I am working on an algorithm that uses the number 6 as part of an equation to find the next number in the Fibonacci series. The fascinating thing about this algorithm lies in the fact that starting from the number 3, we can multiply an existing Fibonacci number by 6, as part of an equation, to return the next number in the series.

This brings up some interesting questions as the Fibonacci series is most often seen in nature. It has been argued that that bee colonies use a hexagon shape for there honeycomb as it is the most efficient to use a six-sided shape when connecting with other shapes. Is there a relation here with the number 6 and the Fibonacci series? It is also known that plant life uses the Fibonacci series.

### Fibonacci Multiplied by 6

An interesting Fibonacci algorithm that multiplies any given Fibonacci number by 6 and to get number(X). X is then subtracted by: (the closest Fibonacci number of lesser value + the Fibonacci number 4 levels previous). The result is the next number in the Fibonacci series.

This equation does not work until the number 3, but it seems apparent that every number after this in the series can be solved with this exact equation. The only exception that has been found is that once we reach the number 55, the Fibonacci number 4 levels previous must be subtracted rather than added.

Fibonacci listed as a sequence:

Fn = n represents a specific value within the sequence.

(F0,0) (F1,1) (F2,1) (F3,2) (F4,3) (F5,5) (F6,8) (F7,13) (F8,21) (F9,34) (F10,55) (F11,89)…

0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711

Given the aforementioned sequence, we can then use the following algorithm:

1. Fn * 6 = X

2. X – (closest Fibonacci number of lesser value + F(n-4)

3. Result is the next number in the series.

Here is how the Fibonacci algorithm output:

// Starting at 3 or F4:

F4 = 3 * 6 – ( 13 + 0) = 5

F5 = 5 * 6 – ( 21 + 1) = 8

F6 = 8 * 6 – ( 34 + 1) = 13

F7 = 13 * 6 – ( 55 + 2) = 21

F8 = 21 * 6 – ( 89 + 3) = 34

F9 = 34 * 6 – ( 144 + 5) = 55

// Now, subtracting F(n-4)

F10 = 55 * 6 – ( 233 – 8) = 89

F11 = 89 * 6 – ( 377 – 13) = 144

F12 = 144 * 6 – ( 610 – 21) = 233

F13 = 233 * 6 – ( 987 – 34) = 377

F14 = 377 * 6 – (1597 – 55) = 610

F15 = 610 * 6 – (2584 – 89) = 987

F16 = 987 * 6 – (4181 – 144) = 1597

F17 = 1597 * 6 – (6765 – 233) = 2584

…

### Fibonacci on GitHub

JUL