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I'm just dabbling here.
#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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Messages In This Thread
I'm just dabbling here. - by ChaoticMC - 07/07/2018, 02:44 AM
RE: I'm just dabbling here. - by sheldonison - 07/09/2018, 02:07 PM
RE: I'm just dabbling here. - by Xorter - 07/09/2018, 08:34 PM
RE: I'm just dabbling here. - by sheldonison - 07/10/2018, 10:34 PM



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