Euler zigzag numbers ( Alternating Permutation )

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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 :
$a(n+1) = \dfrac{\sum_{k=0}^{n} (\binom{N}{k}*a(k)*a(n-k))}{2} \\$
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 <bits/stdc++.h>
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

<script>
 
// Javascript program to find zigzag sequence
 
// Function to print first n zigzag numbers
function ZigZag(n)
{
    // To store factorial and n'th zig zag number
    var fact = Array(n+1).fill(0);
    var zig = Array(n+1).fill(0);
 
    // Initialize factorial upto n
    fact[0] = 1;
    for (var i = 1; i <= n; i++)
        fact[i] = fact[i - 1] * i;
 
    // Set first two zig zag numbers
    zig[0] = 1;
    zig[1] = 1;
 
    document.write( "zig zag numbers: ");
 
    // Print first two zig zag number
    document.write( zig[0] + " " + zig[1] + " ");
 
    // Print the rest zig zag numbers
    for (var i = 2; i < n; i++) 
    {
        var sum = 0;
 
        for (var k = 0; k <= i - 1; k++) 
        {
            // Binomial(n, k)*a(k)*a(n-k)
            sum += parseInt(fact[i - 1]/(fact[i - 1 - k]*fact[k])) 
                                 *zig[k] * zig[i - 1 - k];
        }
         
        // Store the value
        zig[i] = parseInt(sum / 2);
 
        // Print the number
        document.write( parseInt(sum / 2) + " ");
    }
}
 
// Driver code
var n = 10;
 
// Function call
ZigZag(n);
 
// This code is contributed by rutvik_56.
</script>

Output:

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

Time Complexity: O(n²)

Auxiliary Space: O(n)

Reference
https://en.wikipedia.org/wiki/Alternating_permutation
https://oeis.org/A000111

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