integer sequences - "rules" The notation (x, y, z, ...) used for infinite sequences is ambiguous (at least wikipedia says so).
So there is always more than one possible "rule", which means it is possible to continue the sequence in different ways?
Is there a proof for that? Is there a proof that there are always infinite possible "solutions"/"rules"?
I would be happy to know the proof or just the name of it.
Thank you in advance
 A: Consider the sequences $$a_n:=(2n+1)_{n=1}^\infty=1, 3, 5, 7, \dots $$ $$b_n:=\left (\frac {18111} {2}n^4 -90555n^3+\frac {633885} {2}n^2-452773n+217331 \right)_{n=1}^\infty=1, 3, 5, 7, \dots$$
You would believe that $a_n=b_n$ for all $n\in \Bbb{N}$ by just reading the first four terms. But you will soon find that $a_5\ne b_5$ as $$a_5=9$$ and $$b_5=217341$$
That is why the notation $a_1,a_2,a_3,\dots $ is ambiguous.
Edit: You can't prove that some notation is ambiguous. You just use examples to show it is not always the best to use this notation. And the notation is not always ambiguous. Most of the time it is quite clear what is meant by writing $a_1,a_2,\dots$ from the context in which it is being written. So the claim that the notation is always ambiguous is just wrong.
A: To show that there are infinitely many possible rules you can use to continue the sequence
$$
1,2,3,\ldots
$$
(or any other sequence where you know the first few entries) requires a definition of "rule". If a rule is "describe what comes next, forever" then here are infinitely many rules:

*

*continue with $1,1, 1, \ldots$ (1's forever)

*continue with $2,2,2,   \ldots$ (2's forever)

*continue with $3,3,3, \ldots$ (3's forever)

*and    so on.

Of course in many particular situations the author writes the first few terms to suggest a particular pattern. On an exam you are supposed to guess the pattern - and that kind of question is always essentially ambiguous.
The pattern
$$
1,2,3, \ldots
$$
usually suggests the positive integers, but in a context where you were studying recursion it might mean the start of a sequence you get by adding the three previous terms. Then it would continue
$$
1,2,3,6,11,20, \ldots \ .
$$
Those are the tribonacci numbers.
