The Fibonacci Series is based on the principle that the succeeding number is the sum of the previous two numbers. Then how is it logical to start with a 0? Shouldn't it start with 1 directly?

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The Fibonacci sequence is $\dots,-8,5,-3,2,-1,1,0,1,1,2,3,5,8,\dots$ and you can "start" it wherever you like, but it seems most natural to set $F_0=0$. – bof Nov 6 '14 at 9:45

One key number-theoretical reason for starting the sequence $(0,1)$ instead of $(1,1)$ is that it makes the divisibility property of the Fibonacci sequence more straightforward to state; i.e., that $F_k$ divides $F_{nk}$ for any $k,n$. If you start with $F_0=1$ instead of $F_0=0$ then this breaks down (for instance, in that numbering $F_2=2$ but $F_4=5$) and a lot of results have to be presented with indices shifted. This has to do, roughly with the representation of $F_n$ as $\frac{1}{\sqrt{5}}\left(\phi^n-\varphi^n\right)$ (with $\varphi=\frac{1}{\phi} = \phi-1$); the fact that the exponents 'match up' with the index leads to straightforward arguments for the various divisibility properties.

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See this answer for more on the number-theoretical viewpoint. – Bill Dubuque Aug 26 '15 at 3:25

It's just a matter of taste. You can start it at 1,1 instead of 0,1 at the cost of having all your indices shifted by one. On the other hand you can never be sure whether the first two terms are denoted $(F_0, F_1)$ or $(F_1,F_2)$ anyway, so this cost is small.

Shifting one step in the other direction, you can also choose to start the sequence at 1,0. Together, the 0,1 and 1,0 sequences provide a convenient basis for the Fibonacci recurrence started at any pair of values (since the recurrence is linear and homogenous).

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Notably, of course, the sequence that starts (1,0) is just the series shifted by 1 the other direction - i.e., if we call that sequence $G_n$, then $G_n$ = $F_{n-1}$. (Since $G_1 = 0 = F_0$ and $G_2 = 1 = F_1$) – Steven Stadnicki Jun 15 '12 at 14:59

I give you Lucas numbers.

$(0,1)$ instead of $(1,1)$ is just for convenience in many formulas, as previously stated. There's nothing about a recurrence relation that tells you where to start.

Note that this also lets you generalize nicely in a different way. I.e, there are Tribonacci numbers, and so on that satisfy higher order recurrence relations. It's a little nicer to say that such sequences start off at $(0,0,0,...,0,1)$ rather than at $(1,1,2,4,...)$ for however many powers of $2$.

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This is a matter of convention. According to Wikipedia, when Leonardo di Pisa (aka Fibonacci) introduced this sequence he did start with $1$, but he also started the indexing at $1$.

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"according to Wikipedia" - not something you'll often read on math.stack! – Ollie Ford Feb 17 '14 at 1:31

$0$ should not be considered as a Fibonacci number. Accordingly to Zeckendorf's theorem, every positive integer can be uniquely written as the sum of one or more distinct non-consecutive Fibonacci numbers. Taking $0$ as a Fibonacci number is therefore like taking $1$ as a prime number!

By the way, the divisibility property still holds taking $F_1=F_2=1$, so I don't agree with the argument given in the first answer.

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