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Old 2006-02-22, 09:23   #2
Orgasmic Troll
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Jul 2003

641 Posts

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.
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