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Miner's Arrow Number is so big that it is (7000↑↑↑↑↑↑↑↑↑↑78)*0% of the size of the Omniverse and bigger than the "OH NO" number!

Sadly, it doesn't end in 120 steps.

This is the formula:

${\displaystyle Ma_0 = 2}$

${\displaystyle Ma_1 = Ma_0 \uparrow \uparrow Ma_0}$

${\displaystyle Ma_2 = Ma_1 \overbrace{\uparrow ... \uparrow}^{ Ma_1 \uparrow} Ma_1}$

${\displaystyle Ma_3 = Ma_2 \overbrace{\uparrow ... \uparrow}^{ Ma_2 \uparrow} Ma_2}$

${\displaystyle Ma_{118} = Ma_{117} \overbrace{\uparrow ... \uparrow}^{ Ma_{117} \uparrow}Ma_{117}}$

${\displaystyle Ma_{119} = Ma_{118} \overbrace{\uparrow ... \uparrow}^{ Ma_{118} \uparrow} Ma_{118}}$

${\displaystyle Ma_{120} = Ma_{119} \overbrace{\uparrow ... \uparrow}^{ Ma_{119} \uparrow} Ma_{119}}$

${\displaystyle Ma_{121} = Ma_{120} \overbrace{\uparrow ... \uparrow}^{ Ma_{120} \uparrow} Ma_{120}}$

${\displaystyle Ma_n = Ma_{n-1} \overbrace{\uparrow ... \uparrow}^{ Ma_{n-1} \uparrow}Ma_{n-1}}$

Conclusion: It is extremely big, and I know and have proof it is bigger than Graham's number. And it's value is ${\displaystyle Ma_{121}}$.

## Proof Ma121 is bigger than Graham's Number

In the second step of Mas, it is 4↑↑↑↑4, which is bigger than g1 (first step of Graham's Number), but as the number of steps is bigger than G, and the up arrow numbers change (G's numbers never change, as it is three). That means it is bigger than G. Is it bigger than TREE(3)? I actually don't know. (No its not because TREE(3) is a legiattic array)