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Primality Test using Lucas-Lehmer Series

In this article, we will discuss the Lucas-Lehmer series which is used to check the primality of prime numbers of the form 2p – 1 where p is an integer.

First, let’s see what is Lucas-Lehmer series. The Lucas-Lehmer series can be expressed as :

Hence the series is:

Term 0: 4,
Term 1: 4*4 - 2 = 14,
Term 2: 14*14 - 2 = 194,
Term 3: 194*194 - 2 = 37634,
Term 4: 37634*37634 - 2 = 1416317954, ... and so on.

Below is the program to find out first n terms of the Lucas-Lehmer series.

C++

 // C++ program to find out Lucas-Lehmer series.#include #include using namespace std;  // Function to find out first n terms// (considering 4 as 0th term) of// Lucas-Lehmer series.void LucasLehmer(int n) {    // the 0th term of the series is 4.  unsigned long long current_val = 4;    // create an array to store the terms.  vector series;    // compute each term and add it to the array.  series.push_back(current_val);  for (int i = 0; i < n; i++) {    current_val = current_val * current_val - 2;    series.push_back(current_val);  }    // print out the terms one by one.  for (int i = 0; i <= n; i++)     cout << "Term " << i << ": "        << series[i] << endl;  }  // Driver programint main() {  int n = 5;  LucasLehmer(n);  return 0;}

Java

 // Java program to find out// Lucas-Lehmer series.import java.util.*;  class GFG {      // Function to find out     // first n terms(considering     // 4 as 0th term) of Lucas-    // Lehmer series.    static void LucasLehmer(int n)     {          // the 0th term of        // the series is 4.        long current_val = 4;          // create an array        // to store the terms.        ArrayList series = new ArrayList<>();           // compute each term         // and add it to the array.        series.add(current_val);        for (int i = 0; i < n; i++)         {            current_val = current_val                    * current_val - 2;            series.add(current_val);        }          // print out the       // terms one by one.        for (int i = 0; i <= n; i++)         {            System.out.println("Term " + i                    + ": " + series.get(i));        }    }      // Driver Code    public static void main(String[] args)     {          int n = 5;        LucasLehmer(n);    }}  // This code has been contributed by 29AjayKumar

Python3

 # Python3 program to find out Lucas-Lehmer series.  # Function to find out first n terms# (considering 4 as 0th term) of# Lucas-Lehmer series.def LucasLehmer(n):    # the 0th term of the series is 4.  current_val = 4;    # create an array to store the terms.  series = []    # compute each term and add it to the array.  series.append(current_val)    for i in range(n):    current_val = current_val * current_val - 2;    series.append(current_val);    # print out the terms one by one.  for i in range(n + 1):      print("Term", i, ":", series[i])  # Driver programif __name__=='__main__':    n = 5;  LucasLehmer(n);    # This code is contributed by pratham76.

C#

 // C# program to find out// Lucas-Lehmer series.using System;using System.Collections.Generic;  class GFG{      // Function to find out // first n terms(considering // 4 as 0th term) of Lucas-// Lehmer series.static void LucasLehmer(int n) {  // the 0th term of// the series is 4.long current_val = 4;  // create an array// to store the terms.List<long> series = new List<long>();  // compute each term // and add it to the array.series.Add(current_val);for (int i = 0; i < n; i++){    current_val = current_val *                   current_val - 2;    series.Add(current_val);}  // print out the// terms one by one.for (int i = 0; i <= n; i++)     Console.WriteLine("Term " + i +                       ": " + series[i]); }  // Driver Codestatic void Main(){    int n = 5;    LucasLehmer(n);}}  // This code is contributed by // ManishShaw(manishshaw1)

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Output

Term 0: 4
Term 1: 14
Term 2: 194
Term 3: 37634
Term 4: 1416317954
Term 5: 2005956546822746114


Time Complexity: O(n)
Auxiliary Space: O(n)

We can use string to store the big numbers of the series.

Now, what is the relation with prime numbers of this Lucas-Lehmer series?

1. First thing is that we can only check the primality of those numbers which we can represent as, x = (2p – 1) where p is an integer.
2. Now we have to find out the (p-1)th term of Lucas-Lehmer series.
3. If this term is a multiple of x, then x is a prime number.
4. When x is large, i.e. p is large then we may find difficulties to find out the (p-1)th term of the series.
Rather what we can do:
1. Start calculating Lucas-Lehmer series from 0th term and rather storing the whole term only store the s[i]%x (i.e. term modulo x).
2. Compute the next number of this modified series using the previous term. s[i] = (s[i-1]2 – 2)%x.
3. Compute up to (p-1)th term.
4. If the (p-1)th term is 0 then x is prime, otherwise not. Hence, s[p-1] has to be 0 to be x = (2p – 1) prime.

