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++

// 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;
}

Java

// 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

Python 3

# 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

C#

// 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

PHP

<?php
// PHP program to find the N-th term
// in 1, 11, 55, 239, 991, ....

// Function to return the decimal
// value of a binary number
function binaryToDecimal($n)
{
    $num = $n;
    $dec_value = 0;

    // Initializing base value
    // to 1, i.e 2^0
    $base = 1;

    $len = strlen($num);
    for ($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
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
?>

Javascript

<script>
// Javascript program to find the N-th term
// in 1, 11, 55, 239, 991, ....

// Function to return the decimal value
// of a binary number
function binaryToDecimal(n)
{
    let num = n;
    let dec_value = 0;

    // Initializing base value to 1, i.e 2^0
    let base = 1;

    let len = num.length;
    for (let 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
function numberSequence(n)
{
    // base case
    if (n == 1)
        return 1;

    // answer string
    let s = "";

    // add n-1 1's
    for (let i = 1; i < n; i++)
        s += '1';

    // add 0
    s += '0';

    // add n 1's at end
    for (let i = 1; i <= n; i++)
        s += '1';

    let num = binaryToDecimal(s);

    return num;
}

// Driver Code
let n = 4;

document.write(numberSequence(n));

// This code is contributed by subhammahato348.
</script>

Output

239

Time Complexity: O(N), as we are using a loop to traverse N times.
Auxiliary Space: O(N), as we are using extra space for string.

Approach-2: 

The series has a general formulae of 4^N-2^N-1 which is used to get the N-th term in series. 
Below is the implementation of the above approach: 

C++

// 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;
}

Java

// 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

Python 3

# 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

C#

// 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.

PHP

<?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
?>

Javascript

<script>

// Javascript 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
    let num = Math.pow(4, n) - Math.pow(2, n) - 1;

    return num;
}

// Driver Code
let n = 4;

document.write(numberSequence(n));

</script>

Output

239

Time Complexity: O(logN), as we are using pow function which will cost logN time.
Auxiliary Space: O(1), as we are not using any extra space.

Approach-3: 

As we can see the pattern in approach 1, here also we use that pattern to calculate N-th term. But instead of string to store, we use bitwise operator(“<<” and “|”) and built-in pow() function to calculate N-th term of the sequence.

Below is the implementation of the above approach: 

C++

// C++ program to find the N-th term
// in 1, 11, 55, 239, 991, ....
#include <bits/stdc++.h>
using namespace std;

// Find N-th number in sequence
int numberSequence(int n)
{
    // base case
    if (n == 1)
        return 1;

    // value after 0
    int q = pow(2, n) - 1;

    // value before 0
    int p = pow(2, (n - 1)) - 1;

    // calculate N-th term using shifting and OR operation
    p = p << (n + 1);
    p = p | q;

    return p;
}

// Driver Code
int main()
{
    int n = 4;
    cout << numberSequence(n);
    return 0;
}

// This code is contributed by Susobhan Akhuli

Java

// Java program to find the N-th term
// in 1, 11, 55, 239, 991, ....
import java.math.BigInteger;

class GFG {

    public static void main(String[] args)
    {
        int n = 4;
        System.out.println(numberSequence(n));
    }

    public static int numberSequence(int n)
    {
        if (n == 1) {
            return 1;
        }

        int q = (int)Math.pow(2, n) - 1;
        int p = (int)Math.pow(2, n - 1) - 1;

        p = p << (n + 1);
        p = p | q;

        return p;
    }
}

// This code is contributed by Susobhan Akhuli

Python3

# Python program to find the N-th term
# in 1, 11, 55, 239, 991, ....
import math


def numberSequence(n: int) -> int:
    # base case
    if n == 1:
        return 1

    # value after 0
    q = pow(2, n) - 1

    # value before 0
    p = pow(2, (n - 1)) - 1

    # calculate N-th term using shifting and OR operation
    p = p << (n + 1)
    p = p | q

    return p


# Driver Code
n = 4
print(numberSequence(n))

# This code is contributed by Susobhan Akhuli

C#

// C# program to find the N-th term
// in 1, 11, 55, 239, 991, ....
using System;

class Program {
    // Find N-th number in sequence
    static int numberSequence(int n)
    {
        // base case
        if (n == 1)
            return 1;

