N-th term in the series 1, 11, 55, 239, 991,….

Given a number N. The task is to write a program to find the N-th term in the series:

1, 11, 55, 239, 991, …

Examples:



Input: N = 3
Output: 55

Input: N = 4 
Output: 239 

Approach-1: On writing down the binary representation of the given numbers, a pattern can be observed.

1 = 1
11 = 1011
55 = 110111
239 = 11101111
.
.
.

Hence for N = 1, the answer will always be one. For N-th term the binary string will be (n-1)*1 + (0) + (n)*1 which is converted to decimal value to get the answer.

Below is the implementation of the above approach:

C++



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// C++ program to find the N-th term
// in 1, 11, 55, 239, 991, ....
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the decimal value
// of a binary number
int binaryToDecimal(string n)
{
    string num = n;
    int dec_value = 0;
  
    // Initializing base value to 1, i.e 2^0
    int base = 1;
  
    int len = num.length();
    for (int i = len - 1; i >= 0; i--) {
        if (num[i] == '1')
            dec_value += base;
        base = base * 2;
    }
  
    return dec_value;
}
  
// find the binary representation
// of the N-th number in sequence
int numberSequence(int n)
{
    // base case
    if (n == 1)
        return 1;
  
    // answer string
    string s = "";
  
    // add n-1 1's
    for (int i = 1; i < n; i++)
        s += '1';
  
    // add 0
    s += '0';
  
    // add n 1's at end
    for (int i = 1; i <= n; i++)
        s += '1';
  
    int num = binaryToDecimal(s);
  
    return num;
}
  
// Driver Code
int main()
{
    int n = 4;
  
    cout << numberSequence(n);
  
    return 0;
}

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Java

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// Java program to find the N-th 
// term in 1, 11, 55, 239, 991, ....
import java.util.*;
  
class GFG
{
  
// Function to return the decimal
// value of a binary number
static int binaryToDecimal(String n)
{
    String num = n;
    int dec_value = 0;
  
    // Initializing base
    // value to 1, i.e 2^0
    int base = 1;
  
    int len = num.length();
    for (int i = len - 1; i >= 0; i--)
    {
        if (num.charAt(i) == '1')
            dec_value += base;
        base = base * 2;
    }
  
    return dec_value;
}
  
// find the binary representation
// of the N-th number in sequence
static int numberSequence(int n)
{
    // base case
    if (n == 1)
        return 1;
  
    // answer string
    String s = "";
  
    // add n-1 1's
    for (int i = 1; i < n; i++)
        s += '1';
  
    // add 0
    s += '0';
  
    // add n 1's at end
    for (int i = 1; i <= n; i++)
        s += '1';
  
    int num = binaryToDecimal(s);
  
    return num;
}
  
// Driver Code
public static void main(String args[])
{
    int n = 4;
  
    System.out.println(numberSequence(n));
}
}
  
// This code is contributed 
// by Arnab Kundu

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Python 3

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# Python 3 program to find the N-th term
# in 1, 11, 55, 239, 991, ....
   
# Function to return the decimal value
# of a binary number
def binaryToDecimal(n):
  
    num = n
    dec_value = 0
   
    # Initializing base value to 1, i.e 2^0
    base = 1
   
    l = len(num)
    for i in range(l - 1,-1, -1):
        if (num[i] == '1'):
            dec_value += base
        base = base * 2
   
    return dec_value
   
# find the binary representation
# of the N-th number in sequence
def numberSequence(n):
      
    # base case
    if (n == 1):
        return 1
   
    # answer string
    s = ""
   
    # add n-1 1's
    for i in range(1, n):
        s += '1'
   
    # add 0
    s += '0'
   
    # add n 1's at end
    for i in range(1,n+1):
        s += '1'
   
    num = binaryToDecimal(s)
   
    return num
   
# Driver Code
if __name__ == "__main__":
      
    n = 4
   
    print(numberSequence(n))
  
# this code is contributed by ChitraNayal

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C#

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// C# program to find the N-th 
// term in 1, 11, 55, 239, 991, ....
using System;
  
class GFG
{
  
// Function to return the decimal
// value of a binary number
static int binaryToDecimal(String n)
{
    String num = n;
    int dec_value = 0;
  
