Tribonacci Numbers

The tribonacci series is a generalization of the Fibonacci sequence where each term is the sum of the three preceding terms.

The Tribonacci Sequence :
0, 0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136, 5768, 10609, 19513, 35890, 66012, 121415, 223317, 410744, 755476, 1389537, 2555757, 4700770, 8646064, 15902591, 29249425, 53798080, 98950096, 181997601, 334745777, 615693474, 1132436852… so on

General Form of Tribonacci number:

a(n) = a(n-1) + a(n-2) + a(n-3) 
with 
a(0) = a(1) = 0, a(2) = 1. 

Given a value N, task is to print first N Tribonacci Numbers.
Examples :

Input : 5
Output : 0, 0, 1, 1, 2

Input : 10
Output : 0, 0, 1, 1, 2, 4, 7, 13, 24, 44

Input : 20
Output : 0, 0, 1, 1, 2, 4, 7, 13, 24, 44,
         81, 149, 274, 504, 927, 1705, 3136, 
          5768, 10609, 19513

A simple solution is to simply follow recursive formula and write recursive code for it,

C++

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// A simple recursive CPP program to print
// first n Tribinacci numbers.
#include <iostream>
using namespace std;
  
int printTribRec(int n)
{
    if (n == 0 || n == 1 || n == 2)
        return 0;
  
    if (n == 3)
        return 1;
    else
        return printTribRec(n - 1) + 
               printTribRec(n - 2) + 
               printTribRec(n - 3);
}
  
void printTrib(int n)
{
    for (int i = 1; i < n; i++)
        cout << printTribRec(i) << " ";
}
  
// Driver code
int main()
{
    int n = 10;
    printTrib(n);
    return 0;
}

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// A simple recursive CPP program
// first n Tribinacci numbers.
import java.io.*;
  
class GFG {
      
    // Recursion Function
    static int printTribRec(int n)
    {
          
        if (n == 0 || n == 1 || n == 2)
            return 0;
              
        if (n == 3)
            return 1;
        else
            return printTribRec(n - 1) + 
                   printTribRec(n - 2) +
                   printTribRec(n - 3);
    }
      
    static void printTrib(int n)
    {
        for (int i = 1; i < n; i++)
            System.out.print(printTribRec(i)
                             +" ");
    }
       
    // Driver code
    public static void main(String args[])
    {
        int n = 10;
  
        printTrib(n);
    }
}
  
// This code is contributed by Nikita tiwari.

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Python

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# A simple recursive CPP program to print
# first n Tribinacci numbers.
  
def printTribRec(n) :
    if (n == 0 or n == 1 or n == 2) :
        return 0
    elif (n == 3) :
        return 1
    else :
        return (printTribRec(n - 1) + 
                printTribRec(n - 2) +
                printTribRec(n - 3))
          
  
def printTrib(n) :
    for i in range(1, n) :
        print( printTribRec(i) , " ", end = "")
          
  
# Driver code
n = 10
printTrib(n)
  
  
# This code is contributed by Nikita Tiwari.

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C#

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// A simple recursive C# program
// first n Tribinacci numbers.
using System;
  
class GFG {
      
    // Recursion Function
    static int printTribRec(int n)
    {
          
        if (n == 0 || n == 1 || n == 2)
            return 0;
              
        if (n == 3)
            return 1;
        else
            return printTribRec(n - 1) + 
                   printTribRec(n - 2) +
                   printTribRec(n - 3);
    }
      
    static void printTrib(int n)
    {
        for (int i = 1; i < n; i++)
            Console.Write(printTribRec(i)
                                    +" ");
    }
      
    // Driver code
    public static void Main()
    {
        int n = 10;
  
        printTrib(n);
    }
}
  
// This code is contributed by vt_m.

