TITLE: Significance of starting the Fibonacci sequence with 0, 1....
QUESTION [9 upvotes]: DISCLAIMER: I do not deal with in-depth mathematics on a daily basis as some of you may, so please pardon my ignorance or lack of coherence on this topic.
QUESTION: What is the significance of starting the Fibonacci sequence with $0,1$ ?
For instance, if I picked any two random integers, say 2 and 7, to start a sequence would I actually be creating some multiple or derivation of the Fibonacci sequence?
Is there a general mathematical explanation for the relationship between any sequence represented by $a[0] = x, a[1] = y, a[n] = a[n-1] + a[n-2]$ and the Fibonacci sequence?
Or, back to my example sequence, is there a general mathematical relationship between:
$2,7,9,16,25,41,66,107,173,280...$
and
$0,1,1,2,3,5,8,13,21,34...$
Perhaps the Golden Ratio explains it somehow?  Any help would be appreciated.

REPLY [5 votes]: Let's call a sequence "fibonacci-like" if it satisfies the recurrence $$s_{n+2} = s_n + s_{n+1}$$ for all $n$.
It's easy to see (or to show) that if $\{s_i\}$ and $\{t_i\}$ are two fibonacci-like sequences, then so is $\{s_i+t_i\}$, and so is $\{cs_i\}$ where $c$ is any constant.  So the collection of fibonacci-like sequences forms a vector space.  The dimension of the vector space is 2, since specifying two elements of the sequence (say $s_0$ and $s_1$) are enough to determine it completely.
So let's abbreviate such a sequence as $[s_0, s_1]$. The standard Fibonacci sequence $0,1,1,2,3,\ldots$ is  written as $[0,1]$ in this notation.  The Lucas sequence $\mathcal L_i = 1,3,4,7,11,\ldots$ is written $[1,3]$.
Since the space of all fibonacci-like sequences is a two-dimensional vector space, any two elements will form a basis for it, unless one is a multiple of the other.  For example, a simple and standard basis for this vector space is the two vectors $[0,1]$ and $[1,0]$.  The former is simply the standard Fibonacci sequence.  The latter is the sequence $1,0,1,1,2,3,5,\ldots$, which is just the standard Fibonacci sequence shifted right by one position; its $i$th element is $f_{i-1}$,  the $(i-1)$th Fibonacci number.
Now consider the general fibonacci-like sequence $[p,q]$:
$$[p,q] = p[1,0] + q[0,1]$$
So the $i$th element of the sequence $[p,q]$ is exactly $$pf_{i-1} + qf_i.$$  For example, the Lucas sequence has $$\mathcal L_i = f_{i-1} + 3f_i.$$  Simiarly your example sequence is $[2,7]$ and is therefore related to the Fibonacci sequence by $$[2,7] = 2[1,0] + 7[0,1] = 2f_{i-1} + 7f_i.$$
Any two sequences form a basis of the space as long as they are not multiples of one another. For example, any fibonacci-like sequence can be expressed in the form $s_i = af_i + b\mathcal L_i$ for some constants $a$ and $b$.  For your $[2,7]$ sequence, we want $[2,7] = a[0,1] + b[1,3] = [b,a+3b]$.  So $b=2$ and $a=1$, and we get $[2,7] = f_i + 2\mathcal L_i$.
Now consider the Fibonacci sequence, but shifted left by $k$ places, whose $i$th element is $f_{i+k}$ for each $i$.  Then the first two terms of the fibonacci-like sequence $\{f_{i+k}\}$ are $f_k$ and $f_{k+1}$, we get $$\{f_{i+k}\} = [f_k, f_{k+1}] = f_k[1,0] + f_{k+1}[0,1] = f_kf_{i-1} + f_{k+1}f_i$$ and we have just proved the sum-of-indices formula for fibonacci numbers. Take $i=k$ in this and we get $$f_{2k} = f_kf_{k-1} + f_{k+1}f_k $$ which is useful for calculating extremely large Fibonacci numbers quickly.