Program to print the series 1, 3, 4, 8, 15, 27, 50… till N terms

Given a number N, the task is to print the first N terms of the following series:

1, 3, 4, 8, 15, 27, 50…

Examples:

Input: N = 7 
Output: 1, 3, 4, 8, 15, 27, 50
Input: N = 3 
Output: 1, 3, 4 
 

Approach: From the given series we can find the formula for Nth term:

1st term = 1, 2nd term = 3, 3rd term = 4 
4th term = 1st term + 2nd term + 3rd term 
5th term = 2nd term + 3rd term + 4th term 
6th term = 3rd term + 4th term + 5th term 


so on 
 



Therefore, the idea is to keep track of the last three terms of the series and find the consecutive terms of the series.

Below is the implementation of above approach:

C++

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// C++ implementation to print the
// N terms of the series whose three
// terms are given
  
#include "bits/stdc++.h"
using namespace std;
  
// Function to print the series
void printSeries(int n, int a,
                 int b, int c)
{
  
    int d;
  
    // Generate the ith term and
    // print it
    if (n == 1) {
        cout << a << " ";
        return;
    }
    if (n == 2) {
        cout << a << " " << b << " ";
        return;
    }
  
    cout << a << " " << b
         << " " << c << " ";
  
    for (int i = 4; i <= n; i++) {
        d = a + b + c;
        cout << d << " ";
        a = b;
        b = c;
        c = d;
    }
}
  
// Driver Code
int main()
{
    int N = 7, a = 1, b = 3;
    int c = 4;
  
    // Function Call
    printSeries(N, a, b, c);
    return 0;
}

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Java

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// Java implementation to print the
// N terms of the series whose three
// terms are given
   
//include "bits/stdJava.h"
import java.util.*;
class GFG{
   
// Function to print the series
static void printSeries(int n, int a,
                         int b, int c)
{
    int d;
   
    // Generate the ith term and
    // print it
    if (n == 1
    {
        System.out.print(a + " ");
        return;
    }
    if (n == 2
    {
        System.out.print(a + " " +  b + " ");
        return;
    }
   
    System.out.print(a + " "
                     b + " " +  
                     c + " ");
   
    for (int i = 4; i <= n; i++) 
    {
        d = a + b + c;
        System.out.print(d + " ");
        a = b;
        b = c;
        c = d;
    }
}
   
// Driver Code
public static void main(String[] args)
{
    int N = 7, a = 1, b = 3;
    int c = 4;
   
    // Function Call
    printSeries(N, a, b, c);
}
}
  
// This code is contributed by sapnasingh4991

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Python3

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# Python3 implementation to print the
# N terms of the series whose three
# terms are given
  
# Function to print the series
def printSeries(n, a, b, c):
  
    # Generate the ith term and
    # print it
    if (n == 1):
        print(a, end = " ");
        return;
      
    if (n == 2):
        print(a, b, end = " ");
        return;
      
    print(a, b, c, end = " ");
  
    for i in range (4, n + 1):
        d = a + b + c;
        print(d, end = " ");
        a = b;
        b = c;
        c = d;
      
# Driver Code
N = 7; a = 1; b = 3;
c = 4;
  
# Function Call
printSeries(N, a, b, c);
  
# This code is contributed by Code_Mech

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C#

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// C# implementation to print the
// N terms of the series whose three
// terms are given
using System;
class GFG{
  
// Function to print the series
static void printSeries(int n, int a,
                        int b, int c)
{
    int d;
  
    // Generate the ith term and
    // print it
    if (n == 1) 
    {
        Console.Write(a + " ");
        return;
    }
    if (n == 2) 
    {
        Console.Write(a + " "
                      b + " ");
        return;
    }
  
    Console.Write(a + " "
                  b + " "
                  c + " ");
  
    for(int i = 4; i <= n; i++) 
    {
       d = a + b + c;
       Console.Write(d + " ");
  
       a = b;
       b = c;
       c = d;
    }
}
  
// Driver Code
public static void Main()
{
    int N = 7, a = 1, b = 3;
    int c = 4;
  
    // Function call
    printSeries(N, a, b, c);
}
}
  
// This code is contributed by rock cool

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Output:

1 3 4 8 15 27 50

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Improved By : rock_cool, Code_Mech

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