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<?php
// A simple recursive PHP program to 
// print first n Tribinacci numbers.
  
function printTribRec($n)
{
    if ($n == 0 || $n == 1 || $n == 2)
        return 0;
  
    if ($n == 3)
        return 1;
    else
        return printTribRec($n - 1) + 
               printTribRec($n - 2) + 
               printTribRec($n - 3);
}
  
function printTrib($n)
{
    for ($i = 1; $i <= $n; $i++)
        echo printTribRec($i), " ";
}
  
    // Driver Code
    $n = 10;
    printTrib($n);
  
// This code is contributed by ajit
?>

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Output :

0 0 1 1 2 4 7 13 24 44 

Time complexity of above solution is exponential.

A better solution is to use Dynamic Programming.

C++

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// A DP based CPP
// program to print
// first n Tribinacci
// numbers.
#include <iostream>
using namespace std;
  
int printTrib(int n)
{
    int dp[n];
    dp[0] = dp[1] = 0;
    dp[2] = 1;
  
    for (int i = 3; i < n; i++)
        dp[i] = dp[i - 1] + 
                dp[i - 2] +
                dp[i - 3];
  
    for (int i = 0; i < n; i++)
        cout << dp[i] << " ";
}
  
// Driver code
int main()
{
    int n = 10;
    printTrib(n);
    return 0;
}

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// A DP based Java program
// to print first n
// Tribinacci numbers.
import java.io.*;
  
class GFG {
      
    static void printTrib(int n)
    {
        int dp[]=new int[n];
        dp[0] = dp[1] = 0;
        dp[2] = 1;
      
        for (int i = 3; i < n; i++)
            dp[i] = dp[i - 1] +
                    dp[i - 2] +
                    dp[i - 3];
      
        for (int i = 0; i < n; i++)
            System.out.print(dp[i] + " ");
    }
      
    // Driver code
    public static void main(String args[])
    {
        int n = 10;
        printTrib(n);
    }
}
  
/* This code is contributed by Nikita Tiwari.*/

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# A DP based
# Python 3 
# program to print
# first n Tribinacci
# numbers.
  
def printTrib(n) :
  
    dp = [0] * n
    dp[0] = dp[1] = 0;
    dp[2] = 1;
  
    for i in range(3,n) :
        dp[i] = dp[i - 1] + dp[i - 2] + dp[i - 3];
  
    for i in range(0,n) :
        print(dp[i] , " ", end="")
          
  
# Driver code
n = 10
printTrib(n)
  
# This code is contributed by Nikita Tiwari.

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// A DP based C# program
// to print first n
// Tribinacci numbers.
using System;
  
class GFG {
      
    static void printTrib(int n)
    {
        int []dp = new int[n];
        dp[0] = dp[1] = 0;
        dp[2] = 1;
      
        for (int i = 3; i < n; i++)
            dp[i] = dp[i - 1] +
                    dp[i - 2] +
                    dp[i - 3];
      
        for (int i = 0; i < n; i++)
        Console.Write(dp[i] + " ");
    }
      
    // Driver code
    public static void Main()
    {
        int n = 10;
        printTrib(n);
    }
}
  
/* This code is contributed by vt_m.*/

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<?php
// A DP based PHP program 
// to print first n 
// Tribonacci numbers.
  
function printTrib($n)
{
  
    $dp[0] = $dp[1] = 0;
    $dp[2] = 1;
  
    for ($i = 3; $i < $n; $i++)
        $dp[$i] = $dp[$i - 1] + 
                  $dp[$i - 2] +
                  $dp[$i - 3];
  
    for ($i = 0; $i < $n; $i++)
        echo $dp[$i] ," ";
}
  
// Driver code
$n = 10;
printTrib($n);
  
// This code is contributed by ajit
?>

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Output :

0 0 1 1 2 4 7 13 24 44 

Time complexity of above is linear, but it requires extra space. We can optimizes space used in above solution using three variables to keep track of previous three numbers.

