Thread: Busted functions View Single Post
2006-02-22, 09:23   #2
Orgasmic Troll
Cranksta Rap Ayatollah

Jul 2003

641 Posts

Quote:
 Originally Posted by Dougy Definition: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function. Then if, for all $m \in \mathbb{N}$, one of the following is true: (i) $f^n(m)$ is not prime for some $n \in \mathbb{N}$. (where $f^1(m)=f(m)$ and $f^n(m)=f(f^{n-1}(m))$, i.e. function composition) (ii) $f^i(m)=f^j(m)$ for distinct $i,j \in \mathbb{N}$. then $f$ is called a busted function. Some functions are easily busted. E.g. $f(x)=a$ for some $a \in \mathbb{N}$, or $f(x)=x$. While some functions are not bustible. E.g. Any function satisfying $f(p_n)=p_{n+1}$, where $p_i$ is the $i^{th}$ prime. Question: What functions are busted? The motivation behind this conjecture is to disprove anyone who claims something like "if $x_1$ is a prime, and $x_n := f(x_{n-1})$, then $(x_n)$ is a sequence of primes". A function becomes 'busted' when it stops producing new primes. Has this been done before? Can anyone easily bust some functions?
condition II can be rewritten as $f^i(m)=m$ for some i

are you sure you mean that $f: \mathbb{R} \rightarrow \mathbb{R}$? if I have $f(x) = ax + \pi, a \in \mathbb{N}$ then that's a busted function.

what does bustible mean? Either a function is busted or it is not, bustible sounds like we can do something to the function to make sure it is busted.