In the last article we covered hyperoperations, which are expressed using up-arrow notation. In this article we will cover the various extensions to up-arrow notation.
Excelguru's Extended Arrow Notation
On a forum elsewhere on the internet, a user made an extension to up-arrow notation, which is defined as follows:
a{^}b = a^^^...^^^a w/b ^s
a{^)^b = a{^}a{^}a...a{^}a{^}a w/b a's
1{^}(any number) is equal to 1 and 2{^}(any number) is equal to 4. So we will begin with:
3{^}1 = 3^3 = 33 = 27
3{^}2 = 3^^3 = 3^3^3 = 327 = 7,625,597,484,987
3{^}3 = 3^^^3 = 3^^(3^^3) = 3^^7,625,597,484,987
3{^}4 = 3^^^^3 = 3^^^(3^^^3) = 3^^^(3^^7,625,597,484,987) = 3^^(3^^(3^^(...(3^^(3^^3))...))) w/3^^7,625,597,484,987 3s
3{^}5 = 3^^^^^3 = 3^^^^(3^^^^3) = 3^^^^(3^^^(3^^^3)) = 3^^^^(3^^^(3^^7,625,597,484,987))
3{^}6 = 3^^^^^^3 = 3^^^^^(3^^^^^3) = 3^^^^^(3^^^^(3^^^^3))
4{^}1 = 4^4 = 44 = 256
4{^}2 = 4^^4 = 4^4^4^4 = 4^4^256 ~ 108.0723*10^153
4{^}3 = 4^^^4 = 4^^(4^^(4^^4))
4{^}4 = 4^^^^4 = 4^^^(4^^^(4^^^4))
3{^}^1 = 3
3{^}^2 = 3{^}3 = 3^^^3 = 3^^7,625,597,484,987
3{^}^3 = 3{^}3{^}3 = 3{^}(3^^7,625,597,484,987) = 3^^^^^^^^^^...^^^^^^^^3 (w/ 3^^^3 ^'s)
To prevent the degeneration to 4 for base 2, I would suggest simply changing the definition like so:
a{^}b = a^^^^^...^^^^^a w/ b ^s, a >= 3
2{^}b = 2^^^^^...^^^^^3 w/ b ^s
We can then have:
2{^}1 = 23 = 8
2{^}2 = 2^^3 = 24 = 16
2{^}3 = 2^^^3 = 2^^4 = 65,536
2{^}4 = 2^^^^3 = 2^^^4 = 2^^65,536
2{^}^2 = 2{^}2 = 16
2{^}^3 = 2{^}16 = 2^^^^^^^^^^^^^^^^3
In 1995, J. H. Conway and R. K. Guy defined Conway chained arrow notation (which we will cover in more detail in a later article), which is defined like so:
a -> b = ab
a -> b -> c = a^^...^^b with c ^s
a -> b -> c -> d = a -> b -> (a -> b -> c-1 -> d) -> d-1
Similar to power towers, if any number in a chained expression is 1, it and all further numbers are ignored.
Here are some examples of chained expressions with 3 terms:
2 -> 3 -> 4 = 2^^^^3 = 2^^^4 = 2^^65,536
3 -> 2 -> 3 = 3^^^2 = 3^^3 = 3^3^3 = 327 = 7,625,597,484,987
5 -> 4 -> 3 = 5^^^4 = 5^^(5^^(5^^5))