# Ulam Number Sequence

Given a positive integer n , the task is to print nth Number of the Ulam Number Sequence

Ulam Number: In Mathematics, An Ulam number is the member of an Integer sequence that starts with the term U1 = 1 and U1 = 2 and then for every n > 2, Un is defined is smallest positive integer greater than Un-1 which can be expressed as sum of two distinct earlier term of the sequence in exactly one way.

For example,

• 3 is an Ulam Number as it can be expressed as sum of two distinct earlier terms in exactly one way
( i.e 1 + 2 )
• 4 is also an Ulam number. Other than ( 1 + 3) we can express 4 as ( 2 + 2 ) but in (2 + 2) terms are not distinct. So we have only one way
• 5 can be expressed as sum of two distinct earlier terms of the sequence in two ways
as ( 1 + 4 ) and ( 2 + 3), So, 5 is not an Ulam Number

The first few terms of the Ulam number sequence are-

1, 2, 3, 4, 6, 8, 11, 13, 16, 18, 26, 28, 36, 38, 47, 48, 53, 57, 62, 69, 72, 77, 82, 87, 97, 99, 102

Examples:

Input  : 5
Output : 6

Input : 9
Output : 16

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A Simple Solution to print nth term of the Ulam number sequence is to generate entire ulam sequence up to n and print nth term, As we can not compute nth term directly.

Approach to generate ulam number sequence:

• As, We have first two term of the sequence as U1=1 and U2=2. we can start our search from i=3 for next Ulam number.
• For every value of ‘i’ traverse the earlier terms of the sequence using two loops and check if on adding two distinct term of sequence we can get sum as i in exactly one way or not
• If Yes, Then ‘i’ is next Ulam number, Store it. and then search for next ulam number
• If not, then increment ‘i’ and repeat the same
• Keep searching ulam number, until we get our nth term

Below is the implementation of the above idea:

## C++

 // CPP code to print nth // Ulam number    #include using namespace std;    #define MAX 10000    // Array to store Ulam Number vector arr;    // function to compute ulam Number void ulam() {     // push First 2 two term of the sequence     // in the array     // for further calculation        arr.push_back(1);        arr.push_back(2);        // loop to generate Ulam number     for (int i = 3; i < MAX; i++) {            int count = 0;            // traverse the array and check if         // i can be reprsented as sum of         // two distinct element of the array            for (int j = 0; j < arr.size() - 1; j++) {                for (int k = j + 1; k < arr.size(); k++) {                    if (arr[j] + arr[k] == i) {                        count++;                 }                 if (count > 1)                     break;             }             if (count > 1)                 break;         }            // If count is 1 that means         // i can be represented as sum of         // two distinct terms of the sequence            if (count == 1) {             // i is ulam number             arr.push_back(i);         }     } }    // Driver code int main() {     // Pre compute Ulam Number sequence     ulam();        int n = 9;        // Print nth Ulam number     cout << arr[n - 1];        return 0; }

## Java

 // JAVA code to print nth // Ulam number    import java.util.*; class GFG {        static final int MAX = 1000;        // Array to store Ulam Number     static Vector arr = new Vector();        // Function to compute ulam Number     static void ulam()     {            // push First 2 two term of the sequence         // in the array         // for further calculation            arr.add(1);            arr.add(2);            // loop to generate Ulam number         for (int i = 3; i < MAX; i++) {                int count = 0;                // traverse the array and check if             // i can be reprsented as sum of             // two distinct element of the array                for (int j = 0; j < arr.size() - 1; j++) {                    for (int k = j + 1; k < arr.size(); k++) {                        if (arr.get(j) + arr.get(k) == i) {                            count++;                     }                     if (count > 1)                         break;                 }                 if (count > 1)                     break;             }                // If count is 2 that means             // i can be represented as sum of             // two distinct terms of the sequence                if (count == 1) {                 // i is ulam number                 arr.add(i);             }         }     }        // Driver code     public static void main(String[] args)     {         // pre compute Ulam Number sequence         ulam();            int n = 9;            // print nth Ulam number         System.out.println(arr.get(n - 1));     } }

## C#

 // C# code to print nth  // Ulam number  using System; using System.Collections.Generic;    class GFG {      static readonly int MAX = 1000;         // Array to store Ulam Number      static List arr = new List();         // Function to compute ulam Number      static void ulam()      {             // push First 2 two term of           // the sequence in the array          // for further calculation             arr.Add(1);          arr.Add(2);             // loop to generate Ulam number          for (int i = 3; i < MAX; i++)          {              int count = 0;                 // traverse the array and check if              // i can be reprsented as sum of              // two distinct element of the array              for (int j = 0; j < arr.Count - 1; j++)              {                     for (int k = j + 1; k < arr.Count; k++)                  {                      if (arr[j] + arr[k] == i)                     {                          count++;                      }                      if (count > 1)                          break;                  }                  if (count > 1)                      break;              }                 // If count is 2 that means              // i can be represented as sum of              // two distinct terms of the sequence              if (count == 1)              {                  // i is ulam number                  arr.Add(i);              }          }      }         // Driver code      public static void Main(String[] args)      {          // pre compute Ulam Number sequence          ulam();             int n = 9;             // print nth Ulam number          Console.WriteLine(arr[n - 1]);      }  }     // This code is contibuted by Rajput-JI

Output:

16

An Efficient solution is to use a hash table to store the earlier number of the sequence so that instead of using two loops to check if ‘i’ can be expressed as the sum of two distinct terms in exactly one way or not, we can do it in a single loop. Searching will be fast if we will use a hash table.

