^{m-(j+1)} a_{d+j+1} x^{d}$$
That's what I have tried:
Initial check:
Before the first iteration of the loop, we have $j=m$ and $k=0$.
The sum $\sum_{d=0}^{m-(m+1)} a_{d+m+1} x^{d}=\sum_{d=0}^{-1} a_{d+m+1} x^{d}$ is zero, so it is equal to $y$, therefore the invariant holds before the first iteration of the loop.