There were debates to certain expansions for certain sequences of the Y sequence, so I decide to analyze and also fix certain parts of it. Unfortunately, on September 3, I abandoned this proposal because it simply doesn't work.
Check out at the Y sequence calculator using the original version
Using the original expansion, it was expanded as (1,2,4,7,4,6,8,10,12,14,...), so I decided to attempt to fix it using Buchholz's function (normal form is used). Here is an approach:
(1,2,4,7) = ψ0(ψ2(0))
(1,2,4,7,4) = ψ0(ψ2(0) + ψ1(0))
(1,2,4,7,4,4) = ψ0(ψ2(0) + ψ1(0) + ψ1(0))
(1,2,4,7,4,5) = ψ0(ψ2(0) + ψ1(ψ0(0)))
(1,2,4,7,4,5,6) = ψ0(ψ2(0) + ψ1(ψ0(ψ0(0))))
(1,2,4,7,4,5,7) = ψ0(ψ2(0) + ψ1(ψ0(ψ1(0))))
(1,2,4,7,4,5,7,9) = ψ0(ψ2(0) + ψ1(ψ0(ψ1(ψ1(0)))))
(1,2,4,7,4,5,7,10) = ψ0(ψ2(0) + ψ1(ψ0(ψ2(0))))
(1,2,4,7,4,5,7,10,7) = ψ0(ψ2(0) + ψ1(ψ0(ψ2(0) + ψ1(0))))
(1,2,4,7,4,6) = ψ0(ψ2(0) + ψ1(ψ1(0)))
(1,2,4,7,4,6,6) = ψ0(ψ2(0) + ψ1(ψ1(0) + ψ1(0)))
(1,2,4,7,4,6,7) = ψ0(ψ2(0) + ψ1(ψ1(ψ0(0))))
(1,2,4,7,4,6,8) = ψ0(ψ2(0) + ψ1(ψ1(ψ1(0))))
(1,2,4,7,4,6,8,10) = ψ0(ψ2(0) + ψ1(ψ1(ψ1(ψ1(0)))))
(1,2,4,7,4,6,8,10,12) = ψ0(ψ2(0) + ψ1(ψ1(ψ1(ψ1(ψ1(0))))))
But (1,2,4,7,4,7) is expanded as (1,2,4,7,4,6,9,6,8,11,8,10,13,10...) instead of (1,2,4,7,4,6,8,10,12,14,16,...), since the ordinal ψ0(ψ2(0) + ψ2(0)) is greater than ψ0(ψ2(0) + ψ1(ψ2(0))), and the latter's expansion is the sequence (1,2,4,7,4,6,9).
And here is the fixed sequences from (1,2,4,7,4,6,9) to (1,2,4,7,4,7):
(1,2,4,7,4,6,9) = ψ0(ψ2(0) + ψ1(ψ2(0)))
(1,2,4,7,4,6,9,6) = ψ0(ψ2(0) + ψ1(ψ2(0) + ψ1(0)))
(1,2,4,7,4,6,9,6,8) = ψ0(ψ2(0) + ψ1(ψ2(0) + ψ1(ψ1(0))))
(1,2,4,7,4,6,9,6,8,10) = ψ0(ψ2(0) + ψ1(ψ2(0) + ψ1(ψ1(ψ1(0)))))
(1,2,4,7,4,6,9,6,8,11) = ψ0(ψ2(0) + ψ1(ψ2(0) + ψ1(ψ2(0))))
(1,2,4,7,4,6,9,6,8,11,8) = ψ0(ψ2(0) + ψ1(ψ2(0) + ψ1(ψ2(0) + ψ1(0))))
(1,2,4,7,4,6,9,6,8,11,8,10) = ψ0(ψ2(0) + ψ1(ψ2(0) + ψ1(ψ2(0) + ψ1(ψ1(0)))))
(1,2,4,7,4,6,9,6,8,11,8,10,13) = ψ0(ψ2(0) + ψ1(ψ2(0) + ψ1(ψ2(0) + ψ1(ψ2(0)))))
(1,2,4,7,4,6,9,6,8,11,8,10,13,10) = ψ0(ψ2(0) + ψ1(ψ2(0) + ψ1(ψ2(0) + ψ1(ψ2(0) + ψ1(0)))))
And here is the fixed fundamental sequence for (1,2,4,7,4,7):
(1,2,4,7,4,7)[0] = (1,2,4,7,4)
(1,2,4,7,4,7)[1] = (1,2,4,7,4,6,9,6)
(1,2,4,7,4,7)[2] = (1,2,4,7,4,6,9,6,8,11,8)
(1,2,4,7,4,7)[3] = (1,2,4,7,4,6,9,6,8,11,8,10,13,10)
(1,2,4,7,4,7)[4] = (1,2,4,7,4,6,9,6,8,11,8,10,13,10,12,15,12)
The similar problem also goes for (1,2,4,7,6,9), which its intended expansion is (1,2,4,7,6,8,11,10,12,15,14,16,19,...) instead of (1,2,4,7,6,8,10,12,14,...), well...
