User blog:MachineGunSuper/The Hyperfactorial Semi-Array notation

This is an extention to the hyperfactorial notation, just to the m!n part

(HSAN)

\(Ϯ (n)\)= n!n!n!n....n!n, n times.

\(Ϯ (3)\) =3!3!3

\(Ϯ (n,m)\) = \(Ϯ (n)\)!\(Ϯ (n)\)!\(Ϯ (n)\)!...!\(Ϯ (n)\), m times.

\(Ϯ (n,m,p)\) = \(Ϯ (n,m)\)!\(Ϯ (n,m)\)!\(Ϯ (n,m)\)....!\(Ϯ (n,m)\), p times

It doesn't matter how many entries we have, the last one will determine how many times it will repeat without it.


 * If the last entry is 1, ignore it

\(Ϯ (n,#)\) = \(Ϯ (n,n,n,n....,n)\), where n repeats n times.

\(Ϯ (n,#,a)\) = \(Ϯ (n,n,n,n....,n(n times),a,a,a,a...,a(a times))\)

\(Ϯ (3,#,4,5)\) = \(Ϯ (3,3,3,4,4,4,4,5,5,5,5,5)\)

When having multiple numbers and a #, it just repeats the \(Ϯ (n,#)\) operation on all of the elements.

\(Ϯ (n,#,#)\) = \(Ϯ (n,#,n,#,n,#,.....,#,n,#)\), where the pair "n,#" repeats n times.

\(Ϯ (3,#,#)\) = \(Ϯ (3,#,3,#,3,#)\) = \(Ϯ (3,3,3,3,3,3,3,3,3)\)

\(Ϯ (n,#,#,#)\) = \(Ϯ (n,#,#,n,#,#,....)\), where the pair "n,#,#" repeats n times.

\(Ϯ (3,#,#,#)\) = \(Ϯ (3,#,#,3,#,#,3,#,#)\) = \(Ϯ (3,3,3,3....,3,3,3)\), with 27 3's

For an arbitrary number x, \(Ϯ (n,#,#,#,#,#,#,#,.....)\) (with x #'s) = \(Ϯ (n,#,#,#.....,#(with x-1 #'s),n,#,#,#,#,#,#...)\), where the pair "n,#,#,#...#,#,#(with x-1 #'s) repeats n times. 

\(Ϯ (3,#,#,#,#,#)\) = \(Ϯ (3,#,#,#,#,3,#,#,#,#,3,#,#,#,#)\)

\(Ϯ (n,@)\) = \(Ϯ (n,#,#,#,#....,#,#,#)\), with n #'s

Same, \(Ϯ (n,@,@)\) = \(Ϯ (n,@,n,@,n,@,....,n,@)\), where the pair "n,@" repeats n times.

Question of the blog post
Try approximating the value of \(Ϯ (10100,@,@,@)\)