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Euler’s Totient Function
• Difficulty Level : Medium
• Last Updated : 01 Apr, 2021

Euler’s Totient function Φ (n) for an input n is the count of numbers in {1, 2, 3, …, n} that are relatively prime to n, i.e., the numbers whose GCD (Greatest Common Divisor) with n is 1.
Examples :

```Φ(1) = 1
gcd(1, 1) is 1

Φ(2) = 1
gcd(1, 2) is 1, but gcd(2, 2) is 2.

Φ(3) = 2
gcd(1, 3) is 1 and gcd(2, 3) is 1

Φ(4) = 2
gcd(1, 4) is 1 and gcd(3, 4) is 1

Φ(5) = 4
gcd(1, 5) is 1, gcd(2, 5) is 1,
gcd(3, 5) is 1 and gcd(4, 5) is 1

Φ(6) = 2
gcd(1, 6) is 1 and gcd(5, 6) is 1, ```

How to compute Φ(n) for an input nΦ
A simple solution is to iterate through all numbers from 1 to n-1 and count numbers with gcd with n as 1. Below is the implementation of the simple method to compute Euler’s Totient function for an input integer n.

## C++

 `// A simple C++ program to calculate``// Euler's Totient Function``#include ``using` `namespace` `std;` `// Function to return gcd of a and b``int` `gcd(``int` `a, ``int` `b)``{``    ``if` `(a == 0)``        ``return` `b;``    ``return` `gcd(b % a, a);``}` `// A simple method to evaluate Euler Totient Function``int` `phi(unsigned ``int` `n)``{``    ``unsigned ``int` `result = 1;``    ``for` `(``int` `i = 2; i < n; i++)``        ``if` `(gcd(i, n) == 1)``            ``result++;``    ``return` `result;``}` `// Driver program to test above function``int` `main()``{``    ``int` `n;``    ``for` `(n = 1; n <= 10; n++)``        ``cout << ``"phi("``<

## C

 `// A simple C program to calculate Euler's Totient Function``#include ` `// Function to return gcd of a and b``int` `gcd(``int` `a, ``int` `b)``{``    ``if` `(a == 0)``        ``return` `b;``    ``return` `gcd(b % a, a);``}` `// A simple method to evaluate Euler Totient Function``int` `phi(unsigned ``int` `n)``{``    ``unsigned ``int` `result = 1;``    ``for` `(``int` `i = 2; i < n; i++)``        ``if` `(gcd(i, n) == 1)``            ``result++;``    ``return` `result;``}` `// Driver program to test above function``int` `main()``{``    ``int` `n;``    ``for` `(n = 1; n <= 10; n++)``        ``printf``(``"phi(%d) = %d\n"``, n, phi(n));``    ``return` `0;``}`

## Java

 `// A simple java program to calculate``// Euler's Totient Function``import` `java.io.*;` `class` `GFG {` `    ``// Function to return GCD of a and b``    ``static` `int` `gcd(``int` `a, ``int` `b)``    ``{``        ``if` `(a == ``0``)``            ``return` `b;``        ``return` `gcd(b % a, a);``    ``}` `    ``// A simple method to evaluate``    ``// Euler Totient Function``    ``static` `int` `phi(``int` `n)``    ``{``        ``int` `result = ``1``;``        ``for` `(``int` `i = ``2``; i < n; i++)``            ``if` `(gcd(i, n) == ``1``)``                ``result++;``        ``return` `result;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `n;` `        ``for` `(n = ``1``; n <= ``10``; n++)``            ``System.out.println(``"phi("` `+ n + ``") = "` `+ phi(n));``    ``}``}` `// This code is contributed by sunnusingh`