(1,2,4,7,4,7) = ψ0(ψ2(0) + ψ2(0))
(1,2,4,7,4,7,4,7) = ψ0(ψ2(0) + ψ2(0) + ψ2(0))
(1,2,4,7,5) = ψ0(ψ2(ψ0(0)))
(1,2,4,7,5,5) = ψ0(ψ2(ψ0(0) + ψ0(0)))
(1,2,4,7,5,6) = ψ0(ψ2(ψ0(ψ0(0))))
(1,2,4,7,5,7) = ψ0(ψ2(ψ0(ψ1(0))))
(1,2,4,7,5,7,9) = ψ0(ψ2(ψ0(ψ1(ψ1(0)))))
(1,2,4,7,5,7,10) = ψ0(ψ2(ψ0(ψ2(0))))
(1,2,4,7,5,7,10,8,10,13) = ψ0(ψ2(ψ0(ψ2(ψ0(ψ2(0))))))
(1,2,4,7,6) = ψ0(ψ2(ψ1(0)))
(1,2,4,7,6,6) = ψ0(ψ2(ψ1(0) + ψ1(0)))
(1,2,4,7,6,7) = ψ0(ψ2(ψ1(ψ0(0))))
(1,2,4,7,6,8) = ψ0(ψ2(ψ1(ψ1(0))))
(1,2,4,7,6,8,10) = ψ0(ψ2(ψ1(ψ1(ψ1(0)))))
(1,2,4,7,6,8,11) = ψ0(ψ2(ψ1(ψ2(0))))
(1,2,4,7,6,8,11,8,10,13) = ψ0(ψ2(ψ1(ψ2(ψ1(ψ2(0))))))
Yes, again, this is because the ordinal ψ0(ψ2(ψ2(0))) is also greater than ψ0(ψ2(ψ1(ψ2(0)))) as well.
Let's find the fundamental sequence for (1,2,4,7,6,9):
(1,2,4,7,6,9)[0] = (1,2,4,7,6)
(1,2,4,7,6,9)[1] = (1,2,4,7,6,8,11,10)
(1,2,4,7,6,9)[2] = (1,2,4,7,6,8,11,10,12,15,14)
(1,2,4,7,6,9)[3] = (1,2,4,7,6,8,11,10,12,15,14,16,19,18)
(1,2,4,7,6,9)[4] = (1,2,4,7,6,8,11,10,12,15,14,16,19,18,20,23,22)
And now we jump to (1,2,4,7,11) = ψ0(ψ3(0)).
