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A379692 a(1) = 1; a(n) is the most recently occurring value not among the last k terms where k is the number of times a(n-1) has occurred; if no such value exists, a(n) is the smallest unused positive integer.

%I A379692 #12 Jan 02 2025 11:55:18
%S A379692 1,2,1,3,1,2,3,1,4,1,5,1,2,4,1,3,2,5,3,1,6,1,7,1,4,6,1,8,1,2,4,6,1,3,
%T A379692 8,1,5,3,6,8,5,1,7,5,6,8,1,4,7,8,6,5,3,2,4,7,5,8,6,1,9,1,10,1,11,1,12,
%U A379692 1,13,1,14,1,15,1,16,1,17,1,18,1,3,15,1,4
%N A379692 a(1) = 1; a(n) is the most recently occurring value not among the last k terms where k is the number of times a(n-1) has occurred; if no such value exists, a(n) is the smallest unused positive integer.
%C A379692 The number preceding and following each first occurrence is the same (with the exception of a(1)=1 which has no predecessor).
%H A379692 Neal Gersh Tolunsky, <a href="/A379692/b379692.txt">Table of n, a(n) for n = 1..10000</a>
%H A379692 Neal Gersh Tolunsky, <a href="/A379692/a379692.png">Graph of 100000 terms</a>
%H A379692 Neal Gersh Tolunsky, <a href="/A379692/a379692_1.png">Ordinal transform of 200000 terms</a>
%H A379692 Neal Gersh Tolunsky, <a href="/A379692/a379692_2.png">Graph of 400000 terms</a>
%e A379692 To find a(14)=4, we see that a(13) = 2 has occurred 3 times in the sequence, so the most recently occurring value before (5, 1, 2) which is not equal to 5,1,2 is a(9) = 4, so a(14) = 4.
%Y A379692 Cf. A358921.
%K A379692 nonn
%O A379692 1,2
%A A379692 _Neal Gersh Tolunsky_, Dec 29 2024