C++

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// A space optimized
// based CPP program to
// print first n
// Tribinacci numbers.
#include <iostream>
using namespace std;
  
void printTrib(int n)
{
    if (n < 1)
        return;
  
    // Initialize first
    // three numbers
    int first = 0, second = 0;
    int third = 1;
  
    cout << first << " ";
    if (n > 1)
        cout << second << " ";
      
    if (n > 2)
        cout << second << " ";
  
    // Loop to add previous 
    // three numbers for
    // each number starting 
    // from 3 and then assign
    // first, second, third
    // to second, third, and 
    // curr to third respectively
    for (int i = 3; i < n; i++) 
    {
        int curr = first + second + third;
        first = second;
        second = third;
        third = curr;
  
        cout << curr << " ";
    }
}
  
// Driver code
int main()
{
    int n = 10;
    printTrib(n);
    return 0;
}

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// A space optimized
// based Java program
// to print first n 
// Tribinacci numbers.
import java.io.*;
  
class GFG {
      
    static void printTrib(int n)
    {
        if (n < 1)
            return;
      
        // Initialize first
        // three numbers
        int first = 0, second = 0;
        int third = 1;
      
        System.out.print(first + " ");
        if (n > 1)
            System.out.print(second + " ");
          
        if (n > 2)
            System.out.print(second + " ");
      
        // Loop to add previous
        // three numbers for
        // each number starting
        // from 3 and then assign
        // first, second, third
        // to second, third, and curr
        // to third respectively
        for (int i = 3; i < n; i++) 
        {
            int curr = first + second + third;
            first = second;
            second = third;
            third = curr;
      
            System.out.print(curr +" ");
        }
    }
      
    // Driver code
    public static void main(String args[])
    {
        int n = 10;
        printTrib(n);
    }
}
  
// This code is contributed by Nikita Tiwari.

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# A space optimized
# based Python 3 
# program to print
# first n Tribinacci 
# numbers.
  
def printTrib(n) :
    if (n < 1) :
        return
   
    # Initialize first
    # three numbers
    first = 0
    second = 0
    third = 1
  
    print( first , " ", end="")
    if (n > 1) :
        print(second, " ",end="")
    if (n > 2) :
        print(second, " ", end="")
  
    # Loop to add previous
    # three numbers for
    # each number starting
    # from 3 and then assign
    # first, second, third
    # to second, third, and curr
    # to third respectively
    for i in range(3, n) :
        curr = first + second + third
        first = second
        second = third
        third = curr
  
        print(curr , " ", end="")
      
      
# Driver code
n = 10
printTrib(n)
  
# This code is contributed by Nikita Tiwari.

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// A space optimized
// based C# program
// to print first n 
// Tribinacci numbers.
using System;
  
class GFG {
      
    static void printTrib(int n)
    {
        if (n < 1)
            return;
      
        // Initialize first
        // three numbers
        int first = 0, second = 0;
        int third = 1;
      
        Console.Write(first + " ");
        if (n > 1)
        Console.Write(second + " ");
          
        if (n > 2)
        Console.Write(second + " ");
      
        // Loop to add previous
        // three numbers for
        // each number starting
        // from 3 and then assign
        // first, second, third
        // to second, third, and curr
        // to third respectively
        for (int i = 3; i < n; i++) 
        {
            int curr = first + second + third;
            first = second;
            second = third;
            third = curr;
      
            Console.Write(curr +" ");
        }
    }
      
    // Driver code
    public static void Main()
    {
        int n = 10;
        printTrib(n);
    }
}
  
// This code is contributed by vt_m.

