Miner's Extreme Arrow Number

This number is so big it is 100% of the size of the Omniverse!

It sadly doesn't end in a number between 1 and 7420 steps.

Formula:

$$MEa_{1}=Ma_{1}=4$$

$$MEa_{2}=MSa_{2}=Ma_{Ma_{Ma_{Ma_2}}}$$

$$MEa_{3}=MHa_{3}=\overbrace{MSa_{MSa_{..._{MSa_{3}}}}}^{MSa_{3} \text{ MSa's}}$$

$$MEa_{4}=MUa_{4}=\overbrace{MHa_{MHa_{..._{MHa_{4}}}}}^{MHa_{4} \text{ MHa's}}$$

$$MEa_{5}=f5_{5}=\overbrace{MUa_{MUa_{..._{MUa_{5}}}}}^{MUa_{4} \text{ MUa's}}$$

$$MEa_{6}=f6_{6}=\overbrace{f5_{f5_{..._{f5_{6}}}}}^{f5_{4} \text{ f5's}}$$

$$\cdots$$

$$MEa_{7419}=f7419_{1419}=\overbrace{f7418_{f7418_{..._{f7418_{4}}}}}^{f7418_{4} \text{ f7417's}}$$

$$MEa_{7420}=f7420_{1420}=\overbrace{f7419_{f7419_{..._{f7419_{4}}}}}^{f7419_{4} \text{ f7419's}}$$

$$MEa_{7421}=f7421_{1421}=\overbrace{f7420_{f7420_{..._{f7420_{4}}}}}^{f7420_{4} \text{ f7420's}}$$

Conclusion: Still, it is really big. Maybe it is bigger than TREE(3)? Only Leftunknown knows. It is also MEa7421. This random number is the product of the first ten prime numbers divided by 1e12 (1 trillion).