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Basic

H<n> = n^n

H<...,n,1> = H<...,n>

H<...,n,m> = H<...,H<...,n,m-1>,m-1>

Extension 1

H<x[1]...[1]n[1]m> = H<[1]...[1]n,n,n,...,n,n,n[1]m-1> with n entries

H<x[1]...[1]...,n,h[1]1> = H<x[1]...[1]...,n,h>

H<x[1]...[1]...,n,1[1]m> = H<x[1]...[1]...,n[1]m>

H<x[1]...[1]...,n,h[1]m> = H<x[1]...[1]...,H<x[1]...[1]n,h-1[1]m>,h-1[1]m>

Extension 2

Rules are same as Extension 1 but H<x[l]...[l]n[l]m> = H<x[l]...[l]n[l-1]...[l-1]n[l]m-1> with n entries

Extension 3

Let & is any array

H<&[0[...[&]...]0]&> = Check rules of basic and extensions 1 and 2 to resolve the array in the dimension (and maybe &)

Extension 4

-- UNDER CONSTRUCTION --

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