## Python3

 `# A simple Python3 program``# to calculate Euler's``# Totient Function` `# Function to return``# gcd of a and b``def` `gcd(a, b):` `    ``if` `(a ``=``=` `0``):``        ``return` `b``    ``return` `gcd(b ``%` `a, a)` `# A simple method to evaluate``# Euler Totient Function``def` `phi(n):` `    ``result ``=` `1``    ``for` `i ``in` `range``(``2``, n):``        ``if` `(gcd(i, n) ``=``=` `1``):``            ``result``+``=``1``    ``return` `result` `# Driver Code``for` `n ``in` `range``(``1``, ``11``):``    ``print``(``"phi("``,n,``") = "``,``           ``phi(n), sep ``=` `"")``           ` `# This code is contributed``# by Smitha`

## C#

 `// A simple C# program to calculate``// Euler's Totient Function``using` `System;` `class` `GFG {` `    ``// Function to return GCD of a and b``    ``static` `int` `gcd(``int` `a, ``int` `b)``    ``{``        ``if` `(a == 0)``            ``return` `b;``        ``return` `gcd(b % a, a);``    ``}` `    ``// A simple method to evaluate``    ``// Euler Totient Function``    ``static` `int` `phi(``int` `n)``    ``{``        ``int` `result = 1;``        ``for` `(``int` `i = 2; i < n; i++)``            ``if` `(gcd(i, n) == 1)``                ``result++;``        ``return` `result;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``for` `(``int` `n = 1; n <= 10; n++)``        ``Console.WriteLine(``"phi("` `+ n + ``") = "` `+ phi(n));``    ``}``}` `// This code is contributed by nitin mittal`

## PHP

 `<Φphp``// PHP program to calculate``// Euler's Totient Function` `// Function to return``// gcd of a and b``function` `gcd(``\$a``, ``\$b``)``{``    ``if` `(``\$a` `== 0)``        ``return` `\$b``;``    ``return` `gcd(``\$b` `% ``\$a``, ``\$a``);``}` `// A simple method to evaluate``// Euler Totient Function``function` `phi(``\$n``)``{``    ``\$result` `= 1;``    ``for` `(``\$i` `= 2; ``\$i` `< ``\$n``; ``\$i``++)``        ``if` `(gcd(``\$i``, ``\$n``) == 1)``            ``\$result``++;``    ``return` `\$result``;``}` `// Driver Code``for` `(``\$n` `= 1; ``\$n` `<= 10; ``\$n``++)``    ``echo` `"phi("` `.``\$n``. ``") ="` `. phi(``\$n``).``"\n"``;` `// This code is contributed by Sam007``Φ>`

## Javascript

 ``

Output :

```phi(1) = 1
phi(2) = 1
phi(3) = 2
phi(4) = 2
phi(5) = 4
phi(6) = 2
phi(7) = 6
phi(8) = 4
phi(9) = 6
phi(10) = 4```

The above code calls gcd function O(n) times. The time complexity of the gcd function is O(h) where “h” is the number of digits in a smaller number of given two numbers. Therefore, an upper bound on the time complexity of the above solution is O(nLogn) [HowΦ there can be at most Log10n digits in all numbers from 1 to n]
Below is a Better Solution. The idea is based on Euler’s product formula which states that the value of totient functions is below the product overall prime factors p of n. The formula basically says that the value of Φ(n) is equal to n multiplied by product of (1 – 1/p) for all prime factors p of n. For example value of Φ(6) = 6 * (1-1/2) * (1 – 1/3) = 2.
We can find all prime factors using the idea used in this post.

```1) Initialize : result = n
2) Run a loop from 'p' = 2 to sqrt(n), do following for every 'p'.
a) If p divides n, then
Set: result = result  * (1.0 - (1.0 / (float) p));
Divide all occurrences of p in n.
3) Return result  ```

Below is the implementation of Euler’s product formula.

