欧拉曲折数(交替排列)

欧拉之字形数字是整数序列，该整数序列是那些数字的排列形式，因此每个条目交替大于或小于前一个条目。

c1，c2，c3，c4是交替排列，其中
c1 c3 c3

之字形数字如下：1、1、1、2、5、16、61、272、1385、7936、50521……

对于给定的整数N。任务是打印最多N个项的序列。

例子：

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

方法 ：
第(n + 1)个之字形数字为：
$a(n+1) = \dfrac{\sum_{k=0}^{n} (\binom{N}{k}*a(k)*a(n-k))}{2} \\$
我们将找到高达n的阶乘并将它们存储在一个数组中，并创建第二个数组以存储第i个字形数，并应用上述公式来查找所有n个字形数。

下面是上述方法的实现：

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 };
  
    // Initilize 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;
  
    // Initilize 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 prfirst 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)]
   
    # Initilize 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;
  
    // Initilize 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

输出：

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

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