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I'm just dabbling here.
#1
I'm only a new user, so I don't know what to do here. I'm only an amateur too.

Anyways, here's how we define hyperoperations.

H(b,x,0) = S(b)
H(b,0,1) = b
H(b,1,r≥2) = b
H(b,x+1,r+1) = H(H(b,x,r+1),b,r)

Alright, so we want the negative hyperoperations.

So let's denote S(b) as S(b,x). This time with 2 inputs.

So, it has to obey the rule that H(b,b,r) = H(b,2,r+1)...right?

Well, if S(b,b) did equal b+2, then would there something wrong?

S(b,b) = b+2 = S(b,b+1), nothing wrong with that.

However, 0+1 = S(0+0,0) = S(0,0) = 2. This is a contradiction.

This is the only instance, however, where this is wrong. Why? Because 0+0=0. It's the identity of addition.

In fact, even if the identity was 2, then 2+3 = S(2+2,2) = S(2,2) = 2. If the identity was 3, then
3+4 = S(3+3,3) = S(3,3) = 2.

-Wait! If the identity was 1... Then 1+2 = 2...which would actually be true. So, in order for S(b,x) to be functional, then the identity of addition has to be 1.

But wait! [Note: Let's denote the inverse of S as P]

0+(3-1) = P((0+3),(0+1)) = P(3,0) = 2

0+(3-1) = 2

0+(3-1) = 0+3 = 3

2=3

And this is when the identity of addition is 1.

So, no matter which identity we choose, there will always be a contradiction. So, that's it. It's just 0.

Meaning that S(b,b) ≠ b+2.

Meaning that S(b,x) = S(b).

Alright, well this does change a bit about the negative hyperoperations, but not completely. Anyways.

Let's denote H(b,x,-1) as ®.

So, S(b,x) just equals S(b) for all x.

So, S(b,S(x)) = S(b,x)®b = S(b)®b

S(b)®b = S(b)

And...this is just it. This is the solution to the hyperoperation rank -1.

We can't go any farther.

Could somebody check my work?
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#2
(07/07/2018, 02:44 AM)ChaoticMC Wrote: here's how we define hyperoperations.

H(b,x,0) = S(b)
H(b,0,1) = b
H(b,1,r≥2) = b
H(b,x+1,r+1) = H(H(b,x,r+1),b,r)

Alright, so we want the negative hyperoperations...
I would try this definition from wikipedia:
https://en.wikipedia.org/wiki/Hyperoperation 
The standard definition then follows, where a is the base
H0(a,b)=b+1 independent of the value of a
H1(a,b)=a+b where H1(a,0)=a 
H2(a,b)=a*b where H2(a,0)=0
H3(a,b)=a^b where for n>=3, H(a,0)=1, and H(a,1)=a

With that definition, surprisingly H{-1}(b)=H0(b)=b+1; this works for H{-1,-2,-3.....}(b)=b+1 as well.  Here is the table for a=3.

Code:
a=3   b+1   b+a   b*a   a^b   a^^b
b     h0(b) h1(b) h2(b) h3(b) h4(b)
====================================
0     1     3     0     1     1
1     2     4     3     3     3
2     3     5     6     9     27
3     4     6     9     27    7625597484987
4     5     7     12    81
5     6     8     15    243
6     7     9     18    729
7     8     10    21    2187
8     9     11    24    6561
- Sheldon
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#3
Hi!

I agree with Sheldon almost at all. If I were you, I would use the following notations:

Hn(a,b) = a[n]b = H(a,n,b) or the Knuth's uparrows, the second is simplest as I think.
If you want iterate an operator, just use the following functional power formula:
y [z+1] x = (y [z] x)^o(y-1) o y
I suppose that z can be  negative, zero, rational, real or complex, too.
But if you follow Sheldon's zeration formula, it will not lead to addition, so I think that is wrong.
But Sheldon has a lot of successful pari/gp programme and ideas for the realizations of from the tetration to heptation, but hard to interpolate with rational or real numbers.

Good luck!  Wink

Xorter
Xorter Unizo
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#4
(07/09/2018, 08:34 PM)Xorter Wrote: Hi!

I agree with Sheldon almost at all. If I were you, I would use the following notations:

Hn(a,b) = a[n]b = H(a,n,b) or the Knuth's uparrows, the second is simplest as I think...

I like the wikipedia hyperoperation definition since it shows the operator sequence.  Why is "tet" the 4th operator?
  •   successor, which is a unary operation


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Knuth's uparrow notation is great except when discussing the hyper operation sequence where uparrow of (0,-1,-2) donesn't make any sense.  Also, the recursive definition is limited to integer values of n, so I like .
- Sheldon
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