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A380106 a(1) = 0; for n >= 1, if there exists an m < n such that a(m) = a(n), take the largest such m and let a(n+1) be the number of runs in the subsequence a(m)..a(n-1). Otherwise, a(n+1) = 0.

%I A380106 #17 Jan 25 2025 09:16:55
%S A380106 0,0,1,0,2,0,2,2,1,5,0,4,0,2,6,0,3,0,2,5,10,0,4,11,0,3,9,0,3,3,1,21,0,
%T A380106 4,10,13,0,4,4,1,8,0,4,4,1,4,2,25,0,6,32,0,3,21,20,0,4,11,31,0,4,4,1,
%U A380106 17,0,4,4,1,4,2,21,15,0,7,0,2,6,25,28,0,5,56
%N A380106 a(1) = 0; for n >= 1, if there exists an m < n such that a(m) = a(n), take the largest such m and let a(n+1) be the number of runs in the subsequence a(m)..a(n-1). Otherwise, a(n+1) = 0.
%C A380106 This is a variant of Van Eck's sequence A181391 in which we ask: how many runs (of consecutive equal terms) ago did we last see a(n)?
%C A380106 The longest run in the sequence has length 2.
%H A380106 Neal Gersh Tolunsky, <a href="/A380106/b380106.txt">Table of n, a(n) for n = 1..10000</a>
%Y A380106 Cf. A380107, A181391, A380037.
%K A380106 nonn
%O A380106 1,5
%A A380106 _Neal Gersh Tolunsky_, Jan 12 2025