# Recaman’s sequence

Given an integer n. Print first n elements of Recaman’s sequence.

Examples:

```Input : n = 6
Output : 0, 1, 3, 6, 2, 7

Input  : n = 17
Output : 0, 1, 3, 6, 2, 7, 13, 20, 12, 21,
11, 22, 10, 23, 9, 24, 8
```

It is basically a function with domain and co-domain as natural numbers and 0. It is recursively defined as below:
Specifically, let a(n) denote the (n+1)-th term. (0 being already there).
The rule says:

```a(0) = 0,
if n > 0 and the number is not
a(n) = a(n - 1) - n
else
a(n) = a(n-1) + n. ```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Below is simple implementation where we store all n Recaman Sequence numbers in an array. We compute next number using recursive formula mentioned above.

```// C++ program to print n-th number in Recaman's
// sequence
#include <bits/stdc++.h>
using namespace std;

// Prints first n terms of Recaman sequence
int recaman(int n)
{
// Create an array to store terms
int arr[n];

// First term of the sequence is always 0
arr[0] = 0;
printf("%d, ", arr[0]);

// Fill remaining terms using recursive
// formula.
for (int i=1; i< n; i++)
{
int curr = arr[i-1] - i;
int j;
for (j = 0; j < i; j++)
{
// If arr[i-1] - i is negative or
if ((arr[j] == curr) || curr < 0)
{
curr = arr[i-1] + i;
break;
}
}

arr[i] = curr;
printf("%d, ", arr[i]);
}
}

// Driver code
int main()
{
int n = 17;
recaman(n);
return 0;
}
```

Output:

```0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9, 24, 8,
```

Time Complexity : O(n2)
Auxiliary Space : O(n)

Optimizations :
We can us hashing to store previously computed values and can make this program work in O(n) time.

```// C++ program to print n-th number in Recaman's
// sequence
#include <bits/stdc++.h>
using namespace std;

// Prints first n terms of Recaman sequence
void recaman(int n)
{
if (n <= 0)
return;

// Print first term and store it in a hash
printf("%d, ", 0);
unordered_set<int> s;
s.insert(0);

// Print remaining terms using recursive
// formula.
int prev = 0;
for (int i=1; i< n; i++)
{
int curr = prev - i;

// If arr[i-1] - i is negative or
if (curr < 0 || s.find(curr) != s.end())
curr = prev + i;

s.insert(curr);

printf("%d, ", curr);
prev = curr;
}
}

// Driver code
int main()
{
int n = 17;
recaman(n);
return 0;
}
```

Output:

```0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9, 24, 8,
```

Time Complexity : O(n)
Auxiliary Space : O(n)

This article is contributed by Kishlay Verma. 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.

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