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A363193 a(1)=1, and thereafter a(n) = number of occurrences of a(k) among terms a(1..k), where k = n-a(n-1).

%I A363193 #31 May 24 2023 19:55:04
%S A363193 1,1,2,2,1,3,2,1,4,1,5,3,5,1,6,5,2,3,3,4,4,3,2,5,2,4,5,5,4,4,5,5,6,6,
%T A363193 5,7,6,8,7,2,2,7,1,7,1,8,2,2,9,8,7,8,8,2,5,5,4,11,10,3,1,9,11,5,6,6,9,
%U A363193 1,10,6,5,3,7,3,7,2,7,13,6,8,6,12,13,13,7,8
%N A363193 a(1)=1, and thereafter a(n) = number of occurrences of a(k) among terms a(1..k), where k = n-a(n-1).
%H A363193 Neal Gersh Tolunsky, <a href="/A363193/b363193.txt">Table of n, a(n) for n = 1..10000</a>
%H A363193 Neal Gersh Tolunsky, <a href="/A363193/a363193.png">Scatterplot of first 30000 terms</a>
%H A363193 Neal Gersh Tolunsky, <a href="/A363193/a363193_2.png">Scatterplot of first 100000 terms</a>
%e A363193 For n=5, a(5) is the number of times a(5 - a(5-1)) = a(5 - a(4)) = a(3) = 2 has occurred in the sequence among a(1..3). It has occurred 1 time up to that point, so a(5)=1.
%o A363193 (Python)
%o A363193 from bisect import bisect
%o A363193 from itertools import count, islice
%o A363193 def agen(): # generator of terms
%o A363193     an, a, locs = 1, [None, 1], {1: [1]}
%o A363193     yield 1
%o A363193     for n in count(2):
%o A363193         k = n-an
%o A363193         an = bisect(locs[a[k]], k) # sum(1 for i in locs[a[k]] if i <= k)
%o A363193         a.append(an)
%o A363193         if an not in locs: locs[an] = []
%o A363193         locs[an].append(n)
%o A363193         yield an
%o A363193 print(list(islice(agen(), 86))) # _Michael S. Branicky_, May 23 2023
%Y A363193 Cf. A354971.
%K A363193 nonn,easy
%O A363193 1,3
%A A363193 _Neal Gersh Tolunsky_, May 20 2023