I like the "talk" section of wikipedia, Talk:Generalizations_of_Fibonacci_numbers. I am quoting from their writeup.
According to Binet's formula,
$$F_n = \frac{\varphi^n-\psi^n}{\varphi-\psi} = \frac{\varphi^n-\psi^n}{\sqrt 5}$$
Since $\psi = -\frac{1}{\varphi}$, this formula can also be written as
$$F_n = \frac{\varphi^n-(-\varphi)^{-n}}{\sqrt 5}$$
Now if you factor the $-1$ out of the $-\varphi$, you get
$$F_n = \frac{\varphi^n-(-1)^{-n}\varphi^{-n}}{\sqrt 5}$$
...
And from Euler's identity, $-1 = e^{i\pi}$, so
$$F_n = \frac{\varphi^n-e^{i\pi n}\varphi^{-n}}{\sqrt 5}$$
Ok, now continuing on my own.... And from Euler's formula, $e^{i\pi n} = \cos (\pi n) + i \sin (\pi n)$, so Binet's solution can be equivalently expressed as $F_1$, to distinguish it from its complex conjugate.
$$F_1(z) = \frac{\varphi^z-(\cos (\pi z) + i \sin (\pi z))\varphi^{-z}}{\sqrt 5}$$
There is an alternative definition, with $-1 = e^{-i\pi}$, which leads to the complex conjugate solution, which I will label $F_2$.
$$F_2(z) = \frac{\varphi^z-(\cos (\pi z) - i \sin (\pi z))\varphi^{-z}}{\sqrt 5}$$
Due to the fact that linear combinations of solutions to the Fibonacci recurrence relation are also solutions, we can average $F_1$ and $F_2$ together, and get the real valued solution, $F_{\text{real}}(z)$, where at the real axis, the imaginary term cancels with its conjugate.
$$F_\text{real}(z) = \frac{F_1(z)+F_2(z)}{2} = \frac{\varphi^z-\cos (\pi z) \varphi^{-z}}{\sqrt 5}$$
I think both solutions are equally valid, and perhaps sometimes it seems more natural to have a real valued solution, and this derivation shows where the "cos" term in that solution comes from. As I noted in earlier comments, there is also some interesting behavior for the "ratio" function for these various solutions, $r(z)=\frac{F(z+1)}{F(z)}$, which I may post later, which connects the complex solution to the formal Schroeder function solution for the ratio function.
In the context of extending the Fibonacci sequence to non-integer
index value the natural formula is
$$ F(n+m\,i) = \frac{\phi^{n + m\,i} − \psi^{n + m\,i}} {\phi − \psi}. $$
Since $\,\psi=-1/\phi<0,\,$ there is no problem with $\,\psi^n\,$ but what
about $\,\psi^{m\,i}?\,$ Use the definition of the power function as in
DLMF to get
$$ \psi^{m\,i} = (-1)^{m\,i}(-\psi)^{m\,i} = (e^{-\pi})^m \phi^{-m\,i}.$$
Thus, a suitable formula is
$$ F(n+m\,i) = \frac{\phi^{n + m\,i} −\
\psi^n (e^{-\pi})^m \phi^{-m\,i}} {\phi − \psi} $$
or some simple variation as needed.
Best Answer
The negative fibbonacci numbers run as $F(-n)=F(n)$ for odd $n$ and $F(-n)=-F(n)$ for even $n$.
So we get for fibonacci numbers
So yes, all of the natural numbers can be expressed as a sum of non-consecutive fibonacci numbers of negative index.
It ought be a straight-forward conversion from the positive to negative schema.
It's possible to go both ways, and there is a carry rule to prevent adjacent columns being filled with more than one counter.