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For instance, 1/3+1/4 = 7/12, so a notation like would represent the number that would now more commonly be written as the mixed number , or simply the improper fraction . Notation of this form can be distinguished from sequences of numerators and denominators sharing a fraction bar by the visible break in the bar.
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.
As a biprime with proper divisors 1, 3 and 7, twenty-one has a prime aliquot sum of 11 within an aliquot sequence containing only one composite number (21, 11, 1, 0); it is the second composite number with an aliquot sum of 11, following 18. 21 is the first member of the second cluster of consecutive discrete semiprimes (21, 22), where the next such cluster is (33, 34, 35).
Let k be defined as an element in F, the array of Fibonacci numbers. n = F m is the array size. If n is not a Fibonacci number, let F m be the smallest number in F that is greater than n. The array of Fibonacci numbers is defined where F k+2 = F k+1 + F k, when k ≥ 0, F 1 = 1, and F 0 = 1. To test whether an item is in the list of ordered ...
The number in the n-th month is the n-th Fibonacci number. [21] The name "Fibonacci sequence" was first used by the 19th-century number theorist Édouard Lucas. [22] Solution to Fibonacci rabbit problem: In a growing idealized population, the number of rabbit pairs form the Fibonacci sequence.
Print/export Download as PDF; Printable version; In other projects ... Pages in category "Fibonacci numbers" The following 48 pages are in this category, out of 48 ...
This characterization is exact: every sequence of complex numbers that can be written in the above form is constant-recursive. [20] For example, the Fibonacci number is written in this form using Binet's formula: [21] =,
If the number subtracted was the i th Fibonacci number F(i), put a 1 in place i − 2 in the code word (counting the left most digit as place 0). Repeat the previous steps, substituting the remainder for N, until a remainder of 0 is reached. Place an additional 1 after the rightmost digit in the code word.