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Dougy · 2006-02-22, 08:01

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?

2006-02-22, 08:01	#1
Dougy Aug 2004 Melbourne, Australia 152₁₀ Posts	Busted functions 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? Last fiddled with by Dougy on 2006-02-22 at 08:03