ArcAN

Arcioko Array Notation [NOT2]

• AKA: ~!@#

Definition

Foundations

{a} = a ^^ . . (a arrows) . . ^^ a [Up-arrow Notation]

{a, b, c, . . . , 1} = {a, b, c, . . .}

Features

{a, b} = {{ . . (b number of layers) . . {{a}} . . }}

Or in another words: {a, 1} = {a}, {a, b+1} = {{a, b}}

For multiple arguments, {First argument, . . . , Last argument+1} = {First argument, . . ., Antepenultimate argument, {First argument, . . . , Last Argument}}

Now, here comes fun:

a~b = {a, a, . . (b-1 number of a's) . . , a, a}

a(~^1)b = a~b

a(~^(c+1))b = a(~^c)(a(~^(c+1))(b-1))

To clarify, the ~ (and ~^n in general) operator is handled from right to left

Now lets go crazy

a~!b = a(~^b)(a~!(b-1)), a~!1 = a~a

a(~!^1)b = a~!b

a~(!^(c+1))b = a~(!^c)(a~!^c(b-1)), a~(!^c)1 = a~a

Did someone say diagonalization?

a~@b = a~(!^(b-1))(a~$(b-1)), a~$1 = a~a

Now we go REALLY crazy

a~#(1)b = a~!a

a(~#(k)^1)b = a~!a

a~(#(k)^(c+1))b = a~(#(k)^c)(a~#(k)^c(b-1)), a~(#(k)^c)1 = a~a

a~#(k+1)b = a~(#(k)^(b-1))(a~#(k)(b-1)), a~#(k)1 = a~a

Hashes are nice, but what about double hashes?

a~##(1)b = a~#(a)a

a(~##(k)^1)b = a#(a)a

a~(##(k)^(c+1))b = a~(##(k)^c)(a~##(k)^c(b-1)), a~(##(k)^c)1 = a~a

a~##(k+1)b = a~(##(k)^(b-1))(a~##(k)(b-1)), a~##(k)1 = a~a

Double hashes are nice, but what about TRIPLE hashes?

a~###(1)b = a~##(a)a

a(~###(k)^1)b = a##(a)a

a~(###(k)^(c+1))b = a~(###(k)^c)(a~###(k)^c(b-1)), a~(###(k)^c)1 = a~a

a~###(k+1)b = a~(###(k)^(b-1))(a~###(k)(b-1)), a~###(k)1 = a~a

You guessed it, lets diagonalize

a~#^c(1)b = a~#^c(k-1)(a)a

a(~#^c(k)^1)b = a#^(c-1)(a)a

a~(#^c(k)^(c+1))b = a~(#^c(k)^c)(a~#^c(k)^c(b-1)), a~(#^c(k)^c)1 = a~a

a~#^c(k+1)b = a~(#^c(k)^(b-1))(a~#^c(k)(b-1)), a~#^c(k)1 = a~a

Now, you may ask: "Where did the damn arrays go?"

Dont worry:

~{a,b,c . . . ,x,y,z} = a~b~c~ . . . ~x~y~z

~!{a,b,c . . . ,x,y,z} = a~!b~!c . . . ~!x~!y~!z

For any two argument function, f(x,y):

f{a,b,c . . . ,x,y,z} = f(a,f(b,f(c, . . . (x,f(y,z))) . . . ))

Now lets introduce, the dollar sign!

$f{a,b,c, . . . x,y,z} =

Let A = {a,b,c, . . . }

Now, let A(n) be equal to A but truncated at n (Ex: If A = {1,2,3,4,5}, A(3) = {1,2,3})

$fA = f(A(1), f(A(2), f(A(3) . . . , f(a)))) . . . )))

Or more formally:

Let F(n) = f(A(N-n), f(n-1)), F(0) = A(N) Where N is the number of entries inside A

$fA = F(N)

Now lets go beyond

First of all, <(Expression)> means that anything there will be converted into an array and

%g(nfA) = f<g(n,A)>

unfinished