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 I'm just dabbling here. sheldonison Long Time Fellow Posts: 631 Threads: 22 Joined: Oct 2008 07/10/2018, 10:34 PM (This post was last modified: 07/11/2018, 02:54 AM by sheldonison.) (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 « Next Oldest | Next Newest »

 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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