25 = 4! + (4/4)[sup]4[/sup]
26 = 4! + (4+4)/4
27 = 4! + (4-(4/4)
28 = 4! + 4[sup]4-4[/sup]
29 = 4! + 4 + (4/4)
Kind of pointless, but this is fun.
umm… way to get off to a bad start, dude. Try 4+4-4-4.
He probably meant (4+4)-(4+4) = 0.
Oops. Patch 28 to read
28 = 4! + 4[sup]4/4[/sup]
Sorry about the radical signs.
Sunspace, a -> 2 -> b is supposed to be
a^(a^(a^…^a…)), where there are b a’s. We can call this a^^b.
a -> 3 -> a is supposed to be a^^(a^^(a^^…^^a…)), where there are b a’s. We can call this a^^^b. Etc. That’s how you do it if there are only 2 arrows.
If there are 3 or more, then you can simplify it by these rules:
[ol][li]If the last number is 1, then you can remove it, along with the arrow.[/li][li]If the second to last number is 1, then you can remove the last two numbers and arrows.[/li][li]If they’re both greater than one, then you do the following:[/li]~Decrease the last number by 1.
~Replace the second to last number with the original expression itself, except with the second to last number decreased by 1.
[/ol]
The simplification rules are mutually exclusive; only one of them can ever be used on a given chained arrow expression at once.
Those of you discussing chained arrow notation might be interested in this thread, where I used a similar recursive notation to construct some very fast-growing functions.
(4-4), [symbol]Ö[/symbol]4, 4, …
Or how about 4444!!! ? (I know, I know, it’d take a lot more to beat ->s, but you get the idea)
Mind-boggling. And intensely interesting. Thanks, all.
And I’m now addicted to Mathworld.
4! = 123*4
but the number 4 itself can just be an elementary definition.
That is, you look at any group of 4 objects and call it one unit and refuse to acknowledge any smaller group as a single unit.
But, with factorials, those smallers units have to come into play.
Do we have 29 yet?
4! + 4 + (4/4)
Oh. Oops.
Not if you allow 4/4, or sqrt(4), or even 4 - 4.
30= 4! + 4 + (sqrt(4))
31= 4 x 4 x (sqrt(4)) - 40
32= 4 x 4 x (sqrt(4)) x 40
Oops…those zeroes are supposed to be superscripted…I thought I knew how to but I guess I was wrong. Can anyone enlighten me on how to do it?
Use [sup] and [/sup].
33= 4! x (sqrt(4)) - 4[sup]2[/sup] + 4[sup]0[/sup]
34= 4 x 4 x (sqrt(4)) + (sqrt(4))
35=
36= 4 x 4 x (sqrt(4)) + 4
Eh. I used 2 roots and 2 and 0 powers so none of those work if you disallow them. Ah well.
This is actually a very old puzzle from rec.puzzles.