HC Notation is a googological notation derived from Orteil's popular online game Cookie Clicker and takes the form \([ a,b ]_{HC}\). After a person resets the game they gain what is called "heavenly chips", the more heavenly chips you gain the more each successive heavenly chip costs.

The formula to find how many cookies is takes to obtain with a given number of heavenly chips is as follows:

\(\text{Cookies needed (total)}=\frac{HC(HC + 1)}{2}\cdot 10^{12}\) HC takes it a step further by adding different "tier's" of heavenly chips that follow a similar formula

\(\text{Tier 1 HC (Regular HC)}=\frac{\text{Tier 2 HC}(\text{Tier 2 HC} +1)}{2}\cdot 10^{12}\)

The Notation [a,b]HC is basically saying how many cookies do I need if I want "a" amount of Tier "b" Heavenly Chips

One way to solve the bracket is to use the recursive method. For example, if you want to solve \([ 4,3 ]_{HC}\) you first calculate how many Tier 2 chips that would require, then Tier 1, then finally the amount of cookies.

The second way is to use this formula:

\([ a,b ]_{HC}=\frac{(a^{2}+a)^{2^{b-1}}\cdot(10^{12})^{2^{b}-1}}{2^{2^{b}-1}}\)


\([ 4,3 ]_{HC}=\text{ 4 Tier 3 HC}=10^{13}\text{ Tier 2 HC}=5*10^{37}\text{ Tier 1 HC}=1.25*10^{87}\text{ Cookies}\)

\([ 4,3 ]_{HC}=\frac{(4^{2}+4)^{2^{3-1}}\cdot(10^{12})^{2^{3}-1}}{2^{2^{3}-1}}=1.25\cdot 10^{87}\)

Multiple EntriesEdit

\([ a,b,c ]_{HC}\) can be simplified to \([ a,[ b,c ]_{HC} ]_{HC}\)

\([ a,b,c,d ]_{HC}=[ a,[ b,c,d ]_{HC} ]_{HC}=[ a,[ b,[ c,d ]_{HC} ]_{HC} ]_{HC}\)

You get the point. You always work from right to left.

Multiple DimensionsEdit

2 Dimensional BracketEdit

\([ a,b (1) c ]_{HC}\) means there are two brackets each on separate 1-dimensional structures, one is \([ a,b ]_{HC}\) and the other is \([ c ]_{HC}\) or \([ c,1 ]_{HC}\)

In order to solve this, solve \([ c,1 ]_{HC}\) and repeat "a,b" \([ c,1 ]_{HC}\) times.


\([ 5,2 (1) 4 ]_{HC}\)

\([ 4,1 ]_{HC}=10^{13}\)

\([ 5,2,5,2,\ldots,5,2]_{HC}\) where the "5,2" is repeated \(10^{13}\) times

3 Dimensional BracketEdit

These guys follow the same rules as the 2-D bracket

\([ a,b (1) c,d (2) e,f (1) g,h ]_{HC}\)

In order to solve a 3-D bracket, locate and solve the last 2-D structure

\([ a,b (1) c,d (2)\mathbf{ e,f (1) g,h} ]_{HC}\)

\([ e,f (1) g,h ]_{HC}=\mathbf{A}\)

\([ a,b (1) c,d (2) \mathbf{A} ]_{HC}\)

You must repeat the preceding 2-D structure A times

\([ a,b (1) c,d (1) a,b (1) c,d (1)\ldots(1) a,b (1) c,d) ]_{HC}\)

Higher dimensionsEdit

To solve an n-dimensional bracket follow these steps

  1. Locate the last (n-1)-dimensional structure
  2. Solve it to receive some number "a"
  3. Repeat the preceding (n-1)-dimensional structure "a" times with an (n-2) dimension separator, you now have an n-dimensional bracket with one less (n-1)-dimensional structure
  4. Repeat until bracket is solved


\([ 1,4 (1) 3,5 (3) 2,4 (1) 6,2 (2) 3,1 ]_{HC}\)

Since this is a 4-D bracket, we need to locate the last 3-D structure

\([ 1,4 (1) 3,5 (3) \mathbf{2,4 (1) 6,2 (2) 3,1} ]_{HC}\)

Solve this 3-D bracket (see above for how to)

\([ 2,4 (1) 6,2 (2) 3,1 ]_{HC}=\mathbf{A}\)

\([ 1,4 (1) 3,5 (3) \mathbf{A} ]_{HC}\)

Now we have to repeat [1,4 (1) 3,5]HC A amount of times with a (2) separator. (4-2=2)

\([ 1,4 (1) 3,5 (2) 1,4 (1) 3,5 (2)\ldots(2) 1,4 (1) 3,5]_{HC}\) with A repetitions

Megadimensions Edit

These take the form of \([a,b ((1)) c,d]_{HC}\)

They can be simplified in the following manner

\([a,b ((1)) c,d]_{HC} = [a,b ([c,d]_{HC}) c,d]_{HC}\)

This can only be done when there is a "1" in the double parentheses, If there is a ((2)) then solve just as your would a 2-D bracket


\([3,1 ((1)) 1]_{HC}\)

\([1,1]_{HC} = 10^{12}\)

\([3,1 (10^{12}) 1]_{HC}\)

When you have a case like this:

\([a,b ((1)) c,d ((1)) e,f]_{HC}\)

\([c,d ((1)) e,f]_{HC} = [c,d ([e,f]_{HC}) e,f]_{HC} = \mathbf{A}\)

\([a,b ((1)) \mathbf{A}]_{HC} = [a,b (\mathbf{A}) \mathbf{A}]_{HC}\)

There can be more than one parentheses

\([a,b (((1))) c,d]_{HC} = [a,b ([c,d]_{HC}) c,d]_{HC}\)

\([a,b ((((1)))) c,d]_{HC} = [a,b ((([c,d]_{HC}))) c,d]_{HC}\)

They can be written with an exponent denoting how many set of parentheses there are

\([a,b (1)^3 c,d]_{HC} = [a,b ([c,d]_{HC})^2 c,d]_{HC}\)

\([a,b (1)^4 c,d]_{HC} = [a,b ([c,d]_{HC})^3 c,d]_{HC}\)

Psi function Edit

Ψ(a,b,c) = ((...((c)_a)_a...)_a)_a, where ()_a is a-th order of the separator

The Psi function has 3 arguments, the first one "a" represents the order of the separator, the second argument represents how many separators are nested in one another, and finally the third argument represents the number inside the separator.

Example Edit



Sources Edit

AlphaLambda website

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