The sequence is generated by following this pseudo-code:

A:  The first term is zero.
    Repeatedly apply:
        If the last term is *new* to the sequence so far then:
B:          The next term is zero.
        Otherwise:
C:          The next term is how far back this last term occured previously.



Example

Using A:

0
Using B:

0 0
Using C:

0 0 1
Using B:

0 0 1 0
Using C: (zero last occurred two steps back - before the one)

0 0 1 0 2
Using B:

0 0 1 0 2 0
Using C: (two last occurred two steps back - before the zero)

0 0 1 0 2 0 2 2
Using C: (two last occurred one step back)

0 0 1 0 2 0 2 2 1
Using C: (one last appeared six steps back)

0 0 1 0 2 0 2 2 1 6
...



Task

 Create a function/procedure/method/subroutine/... to generate the Van Eck sequence of numbers.
 Use it to display here, on this page:
 The first ten terms of the sequence.
 Terms 991 - to - 1000 of the sequence.


References

 Don't Know (the Van Eck Sequence) - Numberphile video.
 Wikipedia Article: Van Eck's Sequence.
  OEIS sequence: A181391.


 
import java.util.HashMap;
import java.util.Map;
 
public class VanEckSequence {
 
    public static void main(String[] args) {
        System.out.println("First 10 terms of Van Eck's sequence:");
        vanEck(1, 10);
        System.out.println("");
        System.out.println("Terms 991 to 1000 of Van Eck's sequence:");
        vanEck(991, 1000);
    }
 
    private static void vanEck(int firstIndex, int lastIndex) {
        Map<Integer,Integer> vanEckMap = new HashMap<>();        
        int last = 0;
        if ( firstIndex == 1 ) {
            System.out.printf("VanEck[%d] = %d%n", 1, 0);
        }
        for ( int n = 2 ; n <= lastIndex ; n++ ) {
            int vanEck = vanEckMap.containsKey(last) ? n - vanEckMap.get(last) : 0;
            vanEckMap.put(last, n);
            last = vanEck;
            if ( n >= firstIndex ) {
                System.out.printf("VanEck[%d] = %d%n", n, vanEck);
            }
        }
 
    }
 
}
 