User blog:Plain'N'Simple/Using triangular numbers to create a Conway-Arrow Level Notation (probably not what you'd expect)

Preliminary note:

A "triangular number" is a number of the form 1+2+3+4+...+k for a given k. We shall notate "the nth triangular number" as Δn. So:

Δ3 = 1+2+3 = 6

Δ10 = 1+2+3+4+5+6+7+8+9+10 = 55

and so on.

And now to define our function:

For any two positive integers n,m we will define f(n,m) like this:

(i) f(n,1)=3

(ii) f(1,m)=3m

(iii) f(Δk+1,m)=f(k+Δm+k,3)

(iv) if n is not a triangular number, then let Δk be the largest triangular number less then n. In this case:

f(n,m)=f(n-k,f(n,m-1))

And that's it! It doesn't seem like much, but this function grows as fast as Conway Arrows (or 4-array entries in BEAF)!

More specifically: f(Δn,k) is roughly on the same scale as as a Conway Chain of length n+1 (or {k,k,k,n} in BEAF).

So how does this work? Basically, by using the following re-arrangement of the natural numbers into an infinite grid:

01, 02, 04, 07, 11, ...

03, 05, 08, 12, 17, ...

06, 09, 13, 18, 24, ...

10, 14, 19, 25, 32, ...

.

.

.

The rules I've given basically do this:

(i) We look for the position of n in the grid.

(ii) If n=1, then we are at the top-left corner so this is our weakest function: f(1,m)=3m

(iii) If n is the first number in its row, then f(n,m)=f(k,3) where k is the n-th number in the previous row.

(iv) If in is not the first number in its row, then f(n,m)=f(k,f(k,f(k,...f(k,3)))) with m k's, where k is the number immediately before n in the grid.

And this is a classic ω2-level recursion. In fact, every number on the grid corresponds to an ordinal below ω2.

Cool, eh?

(Note to the pros: I'll appreciate if you'll double check my work for silly mistakes that might have crept in)