Wedderburn–Etherington number

The Nth term in the Wedderburn–Etherington number sequence (starting with the number 0 for n = 0) counts the number of unordered rooted trees with n leaves in which all nodes including the root have either zero or exactly two children.

For a given N. The task is to find first N terms of the sequence.

Sequence:



0, 1, 1, 1, 2, 3, 6, 11, 23, 46, 98, 207, 451, 983, 2179, 4850, 10905, 24631, 56011, ….


Trees with 0 or 2 childs:

Examples:

Input : N = 10
Output : 0, 1, 1, 1, 2, 3, 6, 11, 23, 46,

Input : N = 20
Output : 0, 1, 1, 1, 2, 3, 6, 11, 23, 46, 98, 207, 451, 983, 2179, 4850, 10905, 24631, 56011, 127912

Approach:
The Recurrence relation to find Nth number is:

  • a(2x-1) = a(1) * a(2x-2) + a(2) * a(2x-3) + … + a(x-1) * a(x)
  • a(2x) = a(1) * a(2x-1) + a(2) * a(2x-2) + … + a(x-1) * a(x+1) + a(x) * (a(x)+1)/2

Using the above relation we can find the ith term of the series. We will start from the 0th term and then store the answer in a map and then use the values in the map to find the i+1 th term of the series. we will also use base cases for the 0th, 1st and 2nd element which are 0, 1, 1 respectively.

Below is the implementation of the above approach :

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP program to find N terms of the sequence 
#include <bits/stdc++.h>
using namespace std;
  
// Stores the Wedderburn Etherington numbers
map<int, int> store;
  
// Function to return the nth
// Wedderburn Etherington numbers
int Wedderburn(int n)
{
    // Base case
    if (n <= 2)
        return store[n];
  
    // If n is even n = 2x
    else if (n % 2 == 0) 
    {
        // get x
        int x = n / 2, ans = 0;
  
        // a(2x) = a(1)a(2x-1) + a(2)a(2x-2) + ... + 
        // a(x-1)a(x+1)
        for (int i = 1; i < x; i++) {
            ans += store[i] * store[n - i];
        }
  
        // a(x)(a(x)+1)/2
        ans += (store[x] * (store[x] + 1)) / 2;
  
        // Store the ans
        store[n] = ans;
          
        // Return the required answer
        return ans;
    }
      
    else 
    {
        // If n is odd
        int x = (n + 1) / 2, ans = 0;
  
        // a(2x-1) = a(1)a(2x-2) + a(2)a(2x-3) + ... + 
        // a(x-1)a(x),
        for (int i = 1; i < x; i++) {
            ans += store[i] * store[n - i];
        }
  
        // Store the ans
        store[n] = ans;
          
        // Return the required answer
        return ans;
    }
}
  
  
// Function to print first N 
// Wedderburn Etherington numbers
void Wedderburn_Etherington(int n)
{
    // Store first 3 numbers
    store[0] = 0;
    store[1] = 1;
    store[2] = 1;
      
    // Print N terms
    for (int i = 0; i < n; i++)
    {
        cout << Wedderburn(i);
        if(i!=n-1)
            cout << ", ";
    }
}
  
// Driver code
int main()
{
    int n = 10;
  
    // function call
    Wedderburn_Etherington(n);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to find N terms of the sequence 
import java.util.*;
  
class GFG 
{
  
// Stores the Wedderburn Etherington numbers
static HashMap<Integer, 
               Integer> store = new HashMap<Integer, 
                                            Integer>();
  
// Function to return the nth
// Wedderburn Etherington numbers
static int Wedderburn(int n)
{
    // Base case
    if (n <= 2)
        return store.get(n);
  
    // If n is even n = 2x
    else if (n % 2 == 0
    {
        // get x
        int x = n / 2, ans = 0;
  
        // a(2x) = a(1)a(2x-1) + a(2)a(2x-2) + ... + 
        // a(x-1)a(x+1)
        for (int i = 1; i < x; i++) 
        {
            ans += store.get(i) * store.get(n - i);
        }
  
        // a(x)(a(x)+1)/2
        ans += (store.get(x) * (store.get(x) + 1)) / 2;
  
        // Store the ans
        store. put(n, ans);
          
        // Return the required answer
        return ans;
    }
    else
    {
        // If n is odd
        int x = (n + 1) / 2, ans = 0;
  
        // a(2x-1) = a(1)a(2x-2) + a(2)a(2x-3) + ... + 
        // a(x-1)a(x),
        for (int i = 1; i < x; i++)
        {
            ans += store.get(i) * store.get(n - i);
        }
  
        // Store the ans
        store. put(n, ans);
          
