Search results
Results from the WOW.Com Content Network
Rowland (2008) proved that this sequence contains only ones and prime numbers. However, it does not contain all the prime numbers, since the terms gcd(n + 1, a n) are always odd and so never equal to 2. 587 is the smallest prime (other than 2) not appearing in the first 10,000 outcomes that are different from 1. Nevertheless, in the same paper ...
The most naïve algorithm would be to cycle through all subsets of n numbers and, for every one of them, check if the subset sums to the right number. The running time is of order O ( 2 n ⋅ n ) {\displaystyle O(2^{n}\cdot n)} , since there are 2 n {\displaystyle 2^{n}} subsets and, to check each subset, we need to sum at most n elements.
A prime sieve or prime number sieve is a fast type of algorithm for finding primes. There are many prime sieves. The simple sieve of Eratosthenes (250s BCE), the sieve of Sundaram (1934), the still faster but more complicated sieve of Atkin [1] (2003), sieve of Pritchard (1979), and various wheel sieves [2] are most common.
The multiples of a given prime are generated as a sequence of numbers starting from that prime, with constant difference between them that is equal to that prime. [1] This is the sieve's key distinction from using trial division to sequentially test each candidate number for divisibility by each prime. [ 2 ]
Mersenne primes and perfect numbers are two deeply interlinked types of natural numbers in number theory. Mersenne primes, named after the friar Marin Mersenne, are prime numbers that can be expressed as 2 p − 1 for some positive integer p. For example, 3 is a Mersenne prime as it is a prime number and is expressible as 2 2 − 1.
The following is pseudocode which combines Atkin's algorithms 3.1, 3.2, and 3.3 [1] by using a combined set s of all the numbers modulo 60 excluding those which are multiples of the prime numbers 2, 3, and 5, as per the algorithms, for a straightforward version of the algorithm that supports optional bit-packing of the wheel; although not specifically mentioned in the referenced paper, this ...
Let {q 1, q 2, …} be successive prime numbers in the interval (B 1, B 2] and d n = q n − q n−1 the difference between consecutive prime numbers. Since typically B 1 > 2, d n are even numbers. The distribution of prime numbers is such that the d n will all be relatively small. It is suggested that d n ≤ ln 2 B 2.
We can test prime p's manually given the formula above. In one case, testing p = 3, we have 17 (3 − 1)/2 = 17 1 ≡ 2 ≡ −1 (mod 3), therefore 17 is not a quadratic residue modulo 3. In another case, testing p = 13, we have 17 (13 − 1)/2 = 17 6 ≡ 1 (mod 13), therefore 17 is a quadratic residue modulo 13.