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Lucas numbers L(n) 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, ... L(n) = L(n − 1) + L(n − 2) for n ≥ 2, with L(0) = 2 and L(1) = 1. A000032: Prime numbers p n: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, ... The prime numbers p n, with n ≥ 1. A prime number is a natural number greater than 1 that is not a product of two smaller natural numbers. A000040 ...
Second edition of the book. Neil Sloane started collecting integer sequences as a graduate student in 1964 to support his work in combinatorics. [8] [9] The database was at first stored on punched cards.
(Because 3(4k + 1) + 1 = 12k + 4 = 4(3k + 1).) In more generality: For all p ≥ 1 and odd h, f p − 1 (2 p h − 1) = 2 × 3 p − 1 h − 1. (Here f p − 1 is function iteration notation.) For all odd h, f(2h − 1) ≤ 3h − 1 / 2 The Collatz conjecture is equivalent to the statement that, for all k in I, there exists an integer n ...
The 142857 number sequence is also found in several decimals in which the denominator has a factor of 7. In the examples below, the numerators are all 1, however there are instances where it does not have to be, such as 2 / 7 (0. 285714). For example, consider the fractions and equivalent decimal values listed below: 1 / 7 = 0 ...
An integer sequence is computable if there exists an algorithm that, given n, calculates a n, for all n > 0. The set of computable integer sequences is countable.The set of all integer sequences is uncountable (with cardinality equal to that of the continuum), and so not all integer sequences are computable.
In mathematics, a generating function is a representation of an infinite sequence of numbers as the coefficients of a formal power series.Generating functions are often expressed in closed form (rather than as a series), by some expression involving operations on the formal series.
Gilbreath observed a pattern while playing with the ordered sequence of prime numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, ... Computing the absolute value of the difference between term n + 1 and term n in this sequence yields the sequence
The number of ways of writing n as an ordered sum in which each term is odd and greater than 1 is equal to P(n − 5). For example, P(6) = 4, and there are 4 ways to write 11 as an ordered sum in which each term is odd and greater than 1: 11 ; 5 + 3 + 3 ; 3 + 5 + 3 ; 3 + 3 + 5. The number of ways of writing n as an ordered sum in which each ...