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<?php
// A space optimized
// based PHP program to
// print first n
// Tribinacci numbers.|
  
function printTrib($n)
{
    if ($n < 1)
        return;
  
    // Initialize first
    // three numbers
    $first = 0; $second = 0;
    $third = 1;
  
    echo $first, " ";
    if ($n > 1)
        echo $second , " ";
      
    if ($n > 2)
        echo $second , " ";
  
    // Loop to add previous 
    // three numbers for
    // each number starting 
    // from 3 and then assign
    // first, second, third
    // to second, third, and 
    // curr to third respectively
    for ($i = 3; $i < $n; $i++) 
    {
        $curr = $first + $second + $third;
        $first = $second;
        $second = $third;
        $third = $curr;
  
        echo $curr , " ";
    }
}
  
    // Driver code
    $n = 10;
    printTrib($n);
  
// This code is contributed by m_kit
?>

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Output :

0 0 0 1 2 4 7 13 24 44  

Below is more efficient solution using matrix exponentiation.

C++

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#include <iostream>
using namespace std;
  
// Program to print first n 
// tribonacci numbers Matrix 
// Multiplication function 
// for 3*3 matrix
void multiply(int T[3][3], int M[3][3])
{
    int a, b, c, d, e, f, g, h, i;
    a = T[0][0] * M[0][0] + 
        T[0][1] * M[1][0] + 
        T[0][2] * M[2][0];
    b = T[0][0] * M[0][1] + 
        T[0][1] * M[1][1] + 
        T[0][2] * M[2][1];
    c = T[0][0] * M[0][2] + 
        T[0][1] * M[1][2] + 
        T[0][2] * M[2][2];
    d = T[1][0] * M[0][0] + 
        T[1][1] * M[1][0] + 
        T[1][2] * M[2][0];
    e = T[1][0] * M[0][1] + 
        T[1][1] * M[1][1] + 
        T[1][2] * M[2][1];
    f = T[1][0] * M[0][2] + 
        T[1][1] * M[1][2] + 
        T[1][2] * M[2][2];
    g = T[2][0] * M[0][0] + 
        T[2][1] * M[1][0] + 
        T[2][2] * M[2][0];
    h = T[2][0] * M[0][1] + 
        T[2][1] * M[1][1] + 
        T[2][2] * M[2][1];
    i = T[2][0] * M[0][2] + 
        T[2][1] * M[1][2] + 
        T[2][2] * M[2][2];
    T[0][0] = a;
    T[0][1] = b;
    T[0][2] = c;
    T[1][0] = d;
    T[1][1] = e;
    T[1][2] = f;
    T[2][0] = g;
    T[2][1] = h;
    T[2][2] = i;
}
  
// Recursive function to raise 
// the matrix T to the power n
void power(int T[3][3], int n)
{
    // base condition.
    if (n == 0 || n == 1)
        return;
    int M[3][3] = {{ 1, 1, 1 }, 
                   { 1, 0, 0 }, 
                   { 0, 1, 0 }};
  
    // recursively call to
    // square the matrix
    power(T, n / 2);
  
    // calculating square 
    // of the matrix T
    multiply(T, T);
  
    // if n is odd multiply 
    // it one time with M
    if (n % 2)
        multiply(T, M);
}
int tribonacci(int n)
{
    int T[3][3] = {{ 1, 1, 1 }, 
                   { 1, 0, 0 },
                   { 0, 1, 0 }};
  
    // base condition
    if (n == 0 || n == 1)
        return 0;
    else
        power(T, n - 2);
  
    // T[0][0] contains the 
    // tribonacci number so
    // return it
    return T[0][0];
}
  
// Driver Code
int main()
{
    int n = 10;
    for (int i = 0; i < n; i++)
        cout << tribonacci(i) << " ";
    cout << endl;
    return 0;
}

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// Java Program to print
// first n tribonacci numbers
// Matrix Multiplication
// function for 3*3 matrix
import java.io.*;
  
