## The Recamán Sequence

### May 10, 2019

I’ve been watching Numberphile again, specifically their episode about the Recamán sequence 0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, … A005132, defined as follows:

The Recamán sequence is an integer sequence R with R0 = 0 and Rn = Rn−1n if Rn = Rn−1n is positive and not already in the sequence and Rn−1 + n otherwise.

Your task is to write a program that generates the Recamán sequence; use your program to compute R1000000 (the one-million-and-first element). When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

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### 3 Responses to “The Recamán Sequence”

1. Zack said

Interesting problem. Here is my take using Julia:

function recaman(n::Int64)
R = Array{Int64}(undef, n)
R = 1

``````for i = 2:n
if i % 10000 == 0; print(i, "\t"); end # something to keep track of the progress made
x = R[i-1] - i

if (x < 2)||(x in R[2:(i-2)])
R[i] = R[i-1] + i
else
R[i] = x
end
end

return R[n]
``````

end

2. Paul said

In Python.

```from itertools import count, islice

def recaman():
S, x = set(), 0
for n in count(1):
yield x
xmin = x - n
x = xmin if xmin > 0 and xmin not in S else x + n

N = 1000000
r = recaman()
next(islice(r, N, N), None)  # skip N terms
print(next(r))  # -> 2057164```
3. Daniel said

Here’s a solution in C++.

```#include <cstdlib>
#include <iostream>
#include <string>
#include <unordered_set>

using namespace std;

int main(int argc, char* argv[]) {
if (argc != 2) return EXIT_FAILURE;
int n = stoi(argv);
unordered_set<int> set({0});
int R = 0;
for (int i = 1; i <= n; ++i) {
R -= i;
if (R < 0 || set.find(R) != set.end())
R += 2 * i;
set.insert(R);
}
cout << R << endl;
return EXIT_SUCCESS;
}
```

Example Usage:

```\$ ./a.out 1000000
2057164
```