User blog:Wythagoras/Dollar Function: new version

Okay, here is a new (complete?) version of Dollar Function.

Symbols
\(\bullet\) can be anything, but it has a higher level than the thing your're expanding.

\(\circle\) is a set of brackets

\(\diamond\) is a bracket ( with content ) with only lower level brackets surrounding it

\(\text{◆}\) is a group of zeroes

\(\text{◈}\) is a subarray with not any non-nested numbers. Example: [[0]2][0] Counterexample: [0]3

\(v(a)\) is the subarray \(a\) after applying a rule to it once. Note that \(v(a)\) has a lower level than \(a\) itself, but a higher level than any other subarray that has a lower level than \(a\)

Bracket Notation
\(a\text{$}b\bullet=(a+b)\text{$}\bullet\)

\(a\text{$}\circle[0]\bullet\circle=a\text{$}\circle a\bullet\circle\)

\(a\text{$}\circle[b+1]\bullet\circle=a\text{$}\circle[b][b]...[b][b]\bullet\circle\) with a b's

Extended Bracket Notation
I've added 2 rules.

\(a\text{$}[0]_{a\bullet}\bullet=a\text{$}[[...[[0]_{(a-1)\bullet}]_{(a-1)\bullet}...]_{(a-1)\bullet}]_{(a-1)\bullet}\bullet\) with a nests

\(a\text{$}\circle[\diamond]\circle=\circle\diamond\diamond...\diamond\diamond\circle\) with a \(\diamond\)s

\(a\text{$}\circle[\diamond]_{b\bullet}\circle=\circle[[...[[\diamond]_{(b-1)\bullet}]_{(b-1)\bullet}...]_{(b-1)\bullet}]_{(b-1)\bullet}\circle\)

\(a\text{$}[c\bullet]_{b\bullet}\bullet=a\text{$}[[...[[[c-1\bullet]_{b\bullet}]_{(b-1)\bullet}]_{(b-1)\bullet}]...]_{(b-1)\bullet}]_{(b-1)\bullet}\bullet\)