Repeated factorial thing?
"n" must be a whole positive number.
\(k(n) = b(n,n)\)
Since this is a recursive function, I will add a definition for b(0,n), to ground it and not make it become like infinity.
If m = 0: \(b(m,n) = n!\)
If m > 0: \(b(m,n) = b(m-1,b(m-1,n))\) where \(m > 0\)
So, \(k(3) = b(3,3) = b(2,b(2,3)) = b(1,b(1,b(1,3))) = b(0,b(0,b(0,b(0,3)))) = b(0,b(0,b(0,6) = b(0,b(0,720)\)
\(= 720!! \approx 2.6\cdot10^{1746}! = oh god\)
Yay finally I didn't use ellipses to make recursion!
Moral of the story? Don't underestimate recursion. As long as the first operator is strong enough, recursion makes it immensely more powerful.
But wait, if we have expofactorials, where are the tetrafactorials and so on?
\(p(n) = n\uparrow^{n}(n-1)\uparrow^{n}(n-2)\cdots2\uparrow^{n}1\)
So, we add a new modified function \(k_1(n)\) and \(b_1(n,m)\)
"n" must be a positive whole number.
"m" must be a positive whole number . \(k_1(n) = b_1(n,m)\) If m = 0: \(b_1(m,n) = p(n)\)
If m > 0:\(b_1(m,n) = b_1(m-1,b_1(m-1,n))\)
So, \(k_1(3) = b_1(3,3) = b_1(2,b_1(2,3)) = b_1(1,b_1(1,b_1(1,3))) = b_1(0,b_1(0,b_1(0,b_1(0,3)))) = b_1(0,b_1(0,b_1(0,\)
\(3\uparrow\uparrow\uparrow2\uparrow\uparrow\uparrow1) = o no\)
So that about concludes this random function.
As of now, there is no \(b_2(n)\). Maybe I'll add it later, when I have some idea.