Read this http://www.mathsisfun.com/algebra/sequences-finding-rule.html and also http://en.wikipedia.org/wiki/The_Oxford_Murders_%28film%29 where the scene about the murder note left behind contains a sequence of numbers, and it is said that the next number can be pretty much anything
a few more googling eventually lead me into this http://en.wikipedia.org/wiki/Wittgenstein_on_Rules_and_Private_Language which said that all rules are subjected to interpretations, and that you cannto be sure which is the correct one
So my question then becomes:
Given a sequence
$$S=a_1,a_2,a_3,a_4 \cdots$$
What is the maximum number of possible rules that can be used to produce the next term, or is it impossible in general for all sequences S to have a finite number of rules that can produce all given terms and the next terms?
A more concrete example:
$$S_1=1,1,2,3 \cdots$$
We knew that one of the rules that fit $S_1$ is the Fibonacci Sequence, but is there a maximum number of rules that can fit $S_1$ but does NOT generate the Fibonacci Sequence?
If we have a function F(S,$\lambda$) which is defined as follows
$$F(S,\lambda)=max(no. of rules for a given S)$$
where S is a sequence and $\lambda$ is the number of known first few terms in the sequence (e.g. $\lambda$=2 means we are given $a_1,a_2$
What is the first starting point in choosing S (so as to make it less ambigurious) such that the behavior of F can be investigated without loss of (too much) generality?
I am pretty sure that $F$ is always infinite. Given sequence $S = a_1, a_2, \dotsc$ with $\lambda = 2$, we can consider these as two points $(1,a_1)$ and $(2,a_2)$. Given these two points, there are a countably infinite number of cubic polynomials that satisfy these two points (one for each integer choice of $a_3$). We can say that our rule is $a_i = fi^2 + gi + h$ for the $f,g,h$ that agree with our first two terms $a_1$ and $a_2$ and produce any chosen value of $a_3$ for $i=3$.
More generally, I suppose I would define a rule to be the ability to produce on demand any given term of a sequence. Given some $S$ with $\lambda = n$, if we are requested to produce the $m^{\mathrm{th}}$ term of $S$, there are infinitely many degree $n+1$ polynomials (one for each possible choice of the $m^{\mathrm{th}}$ term) that can serve as a rule to our $S$. Then if we are requested to produce some other term, we can choose any of the infinitely many degree $n+2$ polynomials that satisfies the terms we have so far and give us then next requested term. Et cetera.
This doesn't quite work, though, if we allow $\lambda$ to be infinite.