## C++

 `// C++ program to calculate Euler's``// Totient Function using Euler's``// product formula``#include ``using` `namespace` `std;` `int` `phi(``int` `n)``{``    ` `    ``// Initialize result as n``    ``float` `result = n;`` ` `    ``// Consider all prime factors of n``    ``// and for every prime factor p,``    ``// multiply result with (1 - 1/p)``    ``for``(``int` `p = 2; p * p <= n; ++p)``    ``{``        ` `        ``// Check if p is a prime factor.``        ``if` `(n % p == 0)``        ``{``            ` `            ``// If yes, then update n and result``            ``while` `(n % p == 0)``                ``n /= p;``                ` `            ``result *= (1.0 - (1.0 / (``float``)p));``        ``}``    ``}`` ` `    ``// If n has a prime factor greater than sqrt(n)``    ``// (There can be at-most one such prime factor)``    ``if` `(n > 1)``        ``result *= (1.0 - (1.0 / (``float``)n));`` ` `    ``return` `(``int``)result;``}`` ` `// Driver code``int` `main()``{``    ``int` `n;``    ` `    ``for``(n = 1; n <= 10; n++)``    ``{``        ``cout << ``"Phi"` `<< ``"("``             ``<< n << ``")"` `<< ``" = "``             ``<< phi(n) <

## C

 `// C program to calculate Euler's Totient Function``// using Euler's product formula``#include ` `int` `phi(``int` `n)``{``    ``float` `result = n; ``// Initialize result as n` `    ``// Consider all prime factors of n and for every prime``    ``// factor p, multiply result with (1 - 1/p)``    ``for` `(``int` `p = 2; p * p <= n; ++p) {``        ` `        ``// Check if p is a prime factor.``        ``if` `(n % p == 0) {``            ` `            ``// If yes, then update n and result``            ``while` `(n % p == 0)``                ``n /= p;``            ``result *= (1.0 - (1.0 / (``float``)p));``        ``}``    ``}` `    ``// If n has a prime factor greater than sqrt(n)``    ``// (There can be at-most one such prime factor)``    ``if` `(n > 1)``        ``result *= (1.0 - (1.0 / (``float``)n));` `    ``return` `(``int``)result;``}` `// Driver program to test above function``int` `main()``{``    ``int` `n;``    ``for` `(n = 1; n <= 10; n++)``        ``printf``(``"phi(%d) = %d\n"``, n, phi(n));``    ``return` `0;``}`

## Java

 `// Java program to calculate Euler's Totient``// Function using Euler's product formula``import` `java.io.*;` `class` `GFG {``    ``static` `int` `phi(``int` `n)``    ``{``        ``// Initialize result as n``        ``float` `result = n;` `        ``// Consider all prime factors of n and for``        ``// every prime factor p, multiply result``        ``// with (1 - 1/p)``        ``for` `(``int` `p = ``2``; p * p <= n; ++p) {``            ``// Check if p is a prime factor.``            ``if` `(n % p == ``0``) {``                ``// If yes, then update n and result``                ``while` `(n % p == ``0``)``                    ``n /= p;``                ``result *= (``1.0` `- (``1.0` `/ (``float``)p));``            ``}``        ``}` `        ``// If n has a prime factor greater than sqrt(n)``        ``// (There can be at-most one such prime factor)``        ``if` `(n > ``1``)``            ``result *= (``1.0` `- (``1.0` `/ (``float``)n));` `        ``return` `(``int``)result;``    ``}` `    ``// Driver program to test above function``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `n;``        ``for` `(n = ``1``; n <= ``10``; n++)``            ``System.out.println(``"phi("` `+ n + ``") = "` `+ phi(n));``    ``}``}` `// This code is contributed by Nikita Tiwari.`

## Python3

 `# Python 3 program to calculate``# Euler's Totient Function``# using Euler's product formula` `def` `phi(n) :` `    ``result ``=` `n   ``# Initialize result as n``     ` `    ``# Consider all prime factors``    ``# of n and for every prime``    ``# factor p, multiply result with (1 - 1 / p)``    ``p ``=` `2``    ``while` `p ``*` `p<``=` `n :` `        ``# Check if p is a prime factor.``        ``if` `n ``%` `p ``=``=` `0` `:` `            ``# If yes, then update n and result``            ``while` `n ``%` `p ``=``=` `0` `:``                ``n ``=` `n ``/``/` `p``            ``result ``=` `result ``*` `(``1.0` `-` `(``1.0` `/` `float``(p)))``        ``p ``=` `p ``+` `1``        ` `        ` `    ``# If n has a prime factor``    ``# greater than sqrt(n)``    ``# (There can be at-most one``    ``# such prime factor)``    ``if` `n > ``1` `:``        ``result ``=` `result ``*` `(``1.0` `-` `(``1.0` `/` `float``(n)))`` ` `    ``return` `int``(result)``    ` `    ` `# Driver program to test above function``for` `n ``in` `range``(``1``, ``11``) :``    ``print``(``"phi("``, n, ``") = "``, phi(n))``   `  `# This code is contributed``# by Nikita Tiwari.`