We also have found the problem with (1,2,4,7,11,4,7,11) in which is expanded as (1,2,4,7,11,4,7,10,13,16,19...) instead of my fixed (1,2,4,7,11,4,7,10,14,6,9,12,16,...), as ψ0(ψ3(0) + ψ3(0)) > ψ0(ψ3(0) + ψ2(ψ3(0))). Here is my (nearly) corrected expansion (by duplicating the last active pattern then adding those entries by 2):
(1,2,4,7,11,4,7,11)[1] = (1,2,4,7,11,4,7,10,14)
(1,2,4,7,11,4,7,11)[2] = (1,2,4,7,11,4,7,10,14,6,9,12,16)
(1,2,4,7,11,4,7,11)[3] = (1,2,4,7,11,4,7,10,14,6,9,12,16,8,11,14,18)
As well as the expansion for (1,2,4,7,11,6,9,13) (by duplicating the last active pattern then adding those entries by 4):
(1,2,4,7,11,6,9,13)[1] = (1,2,4,7,11,6,9,12,16)
(1,2,4,7,11,6,9,13)[2] = (1,2,4,7,11,6,9,12,16,10,13,16,20)
(1,2,4,7,11,6,9,13)[3] = (1,2,4,7,11,6,9,12,16,10,13,16,20,14,17,20,24)
Wait...
The solution is have to be creating the "partition" of the pattern by using additive booster applied to the previous active entries, rather than presuming the rightmost entries while ignoring the intermediate active patterns before them.
Or maybe analyze backwards?
(1,2,4,8) = ψ0(ψω(0))
(1,2,4,7,11) = ψ0(ψ3(0))
(1,2,4,7,10) = ψ0(ψ2(ψ2(0)))
(1,2,4,7,9) = ψ0(ψ2(ψ1(0)))
(1,2,4,7,8) = ψ0(ψ2(ψ0(0)))
(1,2,4,7,7) = ψ0(ψ2(0) + ψ2(0))
(1,2,4,7,6,9) = ψ0(ψ2(0) + ψ1(ψ2(0)))
...
(1,2,4,7,4) = ψ0(ψ2(0) + ψ1(0))
What the hell? Let's go back to my original fix (red: n = 0, green: n = 1, blue: n = 2, magenta: n = 3, colored for active patterns).
(1,2,4,7,6,9) = (1,2,4,7,6,8,11,10,12,15,14,16,19,18,...) = ψ0(ψ2(ψ2(0)))
(1,2,4,7,6,9,6,9) = (1,2,4,7,6,9,6,8,11,10,13,10,12,15,14,17,14,16,19,18,21,18,...) = ψ0(ψ2(ψ2(0) + ψ2(0)))
(1,2,4,7,6,9,7) = (1,2,4,7,6,9,6,9,6,9,6,9,6,9,...) = ψ0(ψ2(ψ2(ψ0(0))))
(1,2,4,7,6,9,8) = (1,2,4,7,6,9,7,9,12,11,14,12,14,17,16,19,17,19,22,21,24,...) = ψ0(ψ2(ψ2(ψ1(0))))
(1,2,4,7,6,9,8,11) = (1,2,4,7,6,9,8,10,13,12,15,14,16,19,18,21,20,22,25,24,27,26,...) = ψ0(ψ2(ψ2(ψ2(0))))
(1,2,4,7,6,9,8,11,10,13) = (1,2,4,7,6,9,8,11,10,12,15,14,17,16,19,18,20,23,22,25,24,27,26,28,31,30,33,32,35,34,...) = ψ0(ψ2(ψ2(ψ2(ψ2(0)))))
(1,2,4,7,7) = (1,2,4,7,6,9,8,11,10,13,...) = ψ0(ψ3(0))
(1,2,4,7,7,4,7,7) = (1,2,4,7,7,4,7,6,9,9,6,9,8,11,11,8,11,10,13,13,10,13,...) = ψ0(ψ3(0) + ψ3(0))
(1,2,4,7,7,5) = (1,2,4,7,7,4,7,7,4,7,7,4,7,7,...) = ψ0(ψ3(ψ0(0)))
(1,2,4,7,7,6) = (1,2,4,7,7,5,7,10,10,8,10,13,13,11,13,16,16,...) = ψ0(ψ3(ψ1(0)))
(1,2,4,7,7,6,9) = (1,2,4,7,7,6,8,11,11,10,12,15,15,14,16,18,18,17,...) = ψ0(ψ3(ψ2(0)))
(1,2,4,7,7,6,9,9) = (1,2,4,7,7,6,9,8,11,11,10,13,12,15,15,14,17,16,18,18,17,21,...) = ψ0(ψ3(ψ3(0)))
(1,2,4,7,7,7) = (1,2,4,7,7,6,9,9,8,11,11,10,13,13,...) = ψ0(ψ4(0))
(1,2,4,7,7,7,7) = (1,2,4,7,7,7,6,9,9,9,8,11,11,11,10,13,13,13,...) = ψ0(ψ5(0))
And now, we finally neared the Buchholz's ordinal (ψ0(Ω_ω)), which is (1,2,4,7,8) in my formulation. Let's go next.