class GFG 
{
    static void multiply(int T[][], int M[][])
    {
        int a, b, c, d, e, f, g, h, i;
        a = T[0][0] * M[0][0] + 
            T[0][1] * M[1][0] + 
            T[0][2] * M[2][0];
        b = T[0][0] * M[0][1] + 
            T[0][1] * M[1][1] + 
            T[0][2] * M[2][1];
        c = T[0][0] * M[0][2] + 
            T[0][1] * M[1][2] + 
            T[0][2] * M[2][2];
        d = T[1][0] * M[0][0] + 
            T[1][1] * M[1][0] + 
            T[1][2] * M[2][0];
        e = T[1][0] * M[0][1] + 
            T[1][1] * M[1][1] + 
            T[1][2] * M[2][1];
        f = T[1][0] * M[0][2] + 
            T[1][1] * M[1][2] + 
            T[1][2] * M[2][2];
        g = T[2][0] * M[0][0] + 
            T[2][1] * M[1][0] + 
            T[2][2] * M[2][0];
        h = T[2][0] * M[0][1] + 
            T[2][1] * M[1][1] + 
            T[2][2] * M[2][1];
        i = T[2][0] * M[0][2] + 
            T[2][1] * M[1][2] + 
            T[2][2] * M[2][2];
        T[0][0] = a;
        T[0][1] = b;
        T[0][2] = c;
        T[1][0] = d;
        T[1][1] = e;
        T[1][2] = f;
        T[2][0] = g;
        T[2][1] = h;
        T[2][2] = i;
    }
      
    // Recursive function to raise 
    // the matrix T to the power n
    static void power(int T[][], int n)
    {
        // base condition.
        if (n == 0 || n == 1)
            return;
        int M[][] = {{ 1, 1, 1 }, 
                     { 1, 0, 0 }, 
                     { 0, 1, 0 }};
      
        // recursively call to 
        // square the matrix
        power(T, n / 2);
      
        // calculating square 
        // of the matrix T
        multiply(T, T);
      
        // if n is odd multiply
        // it one time with M
        if (n % 2 != 0)
            multiply(T, M);
    }
    static int tribonacci(int n)
    {
        int T[][] = {{ 1, 1, 1 }, 
                     { 1, 0, 0 },
                     { 0, 1, 0 }};
      
        // base condition
        if (n == 0 || n == 1)
            return 0;
        else
            power(T, n - 2);
      
        // T[0][0] contains the 
        // tribonacci number so
        // return it
        return T[0][0];
    }
      
    // Driver Code
    public static void main(String args[])
    {
        int n = 10;
        for (int i = 0; i < n; i++)
        System.out.print(tribonacci(i) + " ");
        System.out.println();
    }
}
  
// This code is contributed by Nikita Tiwari.

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# Program to print first n tribonacci 
# numbers Matrix Multiplication
# function for 3*3 matrix
def multiply(T, M):
      
    a = (T[0][0] * M[0][0] + T[0][1] *
         M[1][0] + T[0][2] * M[2][0])             
    b = (T[0][0] * M[0][1] + T[0][1] * 
         M[1][1] + T[0][2] * M[2][1]) 
    c = (T[0][0] * M[0][2] + T[0][1] * 
         M[1][2] + T[0][2] * M[2][2])
    d = (T[1][0] * M[0][0] + T[1][1] * 
         M[1][0] + T[1][2] * M[2][0]) 
    e = (T[1][0] * M[0][1] + T[1][1] * 
         M[1][1] + T[1][2] * M[2][1])
    f = (T[1][0] * M[0][2] + T[1][1] *
         M[1][2] + T[1][2] * M[2][2])
    g = (T[2][0] * M[0][0] + T[2][1] * 
         M[1][0] + T[2][2] * M[2][0])
    h = (T[2][0] * M[0][1] + T[2][1] * 
         M[1][1] + T[2][2] * M[2][1])
    i = (T[2][0] * M[0][2] + T[2][1] * 
         M[1][2] + T[2][2] * M[2][2])
              
    T[0][0] = a
    T[0][1] = b
    T[0][2] = c
    T[1][0] = d
    T[1][1] = e
    T[1][2] = f
    T[2][0] = g
    T[2][1] = h
    T[2][2] = i
  
