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Copy notation
TypeLinear
Based onrepeating digits
Growth rate\(f_{\omega3}(n)\)

Copy notation is a notation created by the Googology Wiki user, "TechKon" (Formerly SpongeTechX) used to define copied or repeated digits/numbers. 

Definition[]

Copy notation simply defines the amount of digits in a number which are all the same. You can simplify the number \(5,555\) with this notation by using \(n[m]\). n represents the digit you are using, and the m represents the amount of them. In this case, \(5,555\) would be equal to \(5[4]\) in copy notation because there are four fives. \(8,888,888\) would be equal to \(8[7]\) because there are seven eights.

Basically, if \(n\) is a value, \(m\) repeated digits of \(n\) = \(n[m]\), or \(n[m]\) = \(m\) \(n\)'s in copy notation.

All of this applies for 2, 3, etc.-digit numbers. \(10[10]\) = \(10,101,010,101,010,101,010\). That is ten tens.

Examples[]

  • 2[4] = 2,222 or four twos
  • 4[8] = 44,444,444 or eight fours
  • 9[2] = 99 or two nines
  • 15[12] = 151,515,151,515,151,515,151,515 or twelve fifteens

Extension[]

SpongeTechX extended it to multiple brackets.

a[[b]] = a[a[...[a[a]]...]] with b a's

a[[[b]]] = a[[a[[...[[a[[a]]]]...]]]] with b a's

a[[[[b]]]] = a[[[a[[[...[[[a[[[a]]]]]]...]]]]]] with b a's

And so on. Now

defines a[b,c] = a[[...[c]...]] with b pairs of brackets.

Then:

a[b,c,1] = a[b,c]

a[b,c,d] = a[a[b,c,d-1],a[b,c,d-1],d-1]

a[b,c,d,1] = a[b,c,d]

a[b,c,d,e] = a[a[b,c,d,e-1],a[b,c,d,e-1],a[b,c,d,e-1],e-1]

And so on. Then one last extension:

a[b#c] = a[b,b,...,b,b] with c b's

a[b##1] = a[a[b#b]#a[b#b]]

a[b##2] = a[a[b##1]#a[b##1]]

a[b##3] = a[a[b##2]#a[b##2]]

a[b##m] = a[a[b##(m-1)]#a[b##(m-1)]]

a[b###1] = a[a[b##b]##a[b##b]]

a[b###2] = a[a[b###1]##a[b###1]]

a[b###3] = a[a[b###2]##a[n###2]]

a[b###m] = a[a[b###(m-1)]##a[b###(m-1)]]

Approximation[]

\(\frac{a+1}{10^{\left\lfloor\log a\right\rfloor}}10^{b(1+\left\lfloor\log a\right\rfloor)-1}\ge{a[b]}\ge\frac{a}{10^{\left\lfloor\log a\right\rfloor}}10^{b(1+\left\lfloor\log a\right\rfloor)-1}\)

\(a\uparrow\uparrow{b}\ge{a[[b]]}\)

\(a\uparrow^c{b}\ge{a[c,b]}\)

Relation with Hypermathematics[]

Copy Notation has a relation with Hypermathematics. That is, a[b] is equal to a*b using Hypermathematics.

Sources[]

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