        // value after 0
        int q = (int)Math.Pow(2, n) - 1;

        // value before 0
        int p = (int)Math.Pow(2, (n - 1)) - 1;

        // calculate N-th term using shifting and OR
        // operation
        p = p << (n + 1);
        p = p | q;

        return p;
    }

    // Driver code
    static void Main(string[] args)
    {
        int n = 4;
        Console.WriteLine(numberSequence(n));
    }
}

// This code is contributed by Susobhan Akhuli

Javascript

// Find N-th number in sequence
function numberSequence(n) {
  // base case
  if (n === 1) {
    return 1;
  }

  // value after 0
  let q = Math.pow(2, n) - 1;

  // value before 0
  let p = Math.pow(2, n - 1) - 1;

  // calculate N-th term using shifting and OR operation
  p = p << (n + 1);
  p = p | q;

  return p;
}

// Driver Code
let n = 4;
console.log(numberSequence(n));

Output

239

Time Complexity: O(logN) [For pow function]
Auxiliary Space: O(1)

Approach-4:

As we can see the pattern in approach 2, here do the same but instead of using pow() function, here we use the math.log2() in combination with the operator “**” to calculate N-th term of the sequence.

Below is the implementation of the above approach:

C++

// CPP program to find N-th term
// in 1, 11, 55, 239, 991, ....

#include <cmath>
#include <iostream>
using namespace std;

// calculates Nth term of series
int numberSequence(int n)
{
    // calculates the N-th term
    int num = pow(2, n * (int)(log2(4)))
              - pow(2, n * (int)(log2(2))) - 1;
    return num;
}

// Driver Code
int main()
{
    int n = 4;
    cout << numberSequence(n) << endl;
    return 0;
}

// This code is contributed by Susobhan Akhuli

Java

// Java program to find N-th term
// in 1, 11, 55, 239, 991, ....
import java.lang.Math;

public class GFG {
    // calculates Nth term of series
    public static int numberSequence(int n)
    {
        // calculates the N-th term
        int num
            = (int)Math.pow(
                  2, n * (int)(Math.log(4) / Math.log(2)))
              - (int)Math.pow(
                  2, n * (int)(Math.log(2) / Math.log(2)))
              - 1;
        return num;
    }

    public static void main(String[] args)
    {
        int n = 4;
        System.out.println(numberSequence(n));
    }
}

// This code is contributed by Susobhan Akhuli

Python3

# Python 3 program to find N-th term
# in 1, 11, 55, 239, 991, ....
import math

# calculate Nth term of series
def numberSequence(n):
    # calculates the N-th term
    num = 2 ** (n * int(math.log2(4))) - 2 ** (n * int(math.log2(2))) - 1
    return num

# Driver Code
if __name__ == "__main__":
    n = 4
    print(numberSequence(n))

# This code is contributed by Susobhan Akhuli

C#

// C# program to find N-th term
// in 1, 11, 55, 239, 991, ....

using System;

class GFG {
    // calculates Nth term of series
    static int NumberSequence(int n)
    {
        // calculates the N-th term
        int num
            = (int)(Math.Pow(2, n * (int)(Math.Log(4, 2))))
              - (int)(Math.Pow(2,
                               n * (int)(Math.Log(2, 2))))
              - 1;
        return num;
    }

    // Driver Code
    static void Main(string[] args)
    {
        int n = 4;
        Console.WriteLine(NumberSequence(n));
    }
}
// This code is contributed by Susobhan Akhuli

Javascript

// calculate Nth term of series
function numberSequence(n) {
// calculates the N-th term
let num = Math.pow(2, n * Math.log2(4)) - Math.pow(2, n * Math.log2(2)) - 1;
return num;
}

// Driver Code
const n = 4;
console.log(numberSequence(n));

Output

239

Time Complexity: O(Log(exponent))
Auxiliary Space: O(1)