What should we analyse the value of (1,2,4,8) first of all? I decide to conjecture that the sequence (1,2,4,8) eventually dominates the countable limit of the extended Buchholz's function, known as the extended Buchholz's ordinal.
(1,2,4,7,8) = (1,2,4,7,7,7,7,...) = ψ0(ψ{ω}(0))
(1,2,4,7,8,4) = (1,2,4,7,8,3,5,8,9,4,6,9,10,5,7,10,11,...) = ψ0(ψ{ω}(0) + ψ1(0))
(1,2,4,7,8,5) = (1,2,4,7,8,4,7,8,4,7,8,4,7,8,...) = ψ0(ψ{ω}(ψ0(0)))
(1,2,4,7,8,6) = (1,2,4,7,8,5,7,10,11,8,10,13,14,11,13,16,17,...) = ψ0(ψ{ω}(ψ1(0)))
(1,2,4,7,8,6,9) = (1,2,4,7,8,6,8,11,12,10,12,15,16,14,16,19,20,18,...) = ψ0(ψ{ω}(ψ2(0)))
(1,2,4,7,8,6,9,9) = (1,2,4,7,8,6,9,8,11,12,10,13,12,15,16,14,17,16,19,20,18,21,...) = ψ0(ψ{ω}(ψ3(0)))
(1,2,4,7,8,6,9,10) = (1,2,4,7,8,6,9,9,9,9,...) = ψ0(ψ{ω}(ψ{ω}(0)))
And now we passed the Takeuti-Feferman-Buchholz ordinal (TFBO). Let's continue...
(1,2,4,7,8,7) = (1,2,4,7,8,6,9,10,8,11,12,10,13,14,...) = ψ0(ψ{ω + 1}(0))
(1,2,4,7,8,7,7) = (1,2,4,7,8,7,6,9,10,9,8,11,12,11,10,13,14,13,...) = ψ0(ψ{ω + 2}(0))
(1,2,4,7,8,7,7,7) = (1,2,4,7,8,7,7,6,9,10,9,9,8,11,12,11,11,10,13,14,13,13,...) = ψ0(ψ{ω + 3}(0))
(1,2,4,7,8,7,8) = (1,2,4,7,8,7,7,7,7,...) = ψ0(ψ{ω2}(0))
(1,2,4,7,8,7,8,7,8) = (1,2,4,7,8,7,8,7,7,7,7,...) = ψ0(ψ{ω3}(0))
(1,2,4,7,8,8) = (1,2,4,7,8,7,8,7,8,7,8,...) = ψ0(ψ{ω^2}(0))
(1,2,4,7,8,8,8) = (1,2,4,7,8,8,7,8,8,7,8,8,7,8,8,...) = ψ0(ψ{ω^3}(0))
(1,2,4,7,8,9) = (1,2,4,7,8,8,8,8,...) = ψ0(ψ{ω^ω}(0))
(1,2,4,7,8,9,10) = (1,2,4,7,8,9,9,9,9,...) = ψ0(ψ{ω^ω^ω}(0))
(1,2,4,7,8,10) = (1,2,4,7,8,9,10,11,...) = ψ0(ψ{ψ0(ψ1(0))}(0))
(1,2,4,7,8,10,12) = (1,2,4,7,8,10,11,13,14,16,17,19,...) = ψ0(ψ{ψ0(ψ1(ψ1(0)))}(0))
(1,2,4,7,8,10,13) = (1,2,4,7,10,12,14,16,...) = ψ0(ψ{ψ0(ψ2(0))}(0))
(1,2,4,7,8,10,13,13) = (1,2,4,7,10,13,12,15,14,17,16,19,...) = ψ0(ψ{ψ0(ψ3(0))}(0))
(1,2,4,7,8,10,13,14) = (1,2,4,7,8,10,13,14,14,14,14,...) = ψ0(ψ{ψ0(ψ{ω}(0))}(0))
(1,2,4,7,8,10,13,14,16,19) = (1,2,4,7,8,10,13,14,16,18,20,22,24,...) = ψ0(ψ{ψ0(ψ{ψ0(ψ2(0))}(0))}(0))
Then we passed the Bird's ordinal...