## C#

 `// C# program to calculate Euler's Totient``// Function using Euler's product formula``using` `System;` `class` `GFG {``    ` `    ``static` `int` `phi(``int` `n)``    ``{``        ` `        ``// Initialize result as n``        ``float` `result = n;` `        ``// Consider all prime factors``        ``// of n and for every prime``        ``// factor p, multiply result``        ``// with (1 - 1 / p)``        ``for` `(``int` `p = 2; p * p <= n; ++p)``        ``{``            ` `            ``// Check if p is a prime factor.``            ``if` `(n % p == 0)``            ``{``                ` `                ``// If yes, then update``                ``// n and result``                ``while` `(n % p == 0)``                    ``n /= p;``                ``result *= (``float``)(1.0 - (1.0 / (``float``)p));``            ``}``        ``}` `        ``// If n has a prime factor``        ``// greater than sqrt(n)``        ``// (There can be at-most``        ``// one such prime factor)``        ``if` `(n > 1)``            ``result *= (``float``)(1.0 - (1.0 / (``float``)n));` `        ``return` `(``int``)result;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``int` `n;``        ``for` `(n = 1; n <= 10; n++)``            ``Console.WriteLine(``"phi("` `+ n + ``") = "` `+ phi(n));``    ``}``}` `// This code is contributed by nitin mittal.`

## PHP

 `<Φphp``// PHP program to calculate``// Euler's Totient Function``// using Euler's product formula``function` `phi(``\$n``)``{``    ``// Initialize result as n``    ``\$result` `= ``\$n``;` `    ``// Consider all prime factors``    ``// of n and for every prime``    ``// factor p, multiply result``    ``// with (1 - 1/p)``    ``for` `(``\$p` `= 2; ``\$p` `* ``\$p` `<= ``\$n``; ++``\$p``)``    ``{``        ` `        ``// Check if p is``        ``// a prime factor.``        ``if` `(``\$n` `% ``\$p` `== 0)``        ``{``            ` `            ``// If yes, then update``            ``// n and result``            ``while` `(``\$n` `% ``\$p` `== 0)``                ``\$n` `/= ``\$p``;``            ``\$result` `*= (1.0 - (1.0 / ``\$p``));``        ``}``    ``}` `    ``// If n has a prime factor greater``    ``// than sqrt(n) (There can be at-most``    ``// one such prime factor)``    ``if` `(``\$n` `> 1)``        ``\$result` `*= (1.0 - (1.0 / ``\$n``));` `    ``return` `intval``(``\$result``);``}` `// Driver Code``for` `(``\$n` `= 1; ``\$n` `<= 10; ``\$n``++)``echo` `"phi("` `.``\$n``. ``") ="` `. phi(``\$n``).``"\n"``;``    ` `// This code is contributed by Sam007``Φ>`

## Javascript

 `// Javascript program to calculate``// Euler's Totient Function``// using Euler's product formula``function` `phi(n)``{``    ``// Initialize result as n``    ``let result = n;` `    ``// Consider all prime factors``    ``// of n and for every prime``    ``// factor p, multiply result``    ``// with (1 - 1/p)``    ``for` `(let p = 2; p * p <= n; ++p)``    ``{``        ` `        ``// Check if p is``        ``// a prime factor.``        ``if` `(n % p == 0)``        ``{``            ` `            ``// If yes, then update``            ``// n and result``            ``while` `(n % p == 0)``                ``n /= p;``            ``result *= (1.0 - (1.0 / p));``        ``}``    ``}` `    ``// If n has a prime factor greater``    ``// than sqrt(n) (There can be at-most``    ``// one such prime factor)``    ``if` `(n > 1)``        ``result *= (1.0 - (1.0 / n));` `    ``return` `parseInt(result);``}` `// Driver Code``for` `(let n = 1; n <= 10; n++)`` ``document.write(`phi(\${n}) = \${phi(n)}
`);``    ` `// This code is contributed by _saurabh_jaiswal`

Output :

```phi(1) = 1
phi(2) = 1
phi(3) = 2
phi(4) = 2
phi(5) = 4
phi(6) = 2
phi(7) = 6
phi(8) = 4
phi(9) = 6
phi(10) = 4```

We can avoid floating-point calculations in the above method. The idea is to count all prime factors and their multiples and subtract this count from n to get the totient function value (Prime factors and multiples of prime factors won’t have gcd as 1)

```1) Initialize result as n
2) Consider every number 'p' (where 'p' varies from 2 to Φn).
