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a(1) = 0; thereafter a(n+1) = ceiling((a(n)+y)/2), where y is the number of numbers m < n such that a(m) = a(n).
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%I #11 Apr 16 2021 00:05:16

%S 0,0,1,1,1,2,1,2,2,2,3,2,3,2,4,2,4,3,3,3,4,3,4,4,4,5,3,5,3,5,4,5,4,6,

%T 3,6,4,6,4,7,4,7,4,8,4,8,5,5,5,6,5,6,5,7,5,7,5,8,5,8,6,6,6,7,6,7,6,8,

%U 6,8,7,7,7,8,7,8,8,8,9,5,9,5,9,6,9,6,9

%N a(1) = 0; thereafter a(n+1) = ceiling((a(n)+y)/2), where y is the number of numbers m < n such that a(m) = a(n).

%C Variant of A340488, with XOR(a,y) replaced by ceiling((a+y)/2).

%C Every number appears, and their first occurrences are in increasing order.

%C Apparently, a(n) >= A343332(n) for all n.

%H Pontus von Brömssen, <a href="/A343333/b343333.txt">Table of n, a(n) for n = 1..10000</a>

%o (Python)

%o def A343333_list(n_max):

%o a=0

%o a_list=[0]

%o count=[]

%o for i in range(n_max-1):

%o if a==len(count): count.append(0)

%o else: count[a]+=1

%o a=(a+count[a]+1)//2

%o a_list.append(a)

%o return a_list

%Y Cf. A340488, A343332, A343334.

%K nonn,look

%O 1,6

%A _Pontus von Brömssen_, Apr 12 2021