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A347564 Record the number of distinct terms seen thus far, then the number of distinct terms seen only once, then twice, and so on until recording a zero; whereupon repeat the count.

%I A347564 #49 Jan 16 2022 13:05:21
%S A347564 0,1,2,0,3,3,2,0,4,2,1,2,1,0,5,2,1,0,6,3,0,7,4,1,1,0,8,4,0,9,5,1,2,0,
%T A347564 10,5,0,11,6,1,3,1,0,12,6,0,13,7,1,3,0,14,7,0,15,8,1,4,1,1,1,0,16,8,0,
%U A347564 17,9,1,4,0,18,9,0,19,10,1,5,1,2,0,20,10,0
%N A347564 Record the number of distinct terms seen thus far, then the number of distinct terms seen only once, then twice, and so on until recording a zero; whereupon repeat the count.
%C A347564 An Inventory sequence counting the occurrences of distinct terms. After every occurrence of a zero term the count of distinct terms seen so far is recorded, then the count of those seen just once, then twice, etc, until a zero term occurs again, whereupon the count is reset. The first reset occurs after a(0), the first zero term. (see A342585, A348016).
%H A347564 Michael S. Branicky, <a href="/A347564/b347564.txt">Table of n, a(n) for n = 0..24999</a>
%e A347564 a(0) must be 0 because at this point no distinct terms have been seen.
%e A347564 Following zero term a(0), we start again, a(1) = 1 since there is only one distinct term in the sequence so far; namely a(0) = 0.
%e A347564 a(2) = 2 because now there are two distinct terms (0,1) each of which have appeared just once.
%e A347564 a(3) = 0 since there are no distinct terms which have appeared twice.
%e A347564 Following zero term a(3) we start again; a(4) = 3, since there are now 3 distinct terms (0,1,2) in the sequence so far.
%e A347564 a(5) = 3 because only three distinct terms (1,2,3) have appeared just once.
%e A347564 a(6) = 2 since there are two terms (0, 3) which have occurred twice.
%e A347564 As an irregular table the sequence starts:
%e A347564 0;
%e A347564 1, 2, 0;
%e A347564 3, 3, 2, 0;
%e A347564 4, 2, 1, 2, 1, 0;
%e A347564 5, 2, 1, 0;
%e A347564 6, 3, 0;
%e A347564 7, 4, 1, 1, 0;
%o A347564 (Python)
%o A347564 from collections import Counter
%o A347564 def aupton(terms):
%o A347564     num, alst, inventory = 0, [0], Counter([0])
%o A347564     for n in range(2, terms+1):
%o A347564         if num == 0:
%o A347564             c = len(inventory)
%o A347564         else:
%o A347564             c = sum(inventory[i] == num for i in inventory)
%o A347564         num = 0 if c == 0 else num + 1
%o A347564         alst.append(c)
%o A347564         inventory.update([c])
%o A347564     return alst
%o A347564 print(aupton(83)) # _Michael S. Branicky_, Oct 06 2021
%Y A347564 Cf. A342585, A348016.
%K A347564 nonn,tabf
%O A347564 0,3
%A A347564 _David James Sycamore_, Sep 29 2021
%E A347564 a(45) and beyond from _Michael S. Branicky_, Oct 06 2021