I have boiled down the recursion of analytic hyper operators into a formula based on their coefficients. If we write these coefficients as follows:
 = a \Delta_{s+1} (b+1))
 s^i)
We can write the recursive formula; without giving a proof for it (it just requires a few series rearrangement); as:
 = \sum_{i=0}^{\infty} \chi_{n+i}(a, a \Delta_{s} b) \frac{(n+i)!}{n! i!}(-1)^i\)
As you can see; this appears very off.
can vary freely and the result on the L.H.S. doesn't change at all. However, it's being summed across an infinite series so that may compensate. But I wonder if declaring, that since 
takes on every value in between 
and 
; at least; we can say over that interval 
 = \text{Constant})
Since we can set
this implies a strict contradiction:
s^i = \sum_{i=0}^{\infty} \chi_{i}(a,c_1)s^i)
This is a contradiction because it implies
is constant and therefore constant for all b in 
. This would imply there is no analytic continuation of hyper operators! At least, not representable by its Taylor series.
I didn't write out the proof because I'm stuck and I'm curious if it's justifiable to do that last move, or if there is some other routine I can go about to prove the constancy of these coefficients.
If hyper operators aren't analytic; and hopefully I can prove not continuous; I have a separate way of defining them that admit a discrete solution with a more number theoretical algebraic approach.
We can write the recursive formula; without giving a proof for it (it just requires a few series rearrangement); as:
As you can see; this appears very off.
Since we can set
This is a contradiction because it implies
I didn't write out the proof because I'm stuck and I'm curious if it's justifiable to do that last move, or if there is some other routine I can go about to prove the constancy of these coefficients.
If hyper operators aren't analytic; and hopefully I can prove not continuous; I have a separate way of defining them that admit a discrete solution with a more number theoretical algebraic approach.