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0,
2, 0,
4, 1, 1, 0,
6, 4, 2, 1, 2, 0,
8, 6, 5, 2, 3, 2, 2, 0,
10, 7, 9, 4, 5, 4, 3, 2, 1, 1, 1, 0,
12, 11, 11, 6, 7, 5, 5, 4, 2, 2, 2, 3, 1, 0,
14, 13, 15, 8, 9, 8, 6, 5, 5, 4, 3, 4, 2, 2, 1, 1, 0,
...
...
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nn = 9; c[_] = 0; i = 1; Do[k = 0; While[c[k] > 0, Set[{a[i], a[i + 1]}, {c[k], k}]; c[a[i]]++; c[a[i + 1]]++; i += 2; k++]; Set[{a[i], a[i + 1]}, {c[k], k}]; c[a[i]]++; c[a[i + 1]]++; i += 2, nn]; Array[a[2 # - 1] &, (i - 1)/2] (* Michael De Vlieger, Sep 25 2022 *)
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(Python)
from collections import Counter
def aupton(terms):
num, alst, inventory = 0, [0, 0], Counter([0, 0])
for n in range(3, 2*terms):
c = [inventory[num], num]
num = 0 if c[0] == 0 else num + 1
alst.extend(c)
inventory.update(c)
return alst[:2*terms:2]
print(aupton(82)) # Michael S. Branicky, Sep 25 2022
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14, 13, 15, 8, 9, 7, 6, 8, 6, 4, 5, 5, 4, 3, 4, 2, 2, 1, 1, 0,