(08/23/2009, 09:45 AM)bo198214 Wrote: While verifying myself I found that the deviation (up to a difference of 1 in the rank) from our operator sequence comes from forming an unnecessary odd initial condition. I dont know why he does, perhaps it is more suitable for his proof of non-primitive recursiveness.
Oh now I found out where this odd initial conditions comes from!
I assert that Ackermann originally wanted to define left-braced hyperoperations!
Then this initial condition \( \alpha(a,n)=a \) for \( n\ge 2 \) makes sense!
Left-braced hyperoperations \( \psi \) would similarly be defined by:
\( \psi(a,b,0)=a+b \)
\( \psi(a,b,n+1)=(x\mapsto \psi(x,a,n))^{[b]}(\alpha(a,n)) \)
here again we have \( \psi(a,b,1)=ab \) and \( \psi(a,b,2)=a^b \).
But the forth operation is not \( a^{a^{b-1}} \) as one would obtain with the initial condition \( \alpha(a,2)=1 \), but it is \( \psi(a,b,3)=a^{a^b} \) due to the initial value \( \alpha(a,2)=a \)!
So this initial condition makes left-braced hyperoperations look simpler, while it makes right-braced hyperoperations looking odd.
I think he started with the left-braced hyperoperations and then switched to the faster growing right-braced hyperoperations, perhaps it was more suitable for his proof of non-primitive recursiveness of \( n\mapsto \phi(n,n,n) \)