(1,2,4,7,9) = (1,2,4,7,8,10,13,14,16,19,20,22,25,...) = ψ0(ψ{ψ1(0)}(0))
(1,2,4,7,9,7) = (1,2,4,7,9,6,9,11,8,11,13,10,13,15,...) = ψ0(ψ{ψ1(0) + ψ0(0)}(0))
(1,2,4,7,9,7,7) = (1,2,4,7,9,7,6,9,11,9,8,11,13,11,10,13,15,13,...) = ψ0(ψ{ψ1(0) + ψ0(0) + ψ0(0)}(0))
(1,2,4,7,9,7,8) = (1,2,4,7,9,7,7,7,7,...) = ψ0(ψ{ψ1(0) + ψ0(ψ0(0))}(0))
(1,2,4,7,9,7,8,10) = (1,2,4,7,9,7,8,9,10,11,...) = ψ0(ψ{ψ1(0) + ψ0(ψ1(0))}(0))
(1,2,4,7,9,7,8,10,13) = (1,2,4,7,9,7,8,10,12,14,16,...) = ψ0(ψ{ψ1(0) + ψ0(ψ2(0))}(0))
(1,2,4,7,9,7,8,10,13,14) = (1,2,4,7,9,7,8,10,13,13,13,13,...) = ψ0(ψ{ψ1(0) + ψ0(ψ{ω}(0))}(0))
(1,2,4,7,9,7,8,10,13,15) = (1,2,4,7,9,7,8,10,13,14,16,19,20,22,25,26,28,31,...) = ψ0(ψ{ψ1(0) + ψ0(ψ{ψ1(0)}(0))}(0))
(1,2,4,7,9,7,9) = (1,2,4,7,9,7,8,10,13,15,13,14,16,19,21,19,20,22,25,27,25,...) = ψ0(ψ{ψ1(0) + ψ1(0)}(0))
(1,2,4,7,9,7,9,7,9) = (1,2,4,7,9,7,9,7,8,10,13,15,13,15,13,14,16,19,21,19,21,19,20,22,25,27,25,27,25,...) = ψ0(ψ{ψ1(0) + ψ1(0) + ψ1(0)}(0))
(1,2,4,7,9,8) = (1,2,4,7,9,7,9,7,9,7,9,...) = ψ0(ψ{ψ1(ψ0(0))}(0))
(1,2,4,7,9,8,8) = (1,2,4,7,9,8,7,9,8,7,9,8,7,9,8,...) = ψ0(ψ{ψ1(ψ0(0) + ψ0(0))}(0))
(1,2,4,7,9,8,9) = (1,2,4,7,9,8,8,8,8,...) = ψ0(ψ{ψ1(ψ0(ψ0(0)))}(0))
(1,2,4,7,9,8,10) = (1,2,4,7,9,8,9,10,11,...) = ψ0(ψ{ψ1(ψ0(ψ1(0)))}(0))
(1,2,4,7,9,8,10,13) = (1,2,4,7,9,8,10,12,14,16,...) = ψ0(ψ{ψ1(ψ0(ψ2(0)))}(0))
(1,2,4,7,9,8,10,13,14) = (1,2,4,7,9,8,10,13,13,13,13,...) = ψ0(ψ{ψ1(ψ0(ψ{ω}(0)))}(0))
(1,2,4,7,9,8,10,13,15) = (1,2,4,7,9,8,10,13,14,16,19,20,22,25,26,28,31,...) = ψ0(ψ{ψ1(ψ0(ψ{ψ1(0)}(0)))}(0))
(1,2,4,7,9,9) = (1,2,4,7,9,8,10,13,15,14,16,19,21,20,22,25,27,...) = ψ0(ψ{ψ1(ψ1(0))}(0))
(1,2,4,7,9,9,9) = (1,2,4,7,9,9,8,10,13,15,15,14,16,19,21,21,20,22,25,27,17,...) = ψ0(ψ{ψ1(ψ1(0) + ψ1(0))}(0))
(1,2,4,7,9,10) = (1,2,4,7,9,9,9,9,...) = ψ0(ψ{ψ1(ψ1(ψ0(0)))}(0))
