- Let $F_n$ be A000045 (i.e., Fibonacci numbers). Here

$$ F_n = F_{n-1} + F_{n-2}, \\ F_0 = 0, F_1 = 1, \\ F_{-n} = (-1)^{n-1}F_n $$

I conjecture that

$$ F_{-n} = \left\lfloor\frac{n+1}{2}\right\rfloor - \sum\limits_{i=1}^{n-1} \left(\left\lfloor\frac{n-i+1}{2}\right\rfloor + 1\right)F_{-i}, \\ F_{-n} = \left\lfloor\frac{2(n+1)}{3}\right\rfloor - \sum\limits_{i=1}^{n-1} \left(2\left\lfloor\frac{n-i+2}{3}\right\rfloor + 1\right)F_{-i} $$

Here is the *PARI/GP* program to check it numerically:

```
test1(n) = fibonacci(-n) == ((n+1)\2 - sum(i=1, n-1, ((n - i + 1)\2 + 1)*fibonacci(-i)))
test2(n) = fibonacci(-n) == (2*(n+1)\3 - sum(i=1, n-1, (2*((n - i + 2)\3) + 1)*fibonacci(-i)))
```

Is there a way to prove it? Is there any other similar formulas?