# Defying conventional calculus!

Which is the only positive integer whose $$k^\text{th}$$ arithmetic derivative (where $$k \in \mathbb N$$ ) is always equal to the number itself?

Details and Assumptions:

The arithmetic derivative function, denoted by $$n'$$ is a function $$n: \mathbb{N} \cup \left\{0 \right\} \to \mathbb{N} \cup \left\{0 \right\}$$ defined by

• $$n'=1$$ for all prime numbers.
• $$(ab)'=a' b+ a b'$$ for $$a, b \in \mathbb{N} \cup \left\{0 \right\}$$.
• $$n\overbrace { '''....'' }^{ { k \ times } } = (((n\overbrace { ')')'....')' }^{ { k \ times } }$$
• $$0'=1'=0$$

Clarification:

• The set of positive integers (or counting numbers or natural numbers) do not contain the member $$0$$ in them.
• To be clear, it is asked to find a positive integer $$n$$ which satisfies $n = n' = n'' = n''' = \cdots = n\overbrace { '''....'' }^{ { k \ times } }$

×