Van Eck sequence

The sequence is generated by following this pseudo-code:

Van Eck sequence is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

1:  The first term is zero.
    Repeatedly apply:
        If the last term is *new* to the sequence so far then:
2:          The next term is zero.
        Otherwise:
3:          The next term is how far back this last term occured previousely.

Example

Using 1:

0

Using 2:

0 0

Using 3:

0 0 1

Using 2:

0 0 1 0

Using 3: (zero last occured two steps back - before the one)

0 0 1 0 2

Using 1:

0 0 1 0 2 0

Using 3: (two last occured two steps back - before the zero)

0 0 1 0 2 0 2 2

Using 3: (two last occured one step back)

0 0 1 0 2 0 2 2 1

Using 3: (one last appeared six steps back)

0 0 1 0 2 0 2 2 1 6

...

Task

1. Create a function/proceedure/method/subroutine/... to generate the Van Eck sequence of numbers.
2. Use it to display here, on this page:

1. The first ten terms of the sequence.
2. Terms 991 - to - 1000 of the sequence.

Reference

• [Don't Know (the Van Eck Sequence) - Numberphile video].

Python

<lang python>def van_eck():

   n, seen, val = 0, {}, 0
   while True:
       yield val
       last = {val: n}
       val = n - seen.get(val, n)
       seen.update(last)
       n += 1

1. %%

if __name__ == '__main__':

   print("Van Eck: first 10 terms:  ", list(islice(van_eck(), 10)))
   print("Van Eck: terms 991 - 1000:", list(islice(van_eck(), 1000))[-10:])</lang>

Output:

Van Eck: first 10 terms:   [0, 0, 1, 0, 2, 0, 2, 2, 1, 6]
Van Eck: terms 991 - 1000: [4, 7, 30, 25, 67, 225, 488, 0, 10, 136]

Python: Alternate

The following stores the sequence so far in a list seen rather than the first example that just stores last occurrences in a dict. <lang python>def van_eck():

   n = 0
   seen = [0]
   val = 0
   while True:
       yield val
       if val in seen[1:]:
           val = seen.index(val, 1)
       else:
           val = 0
       seen.insert(0, val)
       n += 1</lang>

Output:

As before.