Are there real extensions of the operations of addition, multiplication, exponentiation, etc in the other direction?

Question

We have $\underbrace{a+a+a...+a}_{n\:times}$ which equals $a \times n$, and also $\underbrace{b \times b \times b.... \times b}_{p\: times}$ is $b^p$, so I was wondering if the generalization would work in the other direction in the realm of all real numbers. That is an operation $\&$ such that $\underbrace{c\:\&\:c\:\&\:c...\&\:c}_{n\:times}$ equals $c+n$. Would this be able to be generalized infinitely? For example the next generalization would be $\%$ such that $\underbrace{d\:\%\:d\:\%\:d ... \%\:d }_{n\:times}$ equals $d\:\&\:n$.

Answer 1

There is an operator called the successor operator which, when repeated $n$ times, gives $c + n$. (This would be the $\&$ operator in your question.) In essence, it simply adds 1 onto the number. It is sometimes denoted by $c'$ and is a unary operator, which means it only takes 1 argument, unlike the operators built from it, like addition and multiplication, which are binary.

An example would be $c' = c + 1$, and thus $c'' = c + 1 + 1$, and so on and so forth until you repeat the successor function $n$ times to get $c + n$, addition.

To my knowledge, I don't think there is any operator which when iterated gives the successor, and even if there were I don't think it would be very interesting. If you consider subtraction as a 'tier down' from addition, though, and each additional operator as a repetition of the previous one (like on this site where division is given as repeated subtraction), then there is an infinite amount of operators in the 'other direction'.

Answer 2

Also, the condition $\underbrace{c\:\&\:c\:\&\:c...\&\:c}_{n\:times} =c+n$ when $n=1$ means that $c=c+1$, which is impossible, so no such $\&$ can exist.