How to resolve the algorithm Wilson primes of order n step by step in the Rust programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Wilson primes of order n step by step in the Rust programming language
Table of Contents
Problem Statement
A Wilson prime of order n is a prime number p such that p2 exactly divides:
If n is 1, the latter formula reduces to the more familiar: (p - n)! + 1 where the only known examples for p are 5, 13, and 563.
Calculate and show on this page the Wilson primes, if any, for orders n = 1 to 11 inclusive and for primes p < 18 or, if your language supports big integers, for p < 11,000.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Wilson primes of order n step by step in the Rust programming language
Source code in the rust programming language
// [dependencies]
// rug = "1.13.0"
use rug::Integer;
fn generate_primes(limit: usize) -> Vec<usize> {
let mut sieve = vec![true; limit >> 1];
let mut p = 3;
let mut sq = p * p;
while sq < limit {
if sieve[p >> 1] {
let mut q = sq;
while q < limit {
sieve[q >> 1] = false;
q += p << 1;
}
}
sq += (p + 1) << 2;
p += 2;
}
let mut primes = Vec::new();
if limit > 2 {
primes.push(2);
}
for i in 1..sieve.len() {
if sieve[i] {
primes.push((i << 1) + 1);
}
}
primes
}
fn factorials(limit: usize) -> Vec<Integer> {
let mut f = vec![Integer::from(1)];
let mut factorial = Integer::from(1);
f.reserve(limit);
for i in 1..limit {
factorial *= i as u64;
f.push(factorial.clone());
}
f
}
fn main() {
let limit = 11000;
let f = factorials(limit);
let primes = generate_primes(limit);
println!(" n | Wilson primes\n--------------------");
let mut s = -1;
for n in 1..=11 {
print!("{:2} |", n);
for p in &primes {
if *p >= n {
let mut num = Integer::from(&f[n - 1] * &f[*p - n]);
num -= s;
if num % ((p * p) as u64) == 0 {
print!(" {}", p);
}
}
}
println!();
s = -s;
}
}
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