(1,2,4,7,9,11) = (1,2,4,7,9,10,12,15,17,18,20,23,25,26,28,31,33,...) = ψ0(ψ{ψ1(ψ1(ψ1(0)))}(0))
(1,2,4,7,9,11,13) = (1,2,4,7,9,11,12,14,17,19,21,22,24,27,29,31,32,34,37,39,41,...) = ψ0(ψ{ψ1(ψ1(ψ1(ψ1(0))))}(0))
(1,2,4,7,9,11,14) = (1,2,4,7,9,11,13,15,17,...) = ψ0(ψ{ψ1(ψ2(0))}(0))
(1,2,4,7,9,11,14,14) = (1,2,4,7,9,11,14,13,16,15,18,17,20,...) = ψ0(ψ{ψ1(ψ3(0))}(0))
(1,2,4,7,9,11,14,15) = (1,2,4,7,9,11,14,14,14,14,...) = ψ0(ψ{ψ1(ψ{ω}(0))}(0))
(1,2,4,7,9,11,14,16) = (1,2,4,7,9,11,14,15,17,20,22,24,27,28,30,33,35,37,40,41,43,46,48,50,53,...) = ψ0(ψ{ψ1(ψ{ψ1(0)}(0))}(0))
(1,2,4,7,9,12) = (1,2,4,7,9,11,14,16,18,21,23,25,28,30,...) = ψ0(ψ{ψ2(0))}(0))
(1,2,4,7,9,12,12) = (1,2,4,7,9,12,11,14,16,19,18,21,23,26,25,28,30,33,...) = ψ0(ψ{ψ3(0))}(0))
(1,2,4,7,9,12,13) = (1,2,4,7,9,12,12,12,12,...) = ψ0(ψ{ψ{ω}(0))}(0))
(1,2,4,7,9,12,13,15) = (1,2,4,7,9,12,13,14,15,16,...) = ψ0(ψ{ψ{ψ0(ψ1(0))}(0))}(0))
(1,2,4,7,9,12,13,15,18) = (1,2,4,7,9,12,13,15,17,19,21,...) = ψ0(ψ{ψ{ψ0(ψ2(0))}(0))}(0))
(1,2,4,7,9,12,13,15,18,20) = (1,2,4,7,9,12,13,15,18,19,21,24,25,27,30,31,33,36,...) = ψ0(ψ{ψ{ψ0(ψ{ψ1(0)}(0))}(0))}(0))
(1,2,4,7,9,12,13,15,18,20,23) = (1,2,4,7,9,12,13,15,18,20,22,24,26,...) = ψ0(ψ{ψ{ψ0(ψ{ψ2(0)}(0))}(0))}(0))
(1,2,4,7,9,12,14) = (1,2,4,7,9,12,13,15,18,20,23,24,26,29,31,34,35,37,40,42,45,...) = ψ0(ψ{ψ{ψ1(0)}(0))}(0))
(1,2,4,7,9,12,14,17) = (1,2,4,7,9,12,14,16,19,21,24,26,28,31,33,36,38,40,43,45,48,50,...) = ψ0(ψ{ψ{ψ2(0)}(0))}(0))
(1,2,4,7,9,12,14,17,18) = (1,2,4,7,9,12,14,17,17,17,17,...) = ψ0(ψ{ψ{ψ{ω}(0)}(0))}(0))
(1,2,4,7,9,12,14,17,19) = (1,2,4,7,9,12,14,17,18,20,23,25,28,30,33,34,36,39,41,44,46,49,50,52,55,57,60,62,65,...) = ψ0(ψ{ψ{ψ{ψ1(0)}(0)}(0))}(0))
