On a finite sequence of numbers Suppose that $a_0, a_1,..., a_{n+1}$ are natural numbers such that $a_0 = a_{n+1} = 1$ and for any $1<i\leq n$, $a_i | a_{i-1} + a_{i+1}$ and $a_i >1$. Is it true that one of these numbers must be equal to $2$?
 A: Answer to Orignal Question:
No.
Consider the simplest case, n = 1. $a_0$ = 1, $a_1$ = 1, and $a_2$ = 1. This satisfies the condition and nothing equals 2.
Answer to Edited Version of the Question:
The complete proof is pretty long, and involves consideration of several branch possibilities for the sequence and modulus arithmetic. Ultimately this is the fact that must be proved: Given $a_0 = 1$ and $a_1 > 2$, it is impossible to proceed back toward $a_{n+1} = 1$ without hitting every positive integer between $a_1$ and 1 (note that this includes 2). In other words, the next new lowest number in the sequence cannot be any less than the current lowest number in the sequence minus 1.
More Explanation as requested in comment:
Consider the following family of simple sequences which satisfy the constraints of the problem:
Let b and m be any integers > 1. Let the first m elements ($a_j$ where $0 \le j < m$) be equal to $bj+1$. Let $a_m$ be $b$. And let each element thereafter be one less than the element preceding it, giving a total of $b + m$ elements. For example, if b is 4 and m is 5, the sequence is 1, 5, 9, 13, 17, 4, 3, 2, 1.
Let's also add to this family the mirror images of these sequences and the one simplest case 1, 2, 1, which is yielded when $m = 0$, regardless of the value of b. Let's call this family of sequences F. Note that every member of F satisfies the constraints and every member of the family contains a 2.
What I argue is that every possible sequence that satisfies the constraints fits into a superset F' which includes the elements of some member of F, in the same order, with other elements betwixt the members of F. In other words, every diversion in the pattern which is not like the sequence represented by the members of F must ultimately return to the F family pattern. The exploration of each possible type of diversion from the pattern and proving that it returns ultimately to an element of the F family pattern is what takes a little more work. In fact, any case where $m > 2$ could be reframed as a diversion from the even simpler set of sequences where $m = 1$. And in fact, all sequences where $m = 1$ could be reframed as a diversion from the simplest case where $m = 0$. The exploration of all such permutations can be split into a manageable finite number of cases with modulus arithmetic. But it's long, and more than I have time for.
