Hyperoperation Golfing [duplicate]

Question

Introduction

In mathematics, the hyperoperation sequence is an infinite sequence of arithmetic operations (called hyperoperations) that starts with the unary operation of successor (n = 0), then continues with the binary operations of addition (n = 1), multiplication (n = 2), and exponentiation (n = 3), after which the sequence proceeds with further binary operations extending beyond exponentiation, using right-associativity.

_(Source)

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Challenge

Your challenge is code this sequence, given 3 inputs, n, a, and b, code a function such that $${\displaystyle H_{n}(a,b)={\begin{cases}b+1&{\text{if }}n=0\\a&{\text{if }}n=1{\text{ and }}b=0\\0&{\text{if }}n=2{\text{ and }}b=0\\1&{\text{if }}n\geq 3{\text{ and }}b=0\\H_{n-1}(a,H_{n}(a,b-1))&{\text{otherwise}}\end{cases}}}$$ _{(Also from Wikipedia.)}

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Input

3 positive decimal integers, n, a, and b, taken from STDIN, function or command line arguments, in any order. Make sure to specify this in your answer

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Output

The result of applying \$H_{n}(a,b)\$ with the inputs

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Example inputs and outputs

Input: 0, 6, 3 Output: 4

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Input: 4 5 2 Output: 3125

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Restrictions

• Your program/function should take input in base 10
• Don't use any built in function that already provides H(n, a, b)
• Standard loopholes apply

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This is code-golf, so shortest code in bytes wins!

Answer 1

Dyalog APL, 35 bytes

{⍺=0:⍵+1⋄⍵=0:⍺⍺⍵1⊃⍨3⌊⍺⋄(⍺-1)∇⍺∇⍵-1}

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Answer 2

JavaScript (ES6),  49  47 bytes

Takes input as (a)(b,n).

a=>g=(b,n)=>n*b?g(g(b-1,n),n-1):-~[b,a-1,-1][n]

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Commented

a =>                  // main function taking a
  g = (b, n) =>       // g = recursive function taking b and n
    n * b ?           // if neither n nor b is equal to 0:
      g(              //   go recursive:
        g(b - 1, n),  //     b = g(b - 1, n)
        n - 1         //     decrement n
      )               //   end of recursive call
    :                 // else:
      -~[             //   this is a leaf node:
        b,            //     if n = 0, return b + 1
        a - 1,        //     if n = 1, return (a - 1) + 1 = a
        -1            //     if n = 2, return -1 + 1 = 0
      ][n]            //     if n > 2, return -~undefined = 1

Answer 3

Elixir, 86 bytes

(assumes functions can take a reference to itself as an argument)

fn 0,_,b,_->b+1
1,a,0,_->a
2,_,0,_->0
_,_,0,_->1
n,a,b,f->f.(n-1,a,f.(n,a,b-1,f),f)end

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