The problem can be equivalently (and even more canonically) posed with argument 0
a [0] 0 = 1, no right neutral element e, x[0]e=x => e=x-1
a [1] 0 = a, neutral element 0
a [2] 0 = 0, neutral element 1
a [3] 0 = 1, right neutral element 1, no left neutral element e[3]x=x => e=x^(1/x)
a [4] 0 = 1
a [3L] 0 = a
And now we also see the pattern. For each operation we choose as initial value the neutral element of the corresponding side of the sub/preceding operation, if there is any. Otherwise we choose a, the left operand, as initial value. (*)
"the corresponding side" means the left side for the left-bracketed/lower hyperoperation and means the right side for the right-bracketed/upper/normal hyperoperation.
To look at some deviant hyper operations, let us define (only for this post, hopefully we elaborate a more refined notation later):
a[nr]0=right neutral elment of [n]
a[nR]0=a
a[nl]0=left neutral element of [n]
a[nL]0=a
If we now start with the initial operation a[0]x=x+1, we have the following hierarchy:
a[1]x=a[0R]x=a+x
a[2]x=a[1r]x=ax
a[3]x=a[2r]x=a^x
a[3L]x=a[3L]x=a^a^x
and the following deviant operations:
a[1R]0=a, a[1R]1=a[1]a=2a, ...
a[1R]n=a(n+1)
There is no left neutral element e, e[1R]x=x => e=x/(x+1).
0 is the right neutral element
a[1Rr]0=0, a[1Rr]1=a[1R]0=a, a[1Rr]2=a[1R]a=a(a+1)=a^2+a, a[1Rr]3=a[1R](a^2+a)=a^3+a^2+a, ...
a[1Rr]n=\( \sum_{i=1}^n a^i \)
There is no left neutral element e, e[1Rr]x=x.
1 is the right neutral element
How can this be extended to the real n?
Deviant tetration
a[1Rrr]0=1, a[1Rrr]1=a[1Rr]1=a, a[1Rrr]2=a[1Rr]a=?
a[2R]0=a
a[2R]1=a a=a^2
a[2R]n=a^(n+1)
There is no left neutral element
0 is the right neutral element
Another deviant tetration:
a[2Rr]0=0, a[2Rr]1=a[2R]0=a, a[2Rr]2=a[2R]a=a^(a+1), a[2Rr]3=a^(1+a^(a+1)), ...
We saw already that the sub operations below [0] are all equal to [0], i.e. increments. And those are unfortunately the operation(s) that dont fit into this scheme. Because the initial value to go from a[-1]x=x+1 to a[0]x=x+1, is a[0]0=1, though neither 1 is a right neutral element of [-1] nor is 1=a.
But this can be reformulated that there are no operations below [0] that obey our general rule (*), i.e. the hierarchy starts at 0.
a [0] 0 = 1, no right neutral element e, x[0]e=x => e=x-1
a [1] 0 = a, neutral element 0
a [2] 0 = 0, neutral element 1
a [3] 0 = 1, right neutral element 1, no left neutral element e[3]x=x => e=x^(1/x)
a [4] 0 = 1
a [3L] 0 = a
And now we also see the pattern. For each operation we choose as initial value the neutral element of the corresponding side of the sub/preceding operation, if there is any. Otherwise we choose a, the left operand, as initial value. (*)
"the corresponding side" means the left side for the left-bracketed/lower hyperoperation and means the right side for the right-bracketed/upper/normal hyperoperation.
To look at some deviant hyper operations, let us define (only for this post, hopefully we elaborate a more refined notation later):
a[nr]0=right neutral elment of [n]
a[nR]0=a
a[nl]0=left neutral element of [n]
a[nL]0=a
If we now start with the initial operation a[0]x=x+1, we have the following hierarchy:
a[1]x=a[0R]x=a+x
a[2]x=a[1r]x=ax
a[3]x=a[2r]x=a^x
a[3L]x=a[3L]x=a^a^x
and the following deviant operations:
a[1R]0=a, a[1R]1=a[1]a=2a, ...
a[1R]n=a(n+1)
There is no left neutral element e, e[1R]x=x => e=x/(x+1).
0 is the right neutral element
a[1Rr]0=0, a[1Rr]1=a[1R]0=a, a[1Rr]2=a[1R]a=a(a+1)=a^2+a, a[1Rr]3=a[1R](a^2+a)=a^3+a^2+a, ...
a[1Rr]n=\( \sum_{i=1}^n a^i \)
There is no left neutral element e, e[1Rr]x=x.
1 is the right neutral element
How can this be extended to the real n?
Deviant tetration
a[1Rrr]0=1, a[1Rrr]1=a[1Rr]1=a, a[1Rrr]2=a[1Rr]a=?
a[2R]0=a
a[2R]1=a a=a^2
a[2R]n=a^(n+1)
There is no left neutral element
0 is the right neutral element
Another deviant tetration:
a[2Rr]0=0, a[2Rr]1=a[2R]0=a, a[2Rr]2=a[2R]a=a^(a+1), a[2Rr]3=a^(1+a^(a+1)), ...
We saw already that the sub operations below [0] are all equal to [0], i.e. increments. And those are unfortunately the operation(s) that dont fit into this scheme. Because the initial value to go from a[-1]x=x+1 to a[0]x=x+1, is a[0]0=1, though neither 1 is a right neutral element of [-1] nor is 1=a.
But this can be reformulated that there are no operations below [0] that obey our general rule (*), i.e. the hierarchy starts at 0.
