In one operation the product of all written numbers increases in $3^5$ times. At the beginning, this product equals $3^{0+1+2+...+19} = 3^{190}$. So, after using it $n$ times, the product will be equal to $3^{190+5n}$. By that time all numbers would have to become equal, and therefore at least $3^{19}$, so in the end the product is at least $3^{20p}$, where $p \ge 19$. Then $3^{20p} = 3^{190+5n}$, $5n + 190 = 20p$ and $n + 38 = 4p$, and as $p \ge 19$, we get $n \ge 38$. Now let's show how to achieve this by using the program 38 times.

$$\begin{aligned} 3^0, 3^1, 3^2, 3^3, 3^4 \quad (15) &\rightarrow 3^{15}, 3^{16}, 3^{17}, 3^{18}, 3^{19}; \\ 3^5, 3^6, 3^7, 3^8, 3^9 \quad (10) &\rightarrow 3^{15}, 3^{16}, 3^{17}, 3^{18}, 3^{19}; \\ 3^{10}, 3^{11}, 3^{12}, 3^{13}, 3^{14} \quad (5) &\rightarrow 3^{15}, 3^{16}, 3^{17}, 3^{18}, 3^{19}. \end{aligned}$$
So after 30 uses we have 4 of each of $3^{15}, 3^{16}, 3^{17}, 3^{18}, 3^{19}$.

$$\begin{aligned} 3^{15}, 3^{16}, 3^{17}, 3^{18} \quad (3) &\rightarrow 3^{18}, 3^{19}, 3^{19}, 3^{19}; \\ 3^{16}, 3^{17}, 3^{18}, 3^{19} \quad (2) &\rightarrow 3^{18}, 3^{19}, 3^{19}, 3^{19}; \\ 3^{17}, 3^{18}, 3^{19}, 3^{20} \quad (1) &\rightarrow 3^{18}, 3^{19}, 3^{20}, 3^{20}. \end{aligned}$$
So after 36 uses we have 10 of $3^{18}, 3^{19}$. In the last two moves we make them all equal to $3^{19}$.