Name for sequence y=0[0]0. y=1[1]1, y=2[2]2, y= n[n]n, y=x[x]x, y=i[i]i, y=z[z]z etc?
#11
I hope so too. See you later.

GFR
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#12
(04/27/2008, 02:59 PM)GFR Wrote: Dear Ivars:

0[0]0 = ??? (I try to avoid troubles ... Wink )
1[1]1 = 1+1 = 2
2[2]2 = 2*2 = 4
3[3]3 = 3^3 = 27
4[4]4 = 4#4 = 4^[4^[4^4]]] .... uuuuuhhhhh !!!
......
oo[w]oo = ......  infinite-omega-infinite (Qickfur said that it is still enumerable! I think he is right!). Actually, we could write it as oo[oo]oo, but I think that the entity between the brackets is an ordinal. Am I wrong? Maybe yes ! Sad Ops, suicide comment: how about s=0, the 0-th hyperop?

GFR
\(4[4]4\approx2.361e+8,072,304,726,028,225,379,382,630,397,085,399,030,071,367,921,738,743,031,867,082,828,418,414,481,568,309,149,198,911,814,701,229,483,451,981,557,574,771,156,496,457,238,535,299,087,481,244,990,261,351,116\)
\(\omega[\omega]\omega=\varphi(\omega,0,0)\) using Bowers' climbing method.
\(n[n]n=\) \(\phi\)\((n,n,\omega^\omega)=n\{\{0\}\}n=\) \(\text{booga}\)\((n)\) \(=\) \((n)\) where the \(\phi\) is using the Wainer Hierarchy system of fundamental sequences, and \(\{\{0\}\}\) is an extended operator with respect to Bowers' old rule set.
Please remember to stay hydrated.
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
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