Below is the Implementation of above idea:

## C++

 // Cpp code to print nth // Ulam number    #include using namespace std;    #define MAX 10000    // Array to store Ulam Number vector arr;    // function to compute ulam Number void ulam() {        // Set to search specific Ulam number efficiently     unordered_set s;        // push First 2 two term of the sequence     // in the array and set     // for further calculation        arr.push_back(1);     s.insert(1);        arr.push_back(2);     s.insert(2);        // loop to generate Ulam number     for (int i = 3; i < MAX; i++) {            int count = 0;            // traverse the array and check if         // i can be reprsented as sum of         // two distinct element of the array            for (int j = 0; j < arr.size(); j++) {                // Check if i-arr[j] exist in the array or not using set             // If yes, Then i can be represented as             // sum of arr[j] + (i- arr[j])                if (s.find(i - arr[j]) != s.end() && arr[j] != (i - arr[j]))                 count++;                // if Count is greater than 2             // break the loop             if (count > 2)                 break;         }            // If count is 2 that means         // i can be represented as sum of         // two distinct terms of the sequence            if (count == 2) {             // i is ulam number             arr.push_back(i);             s.insert(i);         }     } }    // Driver code int main() {     // pre compute Ulam Number sequence     ulam();        int n = 9;        // print nth Ulam number     cout << arr[n - 1];        return 0; }

## Java

 // JAVA code to print nth // Ulam number    import java.util.*; class GFG {        static final int MAX = 10000;        // Array to store Ulam Number     static Vector arr = new Vector();        // function to compute ulam Number     static void ulam()     {            // Set to search specific Ulam number efficiently         Set s = new HashSet();            // push First 2 two term of the sequence         // in the array and set         // for further calculation            arr.add(1);         s.add(1);            arr.add(2);         s.add(2);            // loop to generate Ulam number         for (int i = 3; i < MAX; i++) {                int count = 0;                // traverse the array and check if             // i can be reprsented as sum of             // two distinct element of the array                for (int j = 0; j < arr.size(); j++) {                    // Check if i-arr[j] exist in the array or not using set                 // If yes, Then i can be represented as                 // sum of arr[j] + (i- arr[j])                    if (s.contains(i - arr.get(j)) && arr.get(j) != (i - arr.get(j)))                     count++;                    // if Count is greater than 2                 // break the loop                 if (count > 2)                     break;             }                // If count is 2 that means             // i can be represented as sum of             // two distinct terms of the sequence                if (count == 2) {                 // i is ulam number                 arr.add(i);                 s.add(i);             }         }     }        // Driver code     public static void main(String[] args)     {         // pre compute Ulam Number sequence         ulam();            int n = 9;            // print nth Ulam number         System.out.println(arr.get(n - 1));     } }

## Python3

 # Python3 code to print nth Ulam number  MAX = 10000    # Array to store Ulam Number  arr = []     # function to compute ulam Number  def ulam():        # Set to search specific Ulam      # number efficiently      s = set()         # push First 2 two term of the      # sequence in the array and set      # for further calculation      arr.append(1)      s.add(1)         arr.append(2)      s.add(2)         # loop to generate Ulam number      for i in range(3, MAX):             count = 0            # traverse the array and check if          # i can be reprsented as sum of          # two distinct element of the array          for j in range(0, len(arr)):                # Check if i-arr[j] exist in the array              # or not using set. If yes, Then i can              # be represented as sum of arr[j] + (i- arr[j])              if (i - arr[j]) in s and arr[j] != (i - arr[j]):                  count += 1                # if Count is greater than 2              # break the loop              if count > 2:                  break                    # If count is 2 that means          # i can be represented as sum of          # two distinct terms of the sequence          if count == 2:             # i is ulam number              arr.append(i)              s.add(i)             # Driver Code     if __name__ == "__main__":        # pre compute Ulam Number sequence      ulam()         n = 9        # print nth Ulam number      print(arr[n - 1])     # This code is contributed by Rituraj Jain

## C#

 // C# code to print nth // Ulam number using System; using System.Collections.Generic;    class GFG  {        static readonly int MAX = 10000;        // Array to store Ulam Number     static List arr = new List();        // function to compute ulam Number     static void ulam()     {            // Set to search specific Ulam number efficiently         HashSet s = new HashSet();            // push First 2 two term of the sequence         // in the array and set         // for further calculation         arr.Add(1);         s.Add(1);            arr.Add(2);         s.Add(2);            // loop to generate Ulam number         for (int i = 3; i < MAX; i++)         {                int count = 0;                // traverse the array and check if             // i can be reprsented as sum of             // two distinct element of the array                for (int j = 0; j < arr.Count; j++)             {                    // Check if i-arr[j] exist in the array                  // or not using set If yes,                 // Then i can be represented as                 // sum of arr[j] + (i- arr[j])                 if (s.Contains(i - arr[j]) &&                      arr[j] != (i - arr[j]))                     count++;                    // if Count is greater than 2                 // break the loop                 if (count > 2)                     break;             }                // If count is 2 that means             // i can be represented as sum of             // two distinct terms of the sequence             if (count == 2)             {                 // i is ulam number                 arr.Add(i);                 s.Add(i);             }         }     }        // Driver code     public static void Main(String[] args)     {         // pre compute Ulam Number sequence         ulam();            int n = 9;            // print nth Ulam number         Console.WriteLine(arr[n - 1]);     } }    // This code is contributed by Rajput-Ji

Output:

16

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

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.

Improved By : rituraj_jain, Rajput-Ji