# Eulerian Number

• Difficulty Level : Medium
• Last Updated : 07 Aug, 2021

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.

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```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 Program``int` `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 code``n ``=` `3``m ``=` `1``print``( 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`

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 Program``int` `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 Program``n ``=` `3``m ``=` `1``print``(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`

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