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In computational number theory, a variety of algorithms make it possible to generate prime numbers efficiently. These are used in various applications, for example hashing, public-key cryptography, and search of prime factors in large numbers. For relatively small numbers, it is possible to just apply trial division to each successive odd ...
A prime number is a natural number that has exactly two distinct natural number divisors: the number 1 and itself. To find all the prime numbers less than or equal to a given integer n by Eratosthenes' method: Create a list of consecutive integers from 2 through n: (2, 3, 4, ..., n). Initially, let p equal 2, the smallest prime number.
A prime number is a natural number that has no natural number divisors other than the number 1 and itself.. To find all the prime numbers less than or equal to a given integer N, a sieve algorithm examines a set of candidates in the range 2, 3, …, N, and eliminates those that are not prime, leaving the primes at the end.
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 ...
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 it was conjectured to contain all odd primes, even though it is rather inefficient.
The remaining numbers are doubled and incremented by one, giving a list of the odd prime numbers (that is, all primes except 2) below 2n + 2. The sieve of Sundaram sieves out the composite numbers just as the sieve of Eratosthenes does, but even numbers are not considered; the work of "crossing out" the multiples of 2 is done by the final ...
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The idea here is to find an m that is divisible by a large prime number q. This prime is a few digits smaller than m (or N) so q will be easier to prove prime than N. Assuming we find a curve which passes the criterion, proceed to calculate mP and kP. If any of the two calculations produce an undefined expression, we can get a non-trivial ...