A so-called "Lucas Number" is, to me, nothing more than a standard Fibonacci progression of $n_2+n_1$ with a different starting point. There are infinitely many similar progressions, so why does Lucas get special credit?
We tend to think of Fibonacci Numbers beginning at $0, 1, 1$ and continuing out by adding the previous two numbers to make the next in line, and Lucas numbers do the exact same thing with the origin of $2$ and $1$.
I would argue that our starting points for both of these values is a misconception, and instead they should begin with the negative values to infinity by adding $n_1$ to the absolute value of $n_2$ e.g. $n_1+\left\lvert n_2\right\rvert$, making a nice mirror of the values left of zero. This allows for $^-1+\left\lvert^-1\right\rvert=0$, $0+\left\lvert^-1\right\rvert=1$, and then the traditional $0+1=1$ and so on for Fibonacci. It also makes Lucas' origin of $2, 1, 3$ make more sense as the progression would look more like $^-3+\left\lvert^-1\right\rvert=^-2$, $^-1+\left\lvert^-2\right\rvert=1$, $^-2+1=^-1$, $1+\left\lvert^-1\right\rvert=2$, $^-1+2=1$, $2+1=3$ and so on.
That being said, why does Lucas get credit for something Fibonacci already discovered when he didn't add anything of value to it? I could as easily begin with $^-2, 3, 1, 4, 5, 9 ...$ does that mean I should get my name attributed to the series?
Can someone please help me understand what's so special about Lucas Numbers that isn't already understood by the Fibonacci sequence?
Thank you!
The Fibonacci and Lucas sequences are related to the quadratic equation $x^2 - Px + Q = 0 $ with $P = 1$ and $Q = -1$. The discriminant is $D = P^2 - 4Q = 5$ and the roots are $ a = \frac{P+\sqrt{D}}2 = \frac{1+\sqrt{5}}2\quad\text{and}\quad b = \frac{P-\sqrt{D}}2 = \frac{1-\sqrt{5}}2$.
The terms of Fibonacci and Lucas sequences can be expressed in terms of the roots $a$ and $b$ as follows
$ F_n = \frac{a^n-b^n}{a-b} = \frac{a^n-b^n}{ \sqrt{D}} = \frac{a^n-b^n}{ \sqrt{5}} $ and
$ L_n = a^n+b^n $.
It follows that $L_0 = 2 \text{ and } L_1 = a + b = P = 1$.
So the values for $L_0$ and $L_1$ are not arbitrary.