User blog comment:Mh314159/Alpha numbers (and beyond)/@comment-35470197-20191007022858/@comment-39585023-20191015022855

Hi! Thanks!

I don't think it is possible to have a more than one zero after a recursion of α[a,b,c], which decrements only c, while the separate m indicator decrements only a. So expressions like α[0,0,0] and α[0,1,0,1] are not possible. It is possible to have α[0,1,1] and then the rule is applied sequentially, so we would first go to α[α[1,0,1],0,1] and then we would apply it to the innermost expression so we would get α[α[1,α[1,1]],0,1] after dropping trailing zeroes, and finally we would get α[ α[1,α[1,1]], α[ α[1,α[1,1]], 1]] again after dropping trailing zeroes so it is made of up of nested two-term strings.

f{1,1,0}(x) simple follows the drop trailing zeroes rule and becomes f{1,1}(x).

There is a rule for terms like f{1,1}(0) if that's what you meant, and I'm currently using:

f‹S›(0) = f‹S›(p) p = f‹S'›(j)

S is any comma-delimited string of nonzero natural numbers

S'' is S with the initial term decreased by one

S''' is S with the final term decreased by one

j is the sum of the terms in S

Example:

f‹2,1,1›(0) = f‹1,1,1›(f‹2,1›(4))