1, 1, 2, 1, 3, 4, 5, 1, 6, 2, 7, 8, 9, 3, 5, 1, 10, 11, 12, 13, 7, 14, 15, 16, 17, 7, 18, 4, 19, 16, 20, 1, 12, 18, 15, 21, 22, 10, 15, 23, 24, 16, 25, 3, 9, 26, 27, 28, 29, 30, 20, 5, 31, 16, 32, 33, 34, 35, 36, 37, 38, 19, 15, 1, 27, 16, 39, 7, 25, 8, 40, 41, 42, 34, 35, 9, 36, 16, 43, 44, 29, 32, 45, 46, 47, 24, 48, 8, 49, 50, 40, 10, 36, 51, 40, 52, 53
Offset: 1
We start by setting a(1) = 1 for A033879(1) = 1. Then, whenever A033879(k) is equal to some A033879(m) with m < k, we set a(k) = a(m). Otherwise (when the value is a new one, not encountered before), we allot for a(k) the least natural number not present among a(1) .. a(k-1).
For n=2, as A033879(2) = 1, which was already present at A033879(1), we set a(2) = a(1) = 1.
For n=3, as A033879(3) = 2, which is a new value not encountered before, we set a(3) = 1 + max(a(1),a(2)) = 2.
For n=4, as A033879(4) = 1, which was already present at n = 2 and n = 1, we set a(4) = a(1) = 1.
For n=5, as A033879(5) = 4, which is a new value not encountered before, we set a(5) = 1 + max(a(1),a(2),a(3),a(4)) = 3.
For n=12, as A033879(12) = -4, which is a new value not encountered before, we set a(12) = 1 + max(a(1),...,a(11)) = 8. Note that the sign matters here; -4 is not equal to +4, which was encountered already at n=5.
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