a[b]c = a ↑ᵇ c
Example: 3[3,0]3 = 3↑↑↑3
a[b,1]c = a followed by a[b]c up arrows to c
a[b,2]c = a followed by a[b,1]c up arrows to c
a[b,3]c = a followed by a[b,2]c up arrows to c
a[b,n]c = a followed by a[b,n - 1]c up arrows to c
Graham's Number = 3[4,63]3
a[b,n,1]c = a[b,x]c, where x is a[b,n]c
a[b,n,2]c = a[b,x]c, where x is a[b,n,1]c
a[b,n,y]c = a[b,x]c, where x is a[b,n,y - 1]c
a[b,n,y,d]c = a[b,n,x]c, where x is a[b,n,y,d - 1]c
a[b,n,y,d,e] = a[b,n,d,x]c, where x is a[b,n,y,d,e - 1]c
a[[b]]c = a[b,b,b,b,b,...b,b]c, with b numbers of b's
a[[[b]]]c = a[[b,b,b,b,b,..b,b]]c with b numbers of b's
a[b/0]c = a[[[[...[[b]]]...]]c with b numbers of brackets
a[b/1]c = a[[[...[b,b,b,b,b,...b]]...]]c with b number of brackets & b's
a[b/2]c = a[[[...[[b,b,b,b,b,...b]]]...]]c with a[b/1]c number of brackets & b's
a[b/n]c = a[[[...[[b,b,b,b,b,...b]]]...]]c with a[b/n - 1]c number of brackets & b's
a[b,1/0]c = a[[[...[[b/b]]]...]]c with b number of brackets & b's
a[b,1/1]c = a[[[...[[b/b]]]...]]c with a[b,1/0]c number of brackets & b's
a[b,2/0]c = a[x/x]c where x is a[b,1/0]c
a[b,n/0]c = a[x/x]c where x is a[b,n - 1/0]
a[b,n,1/0]c = a[x,x - 1/x] where x is a[b,n/y]c
a[b\0]c = a[b,b,b,b,...b/b]c with b number of b's
a[b\1]c = a[b,b,b,b,...b\0]c with b number of b's
a[b\n]c = a[b,b,b,b,...b\n - 1]c with b number of b's
a[b,1\n]c = a[x\n] where x is a[b\n]c
a[b,2\n]c = a[x\n] where x is a[b,1\n]c
a[b,y\n]c = a[x\n] where x is a[b,y - 1\n]c
a[b,y,1\n]c = a[x,x\n] where x is a[b,y - 1\n]c
a[b,y,d\n]c = a[x,x\n] where x is a[b,y,d - 1\n]c
a[b$0]c = a[b,b,b,b,...b\b]c with b numbers of b's
a[b$1]c = a[x$0]c where x is a[b$0]c
a[b$n]c = a[x$n - 1]c where x is a[b$n - 1]c
a[b,1$0]c = a[b$x]c where x is a[b$b]c
a[b,n$0]c = a[b$x]c where x is a[b,n - 1$b]c
a[b,n,1$0] = a[b,x$0]c where x is a[b,n$0]c
a[b@0]c = a[b,b,b,b,...b$0] with b number of b's
Now, let's define another function.
f(1) = a[b]c
f(2) = a[[b]]c
f(3) = a[b/1]c
f(4) = a[b\0]c
f(5) = a[b$0]c
f(6) = a[b@0]c
f(7) = a[b,b,b,b,...b@0]c with b number of b's
a[b`0]c = f(b)
a[b~0]c = f(x) where x is f(b)
a[b;0]c = f(x) where x is a[b~0]c
Now, let's define yet another function:
f(1) = a[b]c
f(2) = a[b`0]c
f(3) = a[b~0]c
f(4) = a[b;0]c
f(5) = f(x) where x is a[b;0]c
a[b:0]c = f(b)
a[b?0]c = f(x) where x is f(b)
Let's Define Another Function:
f(1) = a[b]c
f(2) = a[b`0]c
f(3) = a[b:0]c
a[b"0]c = f(b)
a[b|0]c = f(x) where x is f(b)
a[b(0)]c = f(x) where x is a[b|0]c
Let's Define Another Function:
f(1) = a[b]c
f(2) = a[b"0]c
f(3) = a[b|0]c
f(4) = a[b(0)]c
f(5) = f(x) in the previous function, where x is a[b(0)]c
a[b←0]c = f(b)
a[b%0]c = f(x) where x is a[b←0]c
If we keep defining functions like this, and then combine them all into one enormous function, we can create much more enormous numbers.
f(1) = a[b]c
f(2) = a[b`0]c
f(3) = a[b:0]c
f(4) = a[b"0]c
f(5) = a[b←0]c
a[b₁]c = f(b)
a[b₂]c = f(f(f(f(f(...f(f(b)))...)) with b number of f's
a[b₃]c = f(f(f(f(f(...f(f(b)))...)) with a[b₂]c number of f's
a[bₙ]c = f(f(f(f(f(...f(f(b)))...)) with a[bₙ ₋ ₁]c number of f's
a[b₁,₁]c = a[bₓ]c where x is a[bb]c
a[b₁,₂]c = a[bₓ]c where x is a[b₁,₁]c
a[b₁,ₙ]c = a[bₓ]c where x is a[b₁,ₙ ₋ ₁]c
a[b₂,₀]c = a[b₁,ₓ]c where x is a[b₁,b]c
a[bₙ,₀]c = a[bₙ ₋ ₁,ₓ]c where x is a[bₙ ₋ ₁,b]c
a[b₁,₁,₁]c = a[bₓ,ₓ]c where x is a[bb,b]c
a[b₁,₁,ₙ]c = a[b₁,ₓ]c where x is a[b₁,₁,ₙ ₋ ₁]c
a[b₁,ₙ,ₙ]c = a[bₓ]c where x is a[b₁,ₙ₋₁,ₙ₋₁]c
a[bₙ,₀,₀]c = a[bₙ ₋ ₁,ₓ,ₓ]c where x is a[bₙ ₋ ₁,b]c
a[b¹]c = a[bb,b,b,b,b,...b]c with b number of subscripted entries
a[b²]c = a[x¹]c where x is a[b¹]c
a[bⁿ]c = a[xⁿ⁻¹]c where x is a[bⁿ⁻¹]c
a[b¹,⁰]c = a[bˣ]c where x is a[bᵇ]c
We can even do a[b¹,⁰,⁰,⁰,⁰,⁰,⁰,⁰,⁰,⁰]c, and that, is the limit of my notation. That limit is approximately Mahlo level.