[Math] The sum of 8 consecutive Fibonacci numbers is not a Fibonacci number

combinatoricsfibonacci-numbers

While studying Fibonacci numbers, I came up with this problem. Of course $F_n = F_{n-1}+F_{n-2}$. I'm sort of stuck with first realizing how to show a number actually isn't a Fibonacci number. I thought that I could somehow rewrite the sum $$F_n+\cdots+F_{n+7}$$
into some sort of rearrangement of $F_n$'s and $F_{n+1}$'s. Could anyone help me show that the sum of 8 consecutive Fibonacci numbers is not a Fibonacci number?

Thanks

Best Answer

Note that $F_{n+k}=F_{n+k+1}-F_{n+k-1}$ by summing this telescoping formula you get

$F_n+\cdots+F_{n+7}=F_{n+9}-F_{n+1}$

But $F_{n+1}<F_{n+7}$ is too small so this number is strictly greater than $F_{n+8}$ and also smaller than $F_{n+9}$.

Related Solutions

[Math] Sum of Fibonacci numbers

Fibonacci numbers can be written as a matrix using:

$$\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n = \begin{bmatrix} F_{n+1} & F_{n} \\ F_{n} & F_{n-1}\end{bmatrix}$$

So that any sum, using $X= \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$, is :

$$\sum_{k=a}^b F_n = \left( \sum_{k=a}^b X^n \right)_{2,1}$$

which is a geometric sum. So you can use geometric sum formula:

$$\begin{align} \sum_{k=a}^b X^n &= \sum_{k=0}^b X^n - \sum_{k=0}^{a-1} X^n \\ &= (X^{b+1} - I)(X-I)^{-1}- (X^{a} - I)(X - I)^{-1} \\ &= \left(X^{b+1} - X^{a}\right)(X - I)^{-1} \\ \end{align}$$

Now $(X - I)^{-1} = X$ for this particular matrix (property of Fibonacci recursion):

$$\begin{align} \sum_{k=a}^b X^n &= \left(X^{b+1} - X^{a}\right)X \\ &= \left(X^{b+2} - X^{a+1}\right) \\ \end{align}$$

$$\sum_{k=a}^b F_n = F_{b+2} - F_{a+1}$$

Zeckendorf with Negative Fibonacci Numbers

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

 -21  13  -8   5  -3   2  -1   1
                          -1
                  -3           1
                  -3
                  -3      -1
          -8           2       1
          -8           2
          -8       t0 12
  -21          5       2       1
  -21     -8       -3     -1

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.

  eg  -21            5     2      1
       21           -5    -2     -1
           13   8   -5    -2    -1
           13           3 -2     -1
           13

It's possible to go both ways, and there is a carry rule to prevent adjacent columns being filled with more than one counter.

Best Answer

Related Solutions

[Math] Sum of Fibonacci numbers

Zeckendorf with Negative Fibonacci Numbers

Related Question