        // Return the required answer
        return ans;
    }
}
  
// Function to print first N 
// Wedderburn Etherington numbers
static void Wedderburn_Etherington(int n)
{
    // Store first 3 numbers
    store. put(0, 0);
    store. put(1, 1);
    store. put(2, 1);
      
    // Print N terms
    for (int i = 0; i < n; i++)
    {
        System.out.print(Wedderburn(i));
        if(i != n - 1)
            System.out.print(" ");
    }
}
  
// Driver code
public static void main(String[] args) 
{
    int n = 10;
  
    // function call
    Wedderburn_Etherington(n);    
}
}
  
// This code is contributed by Princi Singh

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program to find N terms 
# of the sequence
  
# Stores the Wedderburn Etherington numbers
store = dict()
  
# Function to return the nth
# Wedderburn Etherington numbers
def Wedderburn(n):
      
    # Base case
    if (n <= 2):
        return store[n]
  
    # If n is even n = 2x
    elif (n % 2 == 0):
          
        # get x
        x = n // 2
        ans = 0
  
        # a(2x) = a(1)a(2x-1) + a(2)a(2x-2) + ... +
        # a(x-1)a(x+1)
        for i in range(1, x):
            ans += store[i] * store[n - i]
  
        # a(x)(a(x)+1)/2
        ans += (store[x] * (store[x] + 1)) // 2
  
        # Store the ans
        store[n] = ans
  
        # Return the required answer
        return ans
    else:
          
        # If n is odd
        x = (n + 1) // 2
        ans = 0
  
        # a(2x-1) = a(1)a(2x-2) + a(2)a(2x-3) + ... +
        # a(x-1)a(x),
        for i in range(1, x):
            ans += store[i] * store[n - i]
  
        # Store the ans
        store[n] = ans
  
        # Return the required answer
        return ans
  
# Function to prfirst N
# Wedderburn Etherington numbers
def Wedderburn_Etherington(n):
  
    # Store first 3 numbers
    store[0] = 0
    store[1] = 1
    store[2] = 1
  
    # PrN terms
    for i in range(n):
        print(Wedderburn(i), end = "")
        if(i != n - 1):
            print(end = ", ")
  
# Driver code
n = 10
  
# function call
Wedderburn_Etherington(n)
  
# This code is contributed by Mohit Kumar

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to find N terms of the sequence 
using System;
using System.Collections.Generic; 
  
class GFG 
{
  
// Stores the Wedderburn Etherington numbers
static Dictionary<int,
                  int> store = new Dictionary<int,
                                              int>();
  
// Function to return the nth
// Wedderburn Etherington numbers
static int Wedderburn(int n)
{
    // Base case
    if (n <= 2)
        return store[n];
  
    // If n is even n = 2x
    else if (n % 2 == 0) 
    {
        // get x
        int x = n / 2, ans = 0;
  
        // a(2x) = a(1)a(2x-1) + a(2)a(2x-2) + ... + 
        // a(x-1)a(x+1)
        for (int i = 1; i < x; i++) 
        {
            ans += store[i] * store[n - i];
        }
  
        // a(x)(a(x)+1)/2
        ans += (store[x] * (store[x] + 1)) / 2;
  
        // Store the ans
        if(store.ContainsKey(n))
        {
            store.Remove(n);
            store.Add(n, ans);
        }
        else
            store.Add(n, ans);
          
        // Return the required answer
        return ans;
    }
    else
    {
        // If n is odd
        int x = (n + 1) / 2, ans = 0;
  
        // a(2x-1) = a(1)a(2x-2) + a(2)a(2x-3) + ... + 
        // a(x-1)a(x),
        for (int i = 1; i < x; i++)
        {
            ans += store[i] * store[n - i];
        }
  
        // Store the ans
        if(store.ContainsKey(n))
        {
            store.Remove(n);
            store.Add(n, ans);
        }
        else
            store.Add(n, ans);
          
        // Return the required answer
        return ans;
    }
}
  
// Function to print first N 
// Wedderburn Etherington numbers
static void Wedderburn_Etherington(int n)
{
    // Store first 3 numbers
    store.Add(0, 0);
    store.Add(1, 1);
    store.Add(2, 1);
      
    // Print N terms
    for (int i = 0; i < n; i++)
    {
        Console.Write(Wedderburn(i));
        if(i != n - 1)
            Console.Write(" ");
    }
}
  
// Driver code
public static void Main(String[] args) 
{
    int n = 10;
  
    // function call
    Wedderburn_Etherington(n); 
}
}
  
// This code is contributed by PrinciRaj1992 

chevron_right



Output:

0, 1, 1, 1, 2, 3, 6, 11, 23, 46


My Personal Notes arrow_drop_up

Second year Department of Information Technology Jadavpur University

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.