If p divides n, then do following
a) Subtract all multiples of p from 1 to n [all multiples of p
will have gcd more than 1 (at least p) with n]
b) Update n by repeatedly dividing it by p.
3) If the reduced n is more than 1, then remove all multiples
of n from result.```

Below is the implementation of the above algorithm.

## C++

 `// C++ program to calculate Euler's``// Totient Function``#include ``using` `namespace` `std;` `int` `phi(``int` `n)``{``    ``// Initialize result as n``    ``int` `result = n;`` ` `    ``// Consider all prime factors of n``    ``// and subtract their multiples``    ``// from result``    ``for``(``int` `p = 2; p * p <= n; ++p)``    ``{``        ` `        ``// Check if p is a prime factor.``        ``if` `(n % p == 0)``        ``{``            ` `            ``// If yes, then update n and result``            ``while` `(n % p == 0)``                ``n /= p;``                ` `            ``result -= result / p;``        ``}``    ``}`` ` `    ``// If n has a prime factor greater than sqrt(n)``    ``// (There can be at-most one such prime factor)``    ``if` `(n > 1)``        ``result -= result / n;``        ` `    ``return` `result;``}`` ` `// Driver code``int` `main()``{``    ``int` `n;``    ``for``(n = 1; n <= 10; n++)``    ``{``        ``cout << ``"Phi"` `<< ``"("``             ``<< n << ``")"` `<< ``" = "``             ``<< phi(n) << endl;``    ``}``    ``return` `0;``}` `// This code is contributed by koulick_sadhu`

## C

 `// C program to calculate Euler's Totient Function``#include ` `int` `phi(``int` `n)``{``    ``int` `result = n; ``// Initialize result as n` `    ``// Consider all prime factors of n and subtract their``    ``// multiples from result``    ``for` `(``int` `p = 2; p * p <= n; ++p) {``        ` `        ``// Check if p is a prime factor.``        ``if` `(n % p == 0) {``            ` `            ``// If yes, then update n and result``            ``while` `(n % p == 0)``                ``n /= p;``            ``result -= result / p;``        ``}``    ``}` `    ``// If n has a prime factor greater than sqrt(n)``    ``// (There can be at-most one such prime factor)``    ``if` `(n > 1)``        ``result -= result / n;``    ``return` `result;``}` `// Driver program to test above function``int` `main()``{``    ``int` `n;``    ``for` `(n = 1; n <= 10; n++)``        ``printf``(``"phi(%d) = %d\n"``, n, phi(n));``    ``return` `0;``}`

## Java

 `// Java program to calculate``// Euler's Totient Function``import` `java.io.*;` `class` `GFG``{``static` `int` `phi(``int` `n)``{``    ``// Initialize result as n``    ``int` `result = n;` `    ``// Consider all prime factors``    ``// of n and subtract their``    ``// multiples from result``    ``for` `(``int` `p = ``2``; p * p <= n; ++p)``    ``{``        ` `        ``// Check if p is``        ``// a prime factor.``        ``if` `(n % p == ``0``)``        ``{``            ` `            ``// If yes, then update``            ``// n and result``            ``while` `(n % p == ``0``)``                ``n /= p;``            ``result -= result / p;``        ``}``    ``}` `    ``// If n has a prime factor``    ``// greater than sqrt(n)``    ``// (There can be at-most``    ``// one such prime factor)``    ``if` `(n > ``1``)``        ``result -= result / n;``    ``return` `result;``}` `// Driver Code``public` `static` `void` `main (String[] args)``{``    ``int` `n;``    ``for` `(n = ``1``; n <= ``10``; n++)``        ``System.out.println(``"phi("` `+ n +``                           ``") = "` `+ phi(n));``}``}` `// This code is contributed by ajit`