(1,2,4,7,9,12,14,17,19,22) = (1,2,4,7,9,12,14,17,19,21,24,26,29,31,34,36,38,41,43,46,48,51,53,55,58,60,63,65,68,70,...) = ψ0(ψ{ψ{ψ{ψ2(0)}(0)}(0))}(0))
(1,2,4,7,9,12,14,17,19,22,24,27) = (1,2,4,7,9,12,14,17,19,22,24,26,29,31,34,36,39,41,44,46,48,51,53,56,58,61,63,66,68,70,73,75,78,80,83,85,88,90...) = ψ0(ψ{ψ{ψ{ψ{ψ2(0)}(0)}(0)}(0))}(0))
And now, we finally reached the extended Buchholz's ordinal! This is equivalent to (1,2,4,7,10) in Y-sequence using my alternative formulation.
This expands: (1,2,4,7,9,12,14,17,19,22,24,27,29,32,...)
Let's go back and analyze the original formulation and see the awkwardness by comparing it with Buchholz's function. The excerpt of the original expansion and Buchholz's function comparison is as follows:
(1,2,4,7,4,7) => (1,2,4,7,4,6,8,10,...) => ψ0(ψ2(0) + ψ1(ψ2(0)))
(1,2,4,7,4,7,4,7) => (1,2,4,7,4,7,4,6,8,10,...) => ψ0(ψ2(0) + ψ1(ψ2(0)) + ψ1(ψ2(0)))
(1,2,4,7,5) => (1,2,4,7,4,7,4,7,4,7,...) => ψ0(ψ2(0) + ψ1(ψ2(0) + ψ0(0)))
(1,2,4,7,5,5) => (1,2,4,7,5,4,7,5,4,7,5,4,7,5,...) => ψ0(ψ2(0) + ψ1(ψ2(0) + ψ0(0) + ψ0(0)))
(1,2,4,7,5,6) => (1,2,4,7,5,5,5,5,...) => ψ0(ψ2(0) + ψ1(ψ2(0) + ψ0(ψ0(0))))
(1,2,4,7,5,7) => (1,2,4,7,5,6,7,8,...) => ψ0(ψ2(0) + ψ1(ψ2(0) + ψ0(ψ1(0))))
(1,2,4,7,5,7,10) => (1,2,4,7,5,7,9,11,13,...) => ψ0(ψ2(0) + ψ1(ψ2(0) + ψ0(ψ2(0))))
(1,2,4,7,6) => (1,2,4,7,5,7,10,8,10,13,11,13,16,...) => ψ0(ψ2(0) + ψ1(ψ2(0) + ψ1(0)))
(1,2,4,7,6,6) => (1,2,4,7,6,5,7,10,9,8,10,13,12,11,13,16,15,...) => ψ0(ψ2(0) + ψ1(ψ2(0) + ψ1(0) + ψ1(0)))
(1,2,4,7,6,7) => (1,2,4,7,6,6,6,6,...) => ψ0(ψ2(0) + ψ1(ψ2(0) + ψ1(ψ0(0))))
(1,2,4,7,6,8) => (1,2,4,7,6,7,9,12,11,12,14,17,16,17,19,22,21,...) => ψ0(ψ2(0) + ψ1(ψ2(0) + ψ1(ψ1(0))))
(1,2,4,7,6,9) => (1,2,4,7,6,8,10,12,...) => ψ0(ψ2(0) + ψ1(ψ2(0) + ψ1(ψ2(0))))
(1,2,4,7,7) => (1,2,4,7,6,9,8,11,10,13,...) => ψ0(ψ2(0) + ψ2(0))
...
