# Eulerian Number

• Difficulty Level : Medium
• Last Updated : 10 Dec, 2022

In combinatorics, the Eulerian Number A(n, m), is the number of permutations of the numbers 1 to n in which exactly m elements are greater than previous element.

For example, there are 4 permutations of the number 1 to 3 in which exactly 1 element is greater than the previous elements.

Examples :

Input : n = 3, m = 1
Output : 4
are 4 permutations where 1 no. is
greater.

Input : n = 4, m = 1
Output : 11

Eulerian Numbers are the coefficients of the Eulerian polynomials described below.

The Eulerian polynomials are defined by the exponential generating function

The Eulerian polynomials can be computed by the recurrence

An explicit formula for A(n, m) is

We can calculate A(n, m) by recurrence relation:

Example:
Suppose, n = 3 and m = 1.
Therefore,
A(3, 1)
= (3 – 1) * A(2, 0) + (1 + 1) * A(2, 1)
= 2 * A(2, 0) + 2 * A(2, 1)
= 2 * 1 + 2 * ( (2 – 1) * A(1, 0) + (1 + 1) * A(1, 1))
= 2 + 2 * (1 * 1 + 2 * ((1 – 1) * A(0, 0) + (1 + 1) * A(0, 1))
= 2 + 2 * (1 + 2 * (0 * 1 + 2 * 0)
= 2 + 2 * (1 + 2 * 0)
= 2 + 2 * 1
= 2 + 2
= 4
We can verify this with example shown above.

Below is the implementation of finding A(n, m):

## C++

 // CPP Program to find Eulerian number A(n, m)#include using namespace std; // Return euleriannumber A(n, m)int eulerian(int n, int m){    if (m >= n || n == 0)        return 0;     if (m == 0)        return 1;     return (n - m) * eulerian(n - 1, m - 1) +           (m + 1) * eulerian(n - 1, m);} // Driven Programint main(){    int n = 3, m = 1;    cout << eulerian(n, m) << endl;    return 0;}

## Java

 // Java program to find Eulerian number A(n, m)import java.util.*; class Eulerian{    // Return eulerian number A(n, m)    public static int eulerian(int n, int m)    {        if (m >= n || n == 0)            return 0;         if (m == 0)            return 1;         return (n - m) * eulerian(n - 1, m - 1) +            (m + 1) * eulerian(n - 1, m);    }         // driver code       public static void main(String[] args)    {        int n = 3, m = 1;        System.out.print( eulerian(n, m) );    }} // This code is contributed by rishabh_jain

## Python3

 # Python3 Program to find Eulerian number A(n, m) # Return euleriannumber A(n, m)def eulerian(n, m):    if (m >= n or n == 0):        return 0;     if (m == 0):        return 1;     return ((n - m) * eulerian(n - 1, m - 1) +            (m + 1) * eulerian(n - 1, m)) # Driver coden = 3m = 1print( eulerian(n, m) ) # This code is contributed by rishabh_jain

## C#

 // C# program to find Eulerian number A(n, m)using System; class Eulerian {         // Return eulerian number A(n, m)    public static int eulerian(int n, int m)    {        if (m >= n || n == 0)            return 0;         if (m == 0)            return 1;         return (n - m) * eulerian(n - 1, m - 1) +                    (m + 1) * eulerian(n - 1, m);    }     // driver code    public static void Main()    {        int n = 3, m = 1;        Console.WriteLine(eulerian(n, m));    }} // This code is contributed by vt_m

## PHP

 = \$n || \$n == 0)        return 0;     if (\$m == 0)        return 1;     return (\$n - \$m) * eulerian(\$n - 1, \$m - 1) +                 (\$m + 1) * eulerian(\$n - 1, \$m);} // Driven Code\$n = 3; \$m = 1;echo eulerian(\$n, \$m); // This code is contributed by anuj_67.?>

## Javascript



Output

4

Time Complexity: O(2n)
Auxiliary Space: O(log(n)), Due, to recursive call stack

Below is the implementation of finding A(n, m) using Dynamic Programming:

