Fibonacci divisibility Is $2051$ a factor of any fibonacci number?
It is not a factor of any perfect number. The prime factors of $2051$ are $7$ and $293$, which are both prime. the $8$th fibonacci number, is the first time that $7$ appears as a factor, and the $147$th fibonacci is the first time that $293$ appears.  $293$ also appears as a factor at the $294$th fibonacci. I don't know if either appear at the $301$st fibonacci. ($300$ being the sum of the factors).
My hypothesis is that $2051$ and it's multiples are a constant to link fibonacci and the perfect numbers- $1,2,4,8,16$.
 A: Yes.
The Fibonacci sequence is periodic modulo any base. This should be clear from the definition $a_n = a_{n-1} + a_{n-2}$ - if any two consecutive values repeat, then the whole sequence repeats as well - and from the pigeonhole principle - there are only so many possible pairs.
Thus, a zero must appear in the sequence. The first two terms are $1, 1$. Find the next occurence of $1, 1$. The previous term in the sequence is a zero.
Any value that is zero modulo $n$ is divisible by n for every positive integer $n$. Also, Fibonacci modulo-n contains a zero for every positive integer $n$.
Thus, for every positive integer $n$, there is a term in the Fibonacci sequence that is divisible by $n$.

this can be generalised to arbitrary integer sequences where


*

*each term can defined recursively from $k$ previous terms using only operations that "work modulo". 

*the same applies backwards - each term can be defined from $k$ following terms using only operations that "work modulo".

*the sequence can be extended backwards to include zero


These operations includes addition, subtraction and multiplication, but it excludes division where the divisor and the base are not coprime. 
Namely:
Any sequence defined by a linear recurrence relation with integer coefficients is eventually periodic modulo every base with the period not greater than the square of the base.
Any such sequence is periodic modulo every base that is coprime with the leading term of the recurrence. Namely, it is periodic modulo every base if the leading term is $1$.
For any such sequence that can be extended backwards to include a zero, for each positive integer $n$ coprime with the leading term, there are infinitely many terms in the sequence divisible by $n$, and the distance between any two such consecutive terms is never greater than $n^2$.
A: From this answer you can see that $7|F_n$ iff $8|n$.
By a similar method, you note that $293|F_{147}$ and $293|F_{294}$, so it will appear when $147|n$.
Clearly, then, we want to consider $8\cdot147 = 1176$.
In particular, re-referencing the linked answer, we have that $7\cdot293 = 2051|F_{1176}$, according with the comment above.
