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# Cunningham Numbers

• Last Updated : 16 Jul, 2021

Cunningham Number is a number N of the form , where a, b >= 2.
Few Cunningham numbers are:

3, 5, 7, 8, 9, 10, 15, 17, 24, 26, 28…

### Check if N is a Cunningham number

Given a number N, the task is to check if N is an Cunningham Number or not. If N is an Cunningham Number then print “Yes” else print “No”.
Examples:

Input: N = 126
Output: Yes
Explanation:
126 = 5^3+1
Input: N = 16
Output: No

Approach: The idea is to solve the equation in a desired form such that checking that the number is a Cunningham Number or not is easy.

// Cunningham Numbers are the
// which can be represented as
=> => Therefore, if or can be expessed in the form of , then the number is cunningham Number.
Below is the implementation of the above approach:

## C++

 // C++ implementation for the// above approach #include using namespace std; // Function to check if a number// can be expressed as a^b.bool isPower(int a){    if (a == 1)        return true;     for (int i = 2; i * i <= a; i++) {        double val = log(a) / log(i);        if ((val - (int)val) < 0.00000001)            return true;    }     return false;} // Function to check if N is a// Cunningham numberbool isCunningham(int n){    return isPower(n - 1) ||           isPower(n + 1);} // Driver Codeint main(){    // Given Number    int n = 126;     // Function Call    if (isCunningham(n))        cout << "Yes";    else        cout << "No";    return 0;}

## Java

 // Java implementation for the above approachimport java.util.*; class GFG{ // Function to check if a number// can be expressed as a^b.static boolean isPower(int a){    if (a == 1)        return true;     for(int i = 2; i * i <= a; i++)    {       double val = Math.log(a) / Math.log(i);       if ((val - (int)val) < 0.00000001)           return true;    }    return false;} // Function to check if N is a// Cunningham numberstatic boolean isCunningham(int n){    return isPower(n - 1) ||           isPower(n + 1);} // Driver Codepublic static void main (String[] args){         // Given Number    int n = 126;     // Function Call    if (isCunningham(n))        System.out.print("Yes");    else        System.out.print("No");}} // This code is contributed by Ritik Bansal

## Python3

 # Python3 implementation for the# above approachimport math # Function to check if a number# can be expressed as a^b.def isPower(a):         if (a == 1):        return True         i = 2    while(i * i <= a):        val = math.log(a) / math.log(i)        if ((val - int(val)) < 0.00000001):            return True        i += 1    return False     # Function to check if N is a# Cunningham numberdef isCunningham(n):    return isPower(n - 1) or isPower(n + 1)     # Driver Code # Given Numbern = 126 # Function Callif (isCunningham(n)):    print("Yes")else:    print("No") # This code is contributed by shubhamsingh10

## C#

 // C# implementation for the// above approachusing System;class GFG{     // Function to check if a number// can be expressed as a^b.static bool isPower(int a){    if (a == 1)        return true;     for(int i = 2; i * i <= a; i++)    {       double val = Math.Log(a) / Math.Log(i);       if ((val - (int)val) < 0.00000001)           return true;    }    return false;} // Function to check if N is a// Cunningham numberstatic bool isCunningham(int n){    return isPower(n - 1) ||           isPower(n + 1);} // Driver Codepublic static void Main (string[] args){         // Given number    int n = 126;     // Function Call    if (isCunningham(n))        Console.Write("Yes");    else        Console.Write("No");}} // This code is contributed by rock_cool

## Javascript

 
Output:
Yes

Time Complexity: O(n1/2)

Auxiliary Space: O(1)

Reference: https://oeis.org/A080262

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