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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).
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%I #21 May 13 2021 08:03:05

%S 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 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 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 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 The sequence has the same set of values as A033627.

%C The sequence has a kind of a "triple rhythm", compare the distribution of zeros to the Cantor set.

%C Conjecture 1:

%C One can calculate a(n) in a following, non-recursive way, using the ternary representation of n.

%C Let n>=0 be an integer. We consider two cases:

%C 1. There is no digit 2 in the ternary representation of n. Then a(n) = 0.

%C 2. There is a digit 2 in the ternary representation of n.

%C 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 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 Conjecture 2:

%C The sequence can be generated in a following way:

%C 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 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 The assumption about primality is essential, for complex lengths of the arithmetic progression the sequence is irregular, see A276205.

%H Michal Urbanski, <a href="/A276204/b276204.txt">Table of n, a(n) for n = 0..199999</a>

%e For n = 6 we have that:

%e a(6)>0, because a(0)+a(3)=0 and 0,3,6 is an arithmetic progression.

%e a(6)>1, because a(4)+a(5)=0 and 4,5,6 is an arithmetic progression.

%e There is no such arithmetic progression k,m,6 that a(k)+a(m) = 2, so a(6) = 2.

%Y Cf. A276205 (length 4), A276206 (length 5), A276207 (any length).

%Y Cf. A033627, A229037.

%K nonn,hear

%O 0,7

%A _Michal Urbanski_, Aug 24 2016