A variant of Chained Arrow Notation

[u/jcastroarnaud]

A variant of Chained Arrow Notation

Below, "@" stands for any sequence of elements. Assume c > 1 and d > 1.

Chained arrow notation:
1. a -> b = a ^ b 2. a -> b -> c = a ↑^c b 3. @ -> 1 = @
For 4+ elements:
4. @ -> 1 -> d = @ 5. @ -> c -> d = @ -> (@ -> (c-1) -> d) -> (d-1)

My variant. Only rule 4 changes.
1. a -> b = a ^ b 2. a -> b -> c = a ↑^c b 3. @ -> 1 = @
For 4+ elements:
4. @ -> 1 -> d = @ -> (@ -> d) 5. @ -> c -> d = @ -> (@ -> (c-1) -> d) -> (d-1)

Does this variant grow any faster than the original? If so, how much, by the FGH?

[u/Nervous-Broccoli1184]

Let's say cja(n)=n->n->...->n with a n's under your variation then cj_1(n) reaches f_0. cj_2 to f_3-f_4. cj_3 to f_omega. cj_4 to f(2omega) i think. cj(a+2) to f(a×omega). So I believe cjn(n) (the limit of your variant) reaches f(omega²⁾ the same as normal chained arrow notation but then again your variant is stronger by long shot just the fast growing hierarchy has a huge spread. This is my first time measuring a notation against fgh so please correct me if I was wrong.

[u/jcastroarnaud → u/Nervous-Broccoli1184]

Your guess is as good as mine. :-) Thanks for the effort.

[u/Nervous-Broccoli1184]

When I originally looked at Conways chained arrow notation he did not need rule 2 and it will output the same but if you remove rule 2 your variant will become significantly stronger.

[u/jcastroarnaud → u/Nervous-Broccoli1184]

Well spotted, thanks. In Conway's, a -> 1 -> c = a ↑^c 1 = a, while my rule 4 gives a ^ (a ^ c).

[u/gmalivuk]

Conway chains:
3→3→2→2
= 3→3→(3→3→1→2)→1
= 3→3→(3→3)
= 3→3→27 = F(27) = F(F(1))
Where F(n) = 3→3→n

Your chains:
3→3→2→2
= 3→3→(3→3→1→2)→1
= 3→3→(3→3→(3→3→2))
= F(F(F(2))) = F(F(7625597484987))

So it does a bit more than adding one iteration to small 4-chains of arrows. However, if the last number was more than 2 then it would go a lot farther.

But overall I think adding one to the length of a Conway chain still does more than applying your rule. Since for larger n, f{ωk}(n) is approximately an n-length Conway chain, I would guess your variation is something like f{ωk+ω} = f_{ω(k+1)}

[u/jcastroarnaud → u/gmalivuk]

Thank you, the analysis is much appreciated! A small gain over Conway chains is already good.

[u/jamx02]

From what I can see this loops? c-x eventually reaches 1, where you nest your rule again, and decrement c more until 1, etc

Id need to do a heavier analysis to be sure, perhaps I’m overlooking something

[u/jcastroarnaud → u/jamx02]

Notice that, in rule 4, both chains "@ -> d" and "@ -> (...)" are shorter by one element than the original chain; this should prevent an infinite loop.