(1,2,4,8) => ψ0(ψ{ω}(0))
(1,2,4,8,4) => ψ0(ψ{ω}(0) + ψ1(0))
(1,2,4,8,4,4) => ψ0(ψ{ω}(0) + ψ1(0) + ψ1(0))
(1,2,4,8,4,5) => ψ0(ψ{ω}(0) + ψ1(ψ0(0)))
(1,2,4,8,4,6) => ψ0(ψ{ω}(0) + ψ1(ψ1(0)))
(1,2,4,8,4,7) => ψ0(ψ{ω}(0) + ψ1(ψ2(0)))
(1,2,4,8,4,8) => ψ0(ψ{ω}(0) + ψ1(ψ{ω}(0)))
(1,2,4,8,5) => ψ0(ψ{ω}(0) + ψ1(ψ{ω}(0) + ψ0(0)))
(1,2,4,8,6) => ψ0(ψ{ω}(0) + ψ1(ψ{ω}(0) + ψ1(0)))
(1,2,4,8,6,6) => ψ0(ψ{ω}(0) + ψ1(ψ{ω}(0) + ψ1(0) + ψ1(0)))
(1,2,4,8,6,7) => ψ0(ψ{ω}(0) + ψ1(ψ{ω}(0) + ψ1(ψ0(0))))
(1,2,4,8,6,8) => ψ0(ψ{ω}(0) + ψ1(ψ{ω}(0) + ψ1(ψ1(0))))
(1,2,4,8,6,9) => ψ0(ψ{ω}(0) + ψ1(ψ{ω}(0) + ψ1(ψ2(0))))
(1,2,4,8,6,10) => ψ0(ψ{ω}(0) + ψ1(ψ{ω}(0) + ψ1(ψ{ω}(0))))
(1,2,4,8,7) => ψ0(ψ{ω}(0) + ψ2(0))
(1,2,4,8,7,7) => ψ0(ψ{ω}(0) + ψ2(0) + ψ2(0))
(1,2,4,8,7,8) => ψ0(ψ{ω}(0) + ψ2(ψ0(0)))
(1,2,4,8,7,9) => ψ0(ψ{ω}(0) + ψ2(ψ1(0)))
(1,2,4,8,7,10) => ψ0(ψ{ω}(0) + ψ2(ψ2(0)))
(1,2,4,8,7,11) => ψ0(ψ{ω}(0) + ψ3(0))
(1,2,4,8,7,12) => ψ0(ψ{ω}(0) + ψ2(ψ{ω}(0)))
(1,2,4,8,8) => ψ0(ψ{ω}(0) + ψ{ω}(0))
(1,2,4,8,9) => ψ0(ψ{ω}(ψ0(0)))
(1,2,4,8,10) => ψ0(ψ{ω}(ψ1(0)))
(1,2,4,8,10,14) => ψ0(ψ{ω}(ψ{0}(0)))
(1,2,4,8,11) => ψ0(ψ{ω+1}(0))
(1,2,4,8,11,16) => ψ0(ψ{ω+2}(0))
(1,2,4,8,12) => ψ0(ψ{ω2}(0))
(1,2,4,8,12,12) => ψ0(ψ{ω3}(0))
(1,2,4,8,12,13) => ψ0(ψ{ω^ω}(0))
(1,2,4,8,12,14) => ψ0(ψ{ψ0(ψ1(0))}(0))
...
WTF?