[IMO-1981] Find f(3, 2023) when f(x, y) satisfy certain rules

Find f(3, 2023) if f(0,y)=y+1, f(x+1,0)=f(x,1), f(x+1,y+1)=f(x,f(x+1,y)).
This is a question from IMO-1981 and the approach is now widely used in a type of computer algorithm.

[@PantheraLeo04]

This was a really great video, you did a great job breaking down the problem. But I have one question. If we hold x constant at some value n like you did in the video, would the non-recursive equation for f(x,y) always involve 2[n]y where [n] is the nth hyperoperation? (Except n=0, which would just be y[0] since the 0th hyperoperation only takes one input)

[@FightingWolferine]

Apparently yes. Thanks for bringing this up. I previously didn't know the fact that this is actually the famous two-argument Ackermann function: en.wikipedia.org/wiki/Ackermann_function