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%I A276204 #21 May 13 2021 08:03:05
%S A276204 0,0,1,0,0,1,2,2,1,0,0,1,0,0,1,2,2,1,4,4,1,4,4,1,2,2,1,0,0,1,0,0,1,2,
%T A276204 2,1,0,0,1,0,0,1,2,2,1,4,4,1,4,4,1,2,2,1,7,7,1,7,7,1,2,2,1,7,7,1,7,7,
%U A276204 1,2,2,1,4,4,1,4,4,1,2,2,1,0,0,1,0,0,1,2,2,1,0,0,1,0,0,1,2,2,1,4,4,1,4,4,1,2,2,1,0,0,1,0,0,1,2,2,1,0,0,1,0
%N A276204 a(0) = a(1) = 0. For n>1 a(n) is the smallest nonnegative integer such that there is no arithmetic progression k,m,n (of length 3) such that a(k)+a(m) = a(n).
%C A276204 The sequence has the same set of values as A033627.
%C A276204 The sequence has a kind of a "triple rhythm", compare the distribution of zeros to the Cantor set.
%C A276204 Conjecture 1:
%C A276204 One can calculate a(n) in a following, non-recursive way, using the ternary representation of n.
%C A276204 Let n>=0 be an integer. We consider two cases:
%C A276204 1. There is no digit 2 in the ternary representation of n. Then a(n) = 0.
%C A276204 2. There is a digit 2 in the ternary representation of n.
%C A276204 Let i be the number of the position (counting from right) of the rightmost digit 2 in ternary representation of n, then a(n) = A033627(i).
%C A276204 For example: let n = 19. The ternary representation of 19 is 201. The rightmost digit 2 in the number 201 is on the third position (counting from right), so a(19) = A033627(3) = 4.
%C A276204 Conjecture 2:
%C A276204 The sequence can be generated in a following way:
%C A276204 Start with a zero. Take three consecutive copies of all you have, replace all zeros in the third copy with the next value of A033627, repeat.
%C A276204 Both of these conjectures can be generalized for similarly defined sequences where the length of the arithmetic progression in the definition (in A276204 it is 3) is a prime number, see A276206.
%C A276204 The assumption about primality is essential, for complex lengths of the arithmetic progression the sequence is irregular, see A276205.
%H A276204 Michal Urbanski, <a href="/A276204/b276204.txt">Table of n, a(n) for n = 0..199999</a>
%e A276204 For n = 6 we have that:
%e A276204 a(6)>0, because a(0)+a(3)=0 and 0,3,6 is an arithmetic progression.
%e A276204 a(6)>1, because a(4)+a(5)=0 and 4,5,6 is an arithmetic progression.
%e A276204 There is no such arithmetic progression k,m,6 that a(k)+a(m) = 2, so a(6) = 2.
%Y A276204 Cf. A276205 (length 4), A276206 (length 5), A276207 (any length).
%Y A276204 Cf. A033627, A229037.
%K A276204 nonn,hear
%O A276204 0,7
%A A276204 _Michal Urbanski_, Aug 24 2016

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