## Python3

 `# Python3 program to calculate``# Euler's Totient Function``def` `phi(n):``    ` `    ``# Initialize result as n``    ``result ``=` `n;` `    ``# Consider all prime factors``    ``# of n and subtract their``    ``# multiples from result``    ``p ``=` `2``;``    ``while``(p ``*` `p <``=` `n):``        ` `        ``# Check if p is a``        ``# prime factor.``        ``if` `(n ``%` `p ``=``=` `0``):``            ` `            ``# If yes, then``            ``# update n and result``            ``while` `(n ``%` `p ``=``=` `0``):``                ``n ``=` `int``(n ``/` `p);``            ``result ``-``=` `int``(result ``/` `p);``        ``p ``+``=` `1``;` `    ``# If n has a prime factor``    ``# greater than sqrt(n)``    ``# (There can be at-most``    ``# one such prime factor)``    ``if` `(n > ``1``):``        ``result ``-``=` `int``(result ``/` `n);``    ``return` `result;` `# Driver Code``for` `n ``in` `range``(``1``, ``11``):``    ``print``(``"phi("``,n,``") ="``, phi(n));``    ` `# This code is contributed``# by mits`

## C#

 `// C# program to calculate``// Euler's Totient Function``using` `System;` `class` `GFG``{``    ` `static` `int` `phi(``int` `n)``{``// Initialize result as n``int` `result = n;` `// Consider all prime ``// factors of n and``// subtract their``// multiples from result``for` `(``int` `p = 2;``         ``p * p <= n; ++p)``{``    ` `    ``// Check if p is``    ``// a prime factor.``    ``if` `(n % p == 0)``    ``{``        ` `        ``// If yes, then update``        ``// n and result``        ``while` `(n % p == 0)``            ``n /= p;``        ``result -= result / p;``    ``}``}` `// If n has a prime factor``// greater than sqrt(n)``// (There can be at-most``// one such prime factor)``if` `(n > 1)``    ``result -= result / n;``return` `result;``}` `// Driver Code``static` `public` `void` `Main ()``{``    ``int` `n;``    ``for` `(n = 1; n <= 10; n++)``        ``Console.WriteLine(``"phi("` `+ n +``                              ``") = "` `+``                              ``phi(n));``}``}` `// This code is contributed``// by akt_mit`

## PHP

 `<Φphp``// PHP program to calculate``// Euler's Totient Function` `function` `phi(``\$n``)``{``    ``// Initialize``    ``// result as n``    ``\$result` `= ``\$n``;` `    ``// Consider all prime``    ``// factors of n and subtract``    ``// their multiples from result``    ``for` `(``\$p` `= 2;``         ``\$p` `* ``\$p` `<= ``\$n``; ++``\$p``)``    ``{``        ` `        ``// Check if p is``        ``// a prime factor.``        ``if` `(``\$n` `% ``\$p` `== 0)``        ``{``            ` `            ``// If yes, then``            ``// update n and result``            ``while` `(``\$n` `% ``\$p` `== 0)``                ``\$n` `= (int)``\$n` `/ ``\$p``;``            ``\$result` `-= (int)``\$result` `/ ``\$p``;``        ``}``    ``}` `    ``// If n has a prime factor``    ``// greater than sqrt(n)``    ``// (There can be at-most``    ``// one such prime factor)``    ``if` `(``\$n` `> 1)``        ``\$result` `-= (int)``\$result` `/ ``\$n``;``    ``return` `\$result``;``}` `// Driver Code``for` `(``\$n` `= 1; ``\$n` `<= 10; ``\$n``++)``    ``echo` `"phi("``, ``\$n``,``") ="``,``          ``phi(``\$n``), ``"\n"``;``    ` `// This code is contributed``// by ajit``Φ>`

