# Euler zigzag numbers ( Alternating Permutation )

• Last Updated : 28 May, 2022

Euler Zigzag numbers is a sequence of integers which is a number of arrangements of those numbers so that each entry is alternately greater or less than the preceding entry.
c1, c2, c3, c4 is Alternating permutation where
c1 < c2
c3 < c2
c3 < c4…
zigzag numbers are as follows 1, 1, 1, 2, 5, 16, 61, 272, 1385, 7936, 50521 ……
For a given integer N. The task is to print sequence up to N terms.
Examples:

Input : N = 10
Output : 1 1 1 2 5 16 61 272 1385 7936
Input : N = 14
Output : 1 1 1 2 5 16 61 272 1385 7936 50521 353792 2702765 22368256

Approach :
The (n+1)th Zigzag number is :

We will find the factorial upto n and store them in an array and also create a second array to store the i th zigzag number and apply the formula stated above to find all the n zigzag numbers.
Below is the implementation of the above approach :

## C++

 // CPP program to find zigzag sequence #include  using namespace std;   // Function to print first n zigzag numbers void ZigZag(int n) {     // To store factorial and n'th zig zag number     long long fact[n + 1], zig[n + 1] = { 0 };       // Initialize factorial upto n     fact[0] = 1;     for (int i = 1; i <= n; i++)         fact[i] = fact[i - 1] * i;       // Set first two zig zag numbers     zig[0] = 1;     zig[1] = 1;       cout << "zig zag numbers: ";       // Print first two zig zag number     cout << zig[0] << " " << zig[1] << " ";       // Print the rest zig zag numbers     for (int i = 2; i < n; i++)      {         long long sum = 0;           for (int k = 0; k <= i - 1; k++)          {             // Binomial(n, k)*a(k)*a(n-k)             sum += (fact[i - 1]/(fact[i - 1 - k]*fact[k]))                                   *zig[k] * zig[i - 1 - k];         }                   // Store the value         zig[i] = sum / 2;           // Print the number         cout << sum / 2 << " ";     } }   // Driver code int main() {     int n = 10;           // Function call     ZigZag(n);       return 0; }

## Java

 // Java program to find zigzag sequence import java.util.*; import java.lang.*; import java.io.*;   class GFG {   // Function to print first n zigzag numbers static void ZigZag(int n) {     // To store factorial and n'th zig zag number     long[] fact= new long[n + 1];     long[] zig = new long[n + 1];     for (int i = 0; i < n + 1; i++)         zig[i] = 0;       // Initialize factorial upto n     fact[0] = 1;     for (int i = 1; i <= n; i++)         fact[i] = fact[i - 1] * i;       // Set first two zig zag numbers     zig[0] = 1;     zig[1] = 1;       System.out.print("zig zag numbers: ");       // Print first two zig zag number     System.out.print(zig[0] + " " + zig[1] + " ");       // Print the rest zig zag numbers     for (int i = 2; i < n; i++)      {         long sum = 0;           for (int k = 0; k <= i - 1; k++)          {             // Binomial(n, k)*a(k)*a(n-k)             sum += (fact[i - 1] / (fact[i - 1 - k] *                      fact[k])) * zig[k] * zig[i - 1 - k];         }                   // Store the value         zig[i] = sum / 2;           // Print the number         System.out.print(sum / 2 + " " );               } }   // Driver code public static void main (String[] args)                throws java.lang.Exception {     int n = 10;           // Function call     ZigZag(n); } }   // This code is contributed by nidhiva

## Python3

 # Python3 program to find zigzag sequence   # Function to print first n zigzag numbers def ZigZag(n):       # To store factorial and      # n'th zig zag number     fact = [0 for i in range(n + 1)]     zig = [0 for i in range(n + 1)]        # Initialize factorial upto n     fact[0] = 1     for i in range(1, n + 1):         fact[i] = fact[i - 1] * i       # Set first two zig zag numbers     zig[0] = 1     zig[1] = 1       print("zig zag numbers: ", end = " ")       # Print first two zig zag number     print(zig[0], zig[1], end = " ")       # Print the rest zig zag numbers     for i in range(2, n):         sum = 0           for k in range(0, i):                           # Binomial(n, k)*a(k)*a(n-k)             sum += ((fact[i - 1] //                     (fact[i - 1 - k] * fact[k])) *                      zig[k] * zig[i - 1 - k])           # Store the value         zig[i] = sum // 2           # Print the number         print(sum // 2, end = " ")   # Driver code n = 10   # Function call ZigZag(n)   # This code is contributed by Mohit Kumar

## C#

 // C# program to find zigzag sequence using System;       class GFG {   // Function to print first n zigzag numbers static void ZigZag(int n) {     // To store factorial and n'th zig zag number     long[] fact= new long[n + 1];     long[] zig = new long[n + 1];     for (int i = 0; i < n + 1; i++)         zig[i] = 0;       // Initialize factorial upto n     fact[0] = 1;     for (int i = 1; i <= n; i++)         fact[i] = fact[i - 1] * i;       // Set first two zig zag numbers     zig[0] = 1;     zig[1] = 1;       Console.Write("zig zag numbers: ");       // Print first two zig zag number     Console.Write(zig[0] + " " + zig[1] + " ");       // Print the rest zig zag numbers     for (int i = 2; i < n; i++)      {         long sum = 0;           for (int k = 0; k <= i - 1; k++)          {             // Binomial(n, k)*a(k)*a(n-k)             sum += (fact[i - 1] / (fact[i - 1 - k] *                      fact[k])) * zig[k] * zig[i - 1 - k];         }                   // Store the value         zig[i] = sum / 2;           // Print the number         Console.Write(sum / 2 + " " );               } }   // Driver code public static void Main (String[] args) {     int n = 10;           // Function call     ZigZag(n); } }   // This code is contributed by 29AjayKumar

## Javascript

 

Output:

zig zag numbers: 1 1 1 2 5 16 61 272 1385 7936

Time Complexity: O(n2)

Auxiliary Space: O(n)

My Personal Notes arrow_drop_up
Recommended Articles
Page :