The Miner's Loading Number is very big, so big it is exactly
cos
TREE
(
3
)
π
2
%
{\displaystyle \cos\frac{\operatorname{TREE}\left(3\right)\pi}{2}\%}
the Box (Leftunknown said he didn't want the Omniverse appearing 😢 ). It uses the Loader Function (Dy (x)).
The formula:
Ml
1
=
D
D
D
⋯
D
1
(
1
)
(
1
)
(
1
)
(
1
)
(
1
)
⏟
what is D(1)
{\displaystyle \operatorname{Ml}_1 = \underbrace{D^{D^{D^{\cdots^{D^{1}(1)}(1)}(1)}(1)}(1)}_{\text{what is D(1)}}}
Ml
2
=
D
D
D
⋯
D
2
(
2
)
(
2
)
(
2
)
(
2
)
(
2
)
⏟
what is D(2)
{\displaystyle \operatorname{Ml}_2 = \underbrace{D^{D^{D^{\cdots^{D^{2}(2)}(2)}(2)}(2)}(2)}_{\text{what is D(2)}}}
Ml
3
=
D
D
D
⋯
D
3
(
3
)
(
3
)
(
3
)
(
3
)
(
3
)
⏟
what is D(3)
{\displaystyle \operatorname{Ml}_3 = \underbrace{D^{D^{D^{\cdots^{D^{3}(3)}(3)}(3)}(3)}(3)}_{\text{what is D(3)}}}
⋮
{\displaystyle \vdots}
Ml
132
=
D
D
D
⋯
D
132
(
132
)
(
132
)
(
132
)
(
132
)
(
132
)
⏟
D(132)
{\displaystyle \operatorname{Ml}_{132} = \underbrace{D^{D^{D^{\cdots^{D^{132}(132)}(132)}(132)}(132)}(132)}_{\text{D(132)}}}
Ml
x
=
D
D
D
⋯
D
x
(
x
)
(
x
)
(
x
)
(
x
)
(
x
)
⏟
D
(
x
)
times
{\displaystyle \operatorname{Ml}_x = \underbrace{D^{D^{D^{\cdots^{D^{x}(x)}(x)}(x)}(x)}(x)}_{D(x) \text{times}}}
Now, let's nest it!
Ml
x
1
=
M
l
x
⏟
M
l
x
{\displaystyle \operatorname{Ml}_x^1 = \underbrace{Ml_x}_{Ml_x}}
Ml
x
2
=
M
l
M
l
x
⏟
M
l
x
{\displaystyle \operatorname{Ml}_x^2 = \underbrace{Ml_{Ml_x}}_{Ml_x}}
Ml
x
3
=
M
l
M
l
M
l
x
⏟
M
l
x
{\displaystyle \operatorname{Ml}_x^3 = \underbrace{Ml_{Ml_{Ml_x}}}_{Ml_x}}
⋮
{\displaystyle \vdots}
The final number is *drumroll sounds*
Ml
2147483648
2147483648
{\displaystyle \operatorname{Ml}_{2147483648}^{2147483648}}
I don't know if it's bigger than Rayo's number (I think it's smaller), but I am g999 % sure it is a whole lot bigger than Loader's number.