## C++

 // CPP Program to find Eulerian number A(n, m)#include using namespace std; // Return euleriannumber A(n, m)int eulerian(int n, int m){    int dp[n + 1][m + 1];     memset(dp, 0, sizeof(dp));     // For each row from 1 to n    for (int i = 1; i <= n; i++) {         // For each column from 0 to m        for (int j = 0; j <= m; j++) {             // If i is greater than j            if (i > j) {                 // If j is 0, then make that                // state as 1.                if (j == 0)                    dp[i][j] = 1;                 // basic recurrence relation.                else                    dp[i][j] = ((i - j) *                     dp[i - 1][j - 1]) +                    ((j + 1) * dp[i - 1][j]);            }        }    }     return dp[n][m];} // Driven Programint main(){    int n = 3, m = 1;    cout << eulerian(n, m) << endl;    return 0;}

## Java

 // Java program to find Eulerian number A(n, m)import java.util.*; class Eulerian{    // Return euleriannumber A(n, m)    public static int eulerian(int n, int m)    {        int[][] dp = new int[n+1][m+1];         // For each row from 1 to n        for (int i = 1; i <= n; i++) {                 // For each column from 0 to m            for (int j = 0; j <= m; j++) {                 // If i is greater than j                if (i > j) {                     // If j is 0, then make                    // that state as 1.                    if (j == 0)                        dp[i][j] = 1;                     // basic recurrence relation.                    else                        dp[i][j] = ((i - j) *                            dp[i - 1][j - 1]) +                        ((j + 1) * dp[i - 1][j]);                }            }        }         return dp[n][m];    }         // driver code    public static void main(String[] args)    {        int n = 3, m = 1;        System.out.print( eulerian(n, m) );    }} // This code is contributed by rishabh_jain

## Python3

 # Python3 Program to find Eulerian# number A(n, m) # Return euleriannumber A(n, m)def eulerian(n, m):    dp = [[0 for x in range(m+1)]             for y in range(n+1)]     # For each row from 1 to n    for i in range(1, n+1):         # For each column from 0 to m        for j in range(0, m+1):             # If i is greater than j            if (i > j):                # If j is 0, then make that                # state as 1.                 if (j == 0):                    dp[i][j] = 1                 # basic recurrence relation.                else :                    dp[i][j] = (((i - j) *                       dp[i - 1][j - 1]) +                       ((j + 1) * dp[i - 1][j]))                         return dp[n][m] # Driven Programn = 3m = 1print(eulerian(n, m)) # This code is contributed by Prasad Kshirsagar

## C#

 // C# program to find Eulerian number A(n, m)using System; class Eulerian {         // Return euleriannumber A(n, m)    public static int eulerian(int n, int m)    {        int[, ] dp = new int[n + 1, m + 1];         // For each row from 1 to n        for (int i = 1; i <= n; i++) {             // For each column from 0 to m            for (int j = 0; j <= m; j++) {                 // If i is greater than j                if (i > j) {                     // If j is 0, then make                    // that state as 1.                    if (j == 0)                        dp[i, j] = 1;                     // basic recurrence relation.                    else                        dp[i, j] = ((i - j) * dp[i - 1, j - 1]) +                                        ((j + 1) * dp[i - 1, j]);                }            }        }         return dp[n, m];    }     // driver code    public static void Main()    {        int n = 3, m = 1;        Console.WriteLine(eulerian(n, m));    }} // This code is contributed by vt_m

## PHP

 \$j)            {                 // If j is 0, then make that                // state as 1.                if (\$j == 0)                    \$dp[\$i][\$j] = 1;                 // basic recurrence relation.                else                    \$dp[\$i][\$j] = ((\$i - \$j) *                                    \$dp[\$i - 1][\$j - 1]) +                                  ((\$j + 1) * \$dp[\$i - 1][\$j]);            }        }    }     return \$dp[\$n][\$m];} // Driver Code\$n = 3 ;\$m = 1;echo eulerian(\$n, \$m) ; // This code is contributed by Ryuga?>

## Javascript



Output

4

Time Complexity: O(n*m)
Auxiliar Space: O(n*m)

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