I'm just dabbling here.
#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?
  • \( H_0(b)=1+b\; \)  successor, which is a unary operation
  • \( H_1(a,b)=a+b \)
  • \( H_2(a,b)=a\cdot b \)
  • \( H_3(a,b)=a\uparrow b=a^b \) 
  • \( H_4(a,b)=a\uparrow\uparrow b=\text{tet}_a(b) \) 
  • \( H_5(a,b)=a\uparrow\uparrow\uparrow b=\text{pent}_a(b) \)
  • \( \forall n>0\,H_{n}(a,b)=H_{n-1}\left(a,H_{n}(a,b-1)\right) \)
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 \( H_n(a,b) \).
- Sheldon


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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