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For generalized Fibonacci sequences (satisfying the same recurrence relation, but with other initial values, e.g. the Lucas numbers) the number of occurrences of 0 per cycle is 0, 1, 2, or 4. The ratio of the Pisano period of n and the number of zeros modulo n in the cycle gives the rank of apparition or Fibonacci entry point of n .
Fibonacci numbers are also strongly related to the golden ratio: Binet's formula expresses the n-th Fibonacci number in terms of n and the golden ratio, and implies that the ratio of two consecutive Fibonacci numbers tends to the golden ratio as n increases.
A repfigit, or Keith number, is an integer such that, when its digits start a Fibonacci sequence with that number of digits, the original number is eventually reached. An example is 47, because the Fibonacci sequence starting with 4 and 7 (4, 7, 11, 18, 29, 47) reaches 47.
An easily programmed alternative using only integer arithmetic is to calculate two large consecutive Fibonacci numbers and divide them. The ratio of Fibonacci numbers F 25001 {\displaystyle F_{25001}} and F 25000 {\displaystyle F_{25000}} , each over 5000 {\displaystyle 5000} digits, yields over 10,000 {\displaystyle ...
where F n is the n th Fibonacci number. Such a sum is called the Zeckendorf representation of N. The Fibonacci coding of N can be derived from its Zeckendorf representation. For example, the Zeckendorf representation of 64 is 64 = 55 + 8 + 1. There are other ways of representing 64 as the sum of Fibonacci numbers 64 = 55 + 5 + 3 + 1 64 = 34 ...
In mathematics, the Fibonomial coefficients or Fibonacci-binomial coefficients are defined as ... F j is the j-th Fibonacci number and n! F is the nth Fibonorial, ...
In the Fibonacci sequence, each number is the sum of the previous two numbers. Fibonacci omitted the "0" and first "1" included today and began the sequence with 1, 2, 3, ... . He carried the calculation up to the thirteenth place, the value 233, though another manuscript carries it to the next place, the value 377.
For Fibonacci numbers starting with F 1 = 0 and F 2 = 1 and with each succeeding Fibonacci number being the sum of the preceding two, one can generate a sequence of Pythagorean triples starting from (a 3, b 3, c 3) = (4, 3, 5) via