User blog comment:Eners49/The secret 0th hyper-operator?/@comment-35470197-20180726231426

The hyper operators are defined by the repetition of the previous hyper operators, as the power is the repetition of the multiplication and the multiplication is the repetition of the addition.

The addition is given as the repetition of the successor \(S\), which sends \(n\) to the next number \(S(n)\) of \(n\). Say, \(n+0 = n, \ n+1 = S(n+0) = S(n), \ n+2 = S(n+1) = S(S(n))\), and so on.

Put \(S(n,m) = S(n)\). Then \(n+m\) with \(m = S(m') > 0\) is described as \(S(n,n+m'\). You can observe similar rules for other operators: \(n \times m = n + (n \times m'), \ n^m = n \times (n^{m-1})\), and so on. Therefore \(S\) is a candidate of the\(0\)-th operator.

The reason why \(S\) behaves in a way different from your observation is because \(S(2,4) = 3 \neq 2+3\) and \(S(3,3) = 4 \neq 5\). So \(S\) does not obey your assumptions \(2 + 3 = 2 ? 4\) and \(3 + 2 = 3 ? 3\). Therefore it might not be what you want to know.