Sylvester’s sequence

In number system, Sylvester’s sequence is an integer sequence in which each member of the sequence is the product of the previous members, plus one. Given a positive integer N. The task is to print the first N member of the sequence.
Since numbers can be very big, use %10^9 + 7.
Examples:

Input : N = 6
Output : 2 3 7 43 1807 3263443

Input : N = 2
Output : 2 3

The idea is to run a loop and take two variables and initialise them as 1 and 2, one to store the product till now and other to store the current number which is nothing but the first number + 1 and for each step multiply both using arithmetic modular operation i.e (a + b)%N = (a%N + b%N)%N where N is a modular number.
Below is the implementation of this approach:

C++

// CPP program to print terms of Sylvester's sequence 
#include <bits/stdc++.h> 
using namespace std; 
#define N 1000000007 
  
void printSequence(int n) 
{ 
    int a = 1; // To store the product. 
    int ans = 2; // To store the current number. 
  
    // Loop till n. 
    for (int i = 1; i <= n; i++) { 
        cout << ans << " "; 
        ans = ((a % N) * (ans % N)) % N; 
        a = ans; 
        ans = (ans + 1) % N; 
    } 
} 
  
// Driven Program 
int main() 
{ 
    int n = 6; 
    printSequence(n); 
    return 0; 
} 

Java

// JAVA Code for Sylvester sequence 
import java.util.*; 
  
class GFG { 
      
    public static void printSequence(int n) 
    { 
        int a = 1; // To store the product. 
        int ans = 2; // To store the current number. 
        int N = 1000000007; 
          
        // Loop till n. 
        for (int i = 1; i <= n; i++) { 
           System.out.print(ans + " "); 
            ans = ((a % N) * (ans % N)) % N; 
            a = ans; 
            ans = (ans + 1) % N; 
        } 
    } 
  
    /* Driver program to test above function */
    public static void main(String[] args)  
    { 
        int n = 6; 
        printSequence(n); 
          
    } 
} 
    
// This code is contributed by Arnav Kr. Mandal. 

Python

# Python Code for Sylvester sequence 
  
def printSequence(n) : 
    a = 1 # To store the product. 
    ans = 2 # To store the current number. 
    N = 1000000007
      
    # Loop till n. 
    i = 1
    while i <= n : 
        print ans, 
        ans = ((a % N) * (ans % N)) % N 
        a = ans 
        ans = (ans + 1) % N 
        i = i + 1
          
  
# Driver program to test above function  
n = 6
printSequence(n) 
  
# This code is contributed by Nikita Tiwari. 

C#

// C# Code for Sylvester sequence 
using System; 
  
class GFG { 
      
    public static void printSequence(int n) 
    { 
         // To store the product. 
        int a = 1; 
          
        // To store the current number. 
        int ans = 2;  
          
        int N = 1000000007; 
          
        // Loop till n. 
        for (int i = 1; i <= n; i++) 
        { 
            Console.Write(ans + " "); 
            ans = ((a % N) * (ans % N)) % N; 
            a = ans; 
            ans = (ans + 1) % N; 
        } 
    } 
  
    // Driver program  
    public static void Main()  
    { 
        int n = 6; 
        printSequence(n); 
          
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// PHP program to print  
// terms of Sylvester's sequence 
  
$N = 1000000007; 
  
function printSequence($n) 
{ 
    global $N; 
      
    // To store  
    // the product. 
    $a = 1;  
      
    // To store the 
    // current number. 
    $ans = 2;  
  
    // Loop till n. 
    for ($i = 1; $i <= $n; $i++) 
    { 
        echo $ans ," "; 
        $ans = (($a % $N) * ($ans % $N)) % $N; 
        $a = $ans; 
        $ans = ($ans + 1) % $N; 
    } 
} 
  
    // Driver Code 
    $n = 6; 
    printSequence($n); 
      
// This code is contributed by anuj_67. 
?> 

Javascript

<script> 
// Javascript program to print 
// terms of Sylvester's sequence 
  
let N = 1000000007; 
  
function printSequence(n) 
{ 
  
    // To store 
    // the product. 
    let a = 1; 
      
    // To store the 
    // current number. 
    let ans = 2; 
  
    // Loop till n. 
    for (let i = 1; i <= n; i++) 
    { 
        document.write(ans + " "); 
        ans = ((a % N) * (ans % N)) % N; 
        a = ans; 
        ans = (ans + 1) % N; 
    } 
} 
  
    // Driver Code 
    let n = 6; 
    printSequence(n); 
      
// This code is contributed by gfgking. 
</script>

Output:

2 3 7 43 1807 3263443

Time complexity : O(n)
Auxiliary Space : O(1)

C++

Java

Python

C#

PHP

Javascript

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