A Summation of Fibonacci Numbers

Level 2

The "Fib function" , $F(n)$ , returns the corresponding Fibonacci number :

$F(0) = 0$ , $F(1) = 1$ , $F(2) = 1$ , $F(3) = 2$ , $F(4) = 3$ , $F(5) = 5$ , $F(6) = 8$ , $F(7) = 13$ , $F(8) = 21$ , ... and so on.

The "SumFib function" , $S(n)$ , returns the corresponding cumulative sum of Fibonacci numbers:

$S(0) = 0$ , $S(1) = 1$ , $S(2) = 2$ , $S(3) = 4$ , $S(4) = 7$ , $S(5) = 12$ , $S(6) = 20$ , $S(7) = 33$ , $S(8) = 54$ , ... and so on.

Question : Aside from the first three distinct values of each function, { $0, 1, 2$ }, are there any other shared values (or outputs) of both of these functions?

It is not currently known No Yes

1 solution

Tj Evert
Sep 26, 2018

If we advance the recursive relation for Fibonacci numbers , $F(n) = F(n-2) + F(n-1)$ , by 2 places, we get $F(n+2) = F(n+1) + F(n)$ .

Rearranging these terms gives us a modified recursive relation, $F(n) = F(n+2) - F(n+1)$ .

Summing these: (Note: I've written $F(n)$ as $F(n-0)$ to simplify the formatting)

$F(n-0) = F(n+2) - \underline{F(n+1)}$

$F(n-1) = \overline{F(n+1)} - \underline{F(n-0)}$

$F(n-1) = \overline{F(n-0)} - \underline{F(n-1)}$

$F(n-1) = \overline{F(n-1)} - \underline{F(n-2)}$

$F(0) = \overline{F(2)} - F(1)$

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ (notice that the "over-lined" terms cancel out the underlined terms)

$S(n) = F(n+2) - 1$

Since $S(n) = F(n+2) - 1$ , then for all $n > 2$ , the summation $S(n)$ can't equal a Fibonacci number.

A Summation of Fibonacci Numbers

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