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A354971 a(1)=1, a(2)=0; for n > 2, a(n) is the number of times a(n - 1 - a(n-1)) has appeared in the sequence.

%I A354971 #54 Sep 06 2022 15:11:12
%S A354971 1,0,1,1,3,1,1,5,5,5,1,3,3,3,6,3,5,5,5,5,1,7,1,1,9,5,9,8,8,9,9,1,4,2,
%T A354971 10,4,10,4,1,3,2,11,4,11,4,2,2,5,5,2,10,5,5,12,2,12,2,7,3,2,2,7,9,2,3,
%U A354971 3,5,3,10,10,10,10,3,7,13,4,7,7,7,7,11
%N A354971 a(1)=1, a(2)=0; for n > 2, a(n) is the number of times a(n - 1 - a(n-1)) has appeared in the sequence.
%C A354971 Conjecture: every positive integer eventually appears in the sequence.
%C A354971 Conjecture: The longest run of same, consecutive a(n) is 4 terms.
%C A354971 A003056 can be generated using the same rules but starting with 0.
%H A354971 Michael De Vlieger, <a href="/A354971/b354971.txt">Table of n, a(n) for n = 1..10000</a>
%H A354971 Michael De Vlieger, <a href="/A354971/a354971.png">Log-log scatterplot of a(n)</a>, n = 1..2^16, showing records in red, first entries of other m in green, and the latest entries of m in blue.
%H A354971 Neal Gersh Tolunsky, <a href="/A354971/a354971_3.png">Illustration of first 100 terms</a>
%e A354971 For n=9, a(9) is the number of times a(9 - 1 - a(9-1)) = a(8 - a(8)) = a(3) = 1 has appeared in the sequence, so a(9)=5.
%t A354971 nn = 2^20; c[_] = 0; a[1] = c[0] = c[1] = 1; a[2] = j = 0; Do[k = c[a[n - j - 1]]; a[n] = j = k; c[k]++, {n, 3, nn}]; Array[a, nn] (* _Michael De Vlieger_, Jun 25 2022 *)
%o A354971 (MATLAB) function a = A354971( max_n )
%o A354971     a = [1 0]; c = zeros(1,max_n); c(1:2) = [1 1];
%o A354971     for n = 3:max_n
%o A354971         m = a(n-1-a(n-1))+1;
%o A354971         a = [a c(m)];
%o A354971         c(c(m)+1) = c(c(m)+1)+1;
%o A354971     end
%o A354971 end % _Thomas Scheuerle_, Jun 15 2022
%o A354971 (PARI) { nb = [0]; for (n=1, #a=vector(81), print1 (a[n] = if (n==1, 1, n==2, 0, nb[1+a[n-1-a[n-1]]])", "); if (#nb < 1+a[n], nb = concat(nb, vector(#nb))); nb[1+a[n]]++) } \\ _Rémy Sigrist_, Jun 18 2022
%o A354971 (Python)
%o A354971 from collections import Counter
%o A354971 from itertools import count, islice
%o A354971 def agen():
%o A354971     a = [None, 1, 0]; inventory = Counter(a); yield from a[1:]
%o A354971     for n in count(3):
%o A354971         c = inventory[a[n-1-a[n-1]]]
%o A354971         a.append(c); inventory.update([c]); yield c
%o A354971 print(list(islice(agen(), 81))) # _Michael S. Branicky_, Jun 18 2022
%Y A354971 Cf. A003056, A329985
%K A354971 nonn,easy
%O A354971 1,5
%A A354971 _Neal Gersh Tolunsky_, Jun 14 2022