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 <bits/stdc++.h>using namespace std;// Function to print first n zigzag numbersvoid 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 codeint main(){ int n = 10; // Function call ZigZag(n); return 0;} |
Java
// Java program to find zigzag sequenceimport java.util.*;import java.lang.*;import java.io.*;class GFG{// Function to print first n zigzag numbersstatic 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 codepublic 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 numbersdef 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 coden = 10# Function callZigZag(n)# This code is contributed by Mohit Kumar |
C#
// C# program to find zigzag sequenceusing System; class GFG{// Function to print first n zigzag numbersstatic 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 codepublic 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 numbersfunction 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 codevar n = 10;// Function callZigZag(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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