Examples:

Is 2^7 - 1 = 127 is a prime?
so here x = 127, p = 7-1 = 6.
Hence the modified Lucas-Lehmer series is:
term 1: 4,
term 2: (4*4 - 2) % 127 = 14,
term 3: (14*14 - 2) % 127 = 67,
term 4: (67*67 - 2) % 127 = 42,
term 5: (42*42 - 2) % 127 = 111,
term 6: (111*111) % 127 = 0.
Here the 6th term is 0 so 127 is a prime number.

Code to check whether 2^p-1 is prime or not

C++

 // CPP program to check for primality using// Lucas-Lehmer series.#include #include using namespace std;  // Function to check whether (2^p - 1)// is prime or not.bool isPrime(int p) {    // generate the number  long long checkNumber = pow(2, p) - 1;    // First number of the series  long long nextval = 4 % checkNumber;    // Generate the rest (p-2) terms  // of the series.  for (int i = 1; i < p - 1; i++)     nextval = (nextval * nextval - 2) % checkNumber;      // now if the (p-1)th term is  // 0 return true else false.  return (nextval == 0);}  // Driver Programint main() {  // Check whether 2^p-1 is prime or not.  int p = 7;    long long checkNumber = pow(2, p) - 1;    if (isPrime(p))    cout << checkNumber << " is Prime.";  else    cout << checkNumber << " is not Prime.";    return 0;}

Java

 // Java program to check for primality using// Lucas-Lehmer series.  class GFG{// Function to check whether (2^p - 1)// is prime or not.static boolean isPrime(int p) {  // generate the numberdouble checkNumber = Math.pow(2, p) - 1;  // First number of the seriesdouble nextval = 4 % checkNumber;  // Generate the rest (p-2) terms// of the series.for (int i = 1; i < p - 1; i++)     nextval = (nextval * nextval - 2) % checkNumber;   // now if the (p-1)th term is// 0 return true else false.return (nextval == 0);}  // Driver Programpublic static void main(String[] args) {// Check whether 2^p-1 is prime or not.int p = 7;double checkNumber = Math.pow(2, p) - 1;  if (isPrime(p))    System.out.println((int)checkNumber+" is Prime.");else    System.out.println((int)checkNumber+" is not Prime.");  }}// This code is contributed by mits

Python3

 # Python3 Program to check for primality # using Lucas-Lehmer series.  # Function to check whether (2^p - 1)# is prime or not.def isPrime(p):      # generate the number    checkNumber = 2 ** p - 1      # First number of the series    nextval = 4 % checkNumber      # Generate the rest (p-2) terms    # of the series    for i in range(1, p - 1):        nextval = (nextval * nextval - 2) % checkNumber      # now if the (p-1) the term is    # 0 return true else false.    if (nextval == 0): return True    else: return False  # Driver Code  # Check whether 2^(p-1)# is prime or not.p = 7checkNumber = 2 ** p - 1  if isPrime(p):    print(checkNumber, 'is Prime.')else:    print(checkNumber, 'is not Prime')  # This code is contributed by egoista.

C#

 // C# program to check for primality using// Lucas-Lehmer series.using System;  class GFG{// Function to check whether (2^p - 1)// is prime or not.static bool isPrime(int p) {  // generate the numberdouble checkNumber = Math.Pow(2, p) - 1;  // First number of the seriesdouble nextval = 4 % checkNumber;  // Generate the rest (p-2) terms// of the series.for (int i = 1; i < p - 1; i++)     nextval = (nextval * nextval - 2) % checkNumber;   // now if the (p-1)th term is// 0 return true else false.return (nextval == 0);}  // Driver Programstatic void Main() {// Check whether 2^p-1 is prime or not.int p = 7;double checkNumber = Math.Pow(2, p) - 1;  if (isPrime(p))    Console.WriteLine((int)checkNumber+" is Prime.");else    Console.WriteLine((int)checkNumber+" is not Prime.");  }}// This code is contributed by mits

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Output

127 is Prime.

The largest prime number at the time of writing this article is (2^(77232917) – 1) (discovered 2017-12-26). It has 23, 249, 425 digits. These prime numbers are found in the same way discussed above. Huge computational power and several months of processing are required to find out this kind of large prime number.

Time Complexity: O(p)
Auxiliary Space: O(1)

An interesting fact is that for checking this many big prime numbers, p is also taken prime. After processing, if it finds that the number x is not prime then p is taken as the next prime number and the same process is run.

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