# Recursive function to raise 
# the matrix T to the power n
def power(T, n):
  
    # base condition.
    if (n == 0 or n == 1):
        return;
    M = [[ 1, 1, 1 ], 
                [ 1, 0, 0 ], 
                [ 0, 1, 0 ]]
  
    # recursively call to
    # square the matrix
    power(T, n // 2)
  
    # calculating square 
    # of the matrix T
    multiply(T, T)
  
    # if n is odd multiply 
    # it one time with M
    if (n % 2):
        multiply(T, M)
  
def tribonacci(n):
      
    T = [[ 1, 1, 1 ], 
        [1, 0, 0 ],
        [0, 1, 0 ]]
  
    # base condition
    if (n == 0 or n == 1):
        return 0
    else:
        power(T, n - 2)
  
    # T[0][0] contains the 
    # tribonacci number so
    # return it
    return T[0][0]
  
# Driver Code
if __name__ == "__main__":
    n = 10
    for i in range(n):
        print(tribonacci(i),end=" ")
    print()
  
# This code is contributed by ChitraNayal

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// C# Program to print
// first n tribonacci numbers
// Matrix Multiplication
// function for 3*3 matrix
using System;
  
class GFG 
{
    static void multiply(int [,]T, 
                         int [,]M)
    {
        int a, b, c, d, e, f, g, h, i;
        a = T[0,0] * M[0,0] + 
            T[0,1] * M[1,0] + 
            T[0,2] * M[2,0];
        b = T[0,0] * M[0,1] + 
            T[0,1] * M[1,1] + 
            T[0,2] * M[2,1];
        c = T[0,0] * M[0,2] + 
            T[0,1] * M[1,2] + 
            T[0,2] * M[2,2];
        d = T[1,0] * M[0,0] + 
            T[1,1] * M[1,0] + 
            T[1,2] * M[2,0];
        e = T[1,0] * M[0,1] + 
            T[1,1] * M[1,1] + 
            T[1,2] * M[2,1];
        f = T[1,0] * M[0,2] + 
            T[1,1] * M[1,2] + 
            T[1,2] * M[2,2];
        g = T[2,0] * M[0,0] + 
            T[2,1] * M[1,0] + 
            T[2,2] * M[2,0];
        h = T[2,0] * M[0,1] + 
            T[2,1] * M[1,1] + 
            T[2,2] * M[2,1];
        i = T[2,0] * M[0,2] + 
            T[2,1] * M[1,2] + 
            T[2,2] * M[2,2];
        T[0,0] = a;
        T[0,1] = b;
        T[0,2] = c;
        T[1,0] = d;
        T[1,1] = e;
        T[1,2] = f;
        T[2,0] = g;
        T[2,1] = h;
        T[2,2] = i;
    }
      
    // Recursive function to raise 
    // the matrix T to the power n
    static void power(int [,]T, int n)
    {
        // base condition.
        if (n == 0 || n == 1)
            return;
        int [,]M = {{ 1, 1, 1 }, 
                    { 1, 0, 0 }, 
                    { 0, 1, 0 }};
      
        // recursively call to 
        // square the matrix
        power(T, n / 2);
      
        // calculating square 
        // of the matrix T
        multiply(T, T);
      
        // if n is odd multiply 
        // it one time with M
        if (n % 2 != 0)
            multiply(T, M);
    }
      
    static int tribonacci(int n)
    {
        int [,]T = {{ 1, 1, 1 }, 
                    { 1, 0, 0 },
                    { 0, 1, 0 }};
      
        // base condition
        if (n == 0 || n == 1)
            return 0;
        else
            power(T, n - 2);
      
        // T[0][0] contains the 
        // tribonacci number so
        // return it
        return T[0,0];
    }
      
    // Driver Code
    public static void Main()
    {
        int n = 10;
        for (int i = 0; i < n; i++)
        Console.Write(tribonacci(i) + " ");
        Console.WriteLine();
    }
}
  
// This code is contributed by vt_m.

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PHP


Output :

0 0 1 1 2 4 7 13 24 44 

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