# Find n-th element from Stern’s Diatomic Series

Given an integer n. we have to find the nth term of Stern’s Diatomic Series.

Stern’s diatomic series is the sequence which generates the following integer sequence 0, 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4, ……. It arises in the Calkin-Wilf tree. It is sometimes also known as the fusc function.

In mathematical terms, the sequence P(n) of Stern’s diatomic series is defined by the recurrence relation.

Examples :

Input : n = 7
Output : 3

Input : n = 15
Output : 4


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

Approach :
We solve this problem with a very simple concept of Dynamic programming which is used in finding fibonacci numbers. After saving the base case of DP[0] = 0, DP[1] = 1, we have to simply traverse from i = 2 to n and compute DP[i] as per explained definition of Stern’s diatomic series. And finally return the value of DP[n].

Algorithm :

 // SET the Base case
DP[0] = 0;
DP[1] = 1;

// Traversing the array from 2nd Element to nth Element
for (int i=2; i<=n; i++)
{
// Case 1: for even n
if (i%2 == 0)
DP[i] = DP[i/2];

// Case 2: for odd n
else
DP[i] = DP[(i-1)/2] + DP[(i+1)/2];
}
return DP[n];


## C++

 // Program to find the nth element   // of Stern's Diatomic Series  #include  using namespace std;     // function to find nth stern'  // diatomic series  int findSDSFunc(int n)  {      // Initializing the DP array      int DP[n+1];         // SET the Base case      DP[0] = 0;      DP[1] = 1;         // Traversing the array from       // 2nd Element to nth Element      for (int i = 2; i <= n; i++) {                     // Case 1: for even n          if (i % 2 == 0)              DP[i] = DP[i / 2];                     // Case 2: for odd n          else             DP[i] = DP[(i - 1) / 2] +                          DP[(i + 1) / 2];      }      return DP[n];  }     // Driver program  int main()  {      int n = 15;          cout << findSDSFunc(n) << endl;          return 0;  }

## Java

 // Java program to find the nth element   // of Stern's Diatomic Series     class GFG {             // function to find nth stern'      // diatomic series      static int findSDSFunc(int n)      {                     // Initializing the DP array          int DP[] = new int[n+1];                 // SET the Base case          DP[0] = 0;          DP[1] = 1;                 // Traversing the array from           // 2nd Element to nth Element          for (int i = 2; i <= n; i++)          {                             // Case 1: for even n              if (i % 2 == 0)                  DP[i] = DP[i / 2];                             // Case 2: for odd n              else                 DP[i] = DP[(i - 1) / 2] +                              DP[(i + 1) / 2];          }                     return DP[n];      }             // Driver program      public static void main(String[] args)      {          int n = 15;                     System.out.println(findSDSFunc(n));      }  }     // This code is contributed by Smita Semwal.

## Python 3

 # Program to find the nth element   # of Stern's Diatomic Series     # function to find nth stern'  # diatomic series  def findSDSFunc(n):         # Initializing the DP array      DP = [0] * (n+1)         # SET the Base case      DP[0] = 0     DP[1] = 1        # Traversing the array from       # 2nd Element to nth Element      for i in range(2, n+1):                      # Case 1: for even n          if (int(i % 2) == 0):              DP[i] = DP[int(i / 2)]                     # Case 2: for odd n          else:              DP[i] = (DP[int((i - 1) / 2)]                    + DP[int((i + 1) / 2)])             return DP[n]        # Driver program  n = 15    print(findSDSFunc(n))     # This code is contribute by  # Smitha Dinesh Semwal

## C#

 // C# program to find the nth element   // of Stern's Diatomic Series  using System;     class GFG  {      // function to find nth      // stern' diatomic series      static int findSDSFunc(int n)      {                     // Initializing the DP array          int []DP = new int[n + 1];                 // SET the Base case          DP[0] = 0;          DP[1] = 1;                 // Traversing the array from           // 2nd Element to nth Element          for (int i = 2; i <= n; i++)          {                             // Case 1: for even n              if (i % 2 == 0)                  DP[i] = DP[i / 2];                             // Case 2: for odd n              else                 DP[i] = DP[(i - 1) / 2] +                          DP[(i + 1) / 2];          }                     return DP[n];      }             // Driver Code      static public void Main ()      {          int n = 15;          Console.WriteLine(findSDSFunc(n));      }  }     // This code is contributed by aj_36

## PHP

 

Output:

4


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