## Javascript

 `// Javascript program to calculate``// Euler's Totient Function` `function` `phi(n)``{``    ``// Initialize``    ``// result as n``    ``let result = n;` `    ``// Consider all prime``    ``// factors of n and subtract``    ``// their multiples from result``    ``for` `(let p = 2;``         ``p * p <= n; ++p)``    ``{``        ` `        ``// Check if p is``        ``// a prime factor.``        ``if` `(n % p == 0)``        ``{``            ` `            ``// If yes, then``            ``// update n and result``            ``while` `(n % p == 0)``                ``n = parseInt(n / p);``            ``result -= parseInt(result / p);``        ``}``    ``}` `    ``// If n has a prime factor``    ``// greater than sqrt(n)``    ``// (There can be at-most``    ``// one such prime factor)``    ``if` `(n > 1)``        ``result -= parseInt(result / n);``    ``return` `result;``}` `// Driver Code``for` `(let n = 1; n <= 10; n++)``    ``document.write(`phi(\${n}) = \${phi(n)}
`);``    ` `// This code is contributed``// by _saurabh_jaiswal`

Output :

```phi(1) = 1
phi(2) = 1
phi(3) = 2
phi(4) = 2
phi(5) = 4
phi(6) = 2
phi(7) = 6
phi(8) = 4
phi(9) = 6
phi(10) = 4```

Let us take an example to understand the above algorithm.

```n = 10.
Initialize: result = 10

2 is a prime factor, so n = n/i = 5, result = 5
3 is not a prime factor.

The for loop stops after 3 as 4*4 is not less than or equal
to 10.

After for loop, result = 5, n = 5
Since n > 1, result = result - result/n = 4```

Some Interesting Properties of Euler’s Totient Function
1) For a prime number p, Φ(p) is p-1. For example Φ(5) is 4, Φ(7) is 6 and Φ(13) is 12. This is obvious, gcd of all numbers from 1 to p-1 will be 1 because p is a prime.
2) For two numbers a and b, if gcd(a, b) is 1, then Φ(ab) = Φ(a) * Φ(b). For example Φ(5) is 4 and Φ(6) is 2, so Φ(30) must be 8 as 5 and 6 are relatively prime.
3) For any two prime numbers p and q, Φ(pq) = (p-1)*(q-1). This property is used in RSA algorithm.
4) If p is a prime number, then Φ(pk) = pk – pk-1. This can be proved using Euler’s product formula.
5) Sum of values of totient functions of all divisors of n is equal to n. For example, n = 6, the divisors of n are 1, 2, 3 and 6. According to Gauss, sum of Φ(1) + Φ(2) + Φ(3) + Φ(6) should be 6. We can verify the same by putting values, we get (1 + 1 + 2 + 2) = 6.
6) The most famous and important feature is expressed in Euler’s theorem

```The theorem states that if n and a are coprime
(or relatively prime) positive integers, then

aΦ(n) Φ 1 (mod n) ```

The RSA cryptosystem is based on this theorem:
In the particular case when m is prime say p, Euler’s theorem turns into the so-called Fermat’s little theorem

`ap-1 Φ 1 (mod p) `

7) Number of generators of a finite cyclic group under modulo n addition is Φ(n).
Related Article:
Euler’s Totient function for all numbers smaller than or equal to n
Optimized Euler Totient Function for Multiple Evaluations
References:
http://e-maxx.ru/algo/euler_function
http://en.wikipedia.org/wiki/Euler%27s_totient_function
This article is contributed by Ankur. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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