    // Initializing base
    // value to 1, i.e 2^0
    int base_ = 1;
  
    int len = num.Length;
    for (int i = len - 1; i >= 0; i--)
    {
        if (num[i] == '1')
            dec_value += base_;
        base_ = base_ * 2;
    }
  
    return dec_value;
}
  
// find the binary representation
// of the N-th number in sequence
static int numberSequence(int n)
{
    // base case
    if (n == 1)
        return 1;
  
    // answer string
    String s = "";
  
    // add n-1 1's
    for (int i = 1; i < n; i++)
        s += '1';
  
    // add 0
    s += '0';
  
    // add n 1's at end
    for (int i = 1; i <= n; i++)
        s += '1';
  
    int num = binaryToDecimal(s);
  
    return num;
}
  
// Driver Code
public static void Main()
{
    int n = 4;
  
    Console.WriteLine(numberSequence(n));
}
}
  
// This code is contributed
// by Subhadeep

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PHP

= 0; $i–)
{
if ($num[$i] == ‘1’)
$dec_value += $base;
$base = $base * 2;
}

return $dec_value;
}

// find the binary representation
// of the N-th number in sequence
function numberSequence($n)
{
// base case
if ($n == 1)
return 1;

// answer string
$s = “”;

// add n-1 1’s
for ($i = 1; $i < $n; $i++) $s .= '1'; // add 0 $s .= '0'; // add n 1's at end for ($i = 1; $i <= $n; $i++) $s .= '1'; $num = binaryToDecimal($s); return $num; } // Driver Code $n = 4; echo numberSequence($n); // This code is contributed by mits ?>

Output:

239

Approach-2: The series has a general formulae of 4N-2N-1 which is used to get the N-th term in series.

Below is the implementation of the above approach:

C++

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// C++ program to find the N-th term
// in 1, 11, 55, 239, 991, ....
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the N-th term
int numberSequence(int n)
{
    // calculates the N-th term
    int num = pow(4, n) - pow(2, n) - 1;
  
    return num;
}
  
// Driver Code
int main()
{
    int n = 4;
  
    cout << numberSequence(n);
  
    return 0;
}

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Java

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// Java program to find the N-th 
// term in 1, 11, 55, 239, 991, ....
class GFG
{
// Function to find the N-th term
static int numberSequence(int n)
{
    // calculates the N-th term
    int num = (int)(Math.pow(4, n) - 
                    Math.pow(2, n)) - 1;
  
    return num;
}
  
// Driver Code
public static void main(String args[])
{
    int n = 4;
  
    System.out.println(numberSequence(n));
}
}
  
// This code is contributed 
// by Arnab Kundu

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Python 3

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# Python 3 program to find N-th term 
# in 1, 11, 55, 239, 991, .... 
  
# calculate Nth term of series
def numberSequence(n) :
  
    # calculates the N-th term 
    num = pow(4, n) - pow(2, n) - 1
  
    return num
  
# Driver Code
if __name__ == "__main__" :
  
    n = 4
      
    print(numberSequence(n))
  
# This code is contributed by ANKITRAI1

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C#

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// C# program to find the N-th 
// term in 1, 11, 55, 239, 991, ....
using System;
  
class GFG
{
// Function to find the N-th term
static int numberSequence(int n)
{
    // calculates the N-th term
    int num = (int)(Math.Pow(4, n) - 
                    Math.Pow(2, n)) - 1;
  
    return num;
}
  
// Driver Code
public static void Main()
{
    int n = 4;
  
    Console.WriteLine(numberSequence(n));
}
}
  
// This code is contributed 
// by chandan_jnu.

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PHP

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<?php
// PHP program to find the N-th term
// in 1, 11, 55, 239, 991, ....
  
// Function to find the N-th term
function numberSequence($n)
{
    // calculates the N-th term
    $num = pow(4, $n) - 
           pow(2, $n) - 1;
  
    return $num;
}
  
// Driver Code
$n = 4;
  
echo numberSequence($n);
  
// This code is contributed by mits
?>

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Output:

239


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