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.

\(\circ\) 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 \(a\). \(v(a)\) should be solved when you have to solve it

Bracket Notation
\(a\$b\bullet=(a+b)\$\bullet\)

\(a\$\circ[0]\bullet\circ=a\$\circ a\bullet\circ\)

\(a\$\circ[b+1]\bullet\circ=a\$\circ[b][b]...[b][b]\bullet\circ\) with a b's

Extended Bracket Notation
I've added 2 rules.

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

\(a\$\circ[\diamond]\circ=a\$\circ\diamond\diamond...\diamond\diamond\circ\) with a \(\diamond\)s

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

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

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

Linear Array Notation
\(\bullet[\bullet0]_\bullet\bullet = \bullet[\bullet]_\bullet\bullet\)

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

\(a\$\bullet[\text{◆},0,b,\bullet]_c\bullet = a\$\bullet[\text{◆},[\text{◆},[\text{◆},...,b-1,\bullet]_c,b-1,\bullet]_c,b-1,\bullet]_c\bullet\) with a nests

Extended Array Notation
\(a\$[0 \rightarrow_{b\bullet}c\bullet]_d = a\$[a\rightarrow_{(b-1)\bullet}a...a\rightarrow_{(b-1)\bullet}a \rightarrow_{b\bullet}c-1\bullet]_d\)