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++

// 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;
}

Java

// 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

Python3

# 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 print first N
# Wedderburn Etherington numbers
def Wedderburn_Etherington(n):

    # Store first 3 numbers
    store[0] = 0
    store[1] = 1
    store[2] = 1

    # Print N 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

C#

// 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

Javascript

<script>
// Javascript program to find N terms of the sequence

// Stores the Wedderburn Etherington numbers
var store = new Map();

// Function to return the nth
// Wedderburn Etherington numbers
function Wedderburn(n)
{
    // Base case
    if (n <= 2)
        return store[n];

    // If n is even n = 2x
    else if (n % 2 == 0)
    {

        // get x
        var x = parseInt(n / 2), ans = 0;

        // a(2x) = a(1)a(2x-1) + a(2)a(2x-2) + ... +
        // a(x-1)a(x+1)
        for (var 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
        var 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 (var 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
function Wedderburn_Etherington(n)
{
    // Store first 3 numbers
    store[0] = 0;
    store[1] = 1;
    store[2] = 1;

    // Print N terms
    for (var i = 0; i < n; i++)
    {
        document.write( Wedderburn(i));
        if(i != n - 1)
            document.write( ", ");
    }
}

// Driver code
var n = 10;

// function call
Wedderburn_Etherington(n);

// This code is contributed by rrrtnx.
</script>

Output:

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

Time Complexity: O(n²logn)

Auxiliary Space: O(n)