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A276206 a(0) = a(1) = a(2) = a(3) = 0. For n>3 a(n) is the smallest nonnegative integer such that there is no arithmetic progression i,j,k,m,n (of length 5) such that a(i)+a(j)+a(k)+a(m) = a(n).

%I A276206 #12 Jun 08 2017 16:33:47
%S A276206 0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,2,2,2,2,1,0,0,0,0,1,0,0,0,0,
%T A276206 1,0,0,0,0,1,0,0,0,0,1,2,2,2,2,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,
%U A276206 0,1,2,2,2,2,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,2,2,2,2,1,4,4,4,4,1,4,4,4,4,1,4,4,4,4,1,4,4,4,4,1,2
%N A276206 a(0) = a(1) = a(2) = a(3) = 0. For n>3 a(n) is the smallest nonnegative integer such that there is no arithmetic progression i,j,k,m,n (of length 5) such that a(i)+a(j)+a(k)+a(m) = a(n).
%C A276206 The sequence has the same set of values as A051039 (4-Stohr sequence)
%C A276206 Conjecture 1:
%C A276206 One can calculate a(n) in a following, non-recursive way, using the quinary representation of n.
%C A276206 Let n>=0 be an integer. We consider two cases:
%C A276206 1 There is no digit 4 in the quinary representation of n
%C A276206 Then a(n)=0
%C A276206 2 There is a digit 4 in the quinary representation of n
%C A276206 Let i be the number of the position (counting from right) of the rightmost digit 4 in quinary representation of n, then a(n)=A051039(i).
%C A276206 For example let n=22. The quinary representation of 22 is 42. The rightmost digit 4 in the number 42 is on the second position (counting from right), so a(22) = A051039(2) = 2
%C A276206 Conjecture 2:
%C A276206 The sequence can be generated in a following way:
%C A276206 Start with a zero. Take five consecutive copies of all you have, replace all zeros in the fifth copy with the next value of A051039, repeat
%C A276206 Both of these conjectures can be generalized for similarly defined sequences where the length of the arithmetic progression in the definition (in A276206 it is 5) is a prime number, see A276204.
%C A276206 The assumption about primality is essential, for complex lengths of the arithmetic progression the sequence is more irregular, see A276205.
%H A276206 Michal Urbanski, <a href="/A276206/b276206.txt">Table of n, a(n) for n = 0..199999</a>
%e A276206 For n = 23 we have that:
%e A276206 a(23)>0, because a(3)+a(8)+a(13)+a(18)=0 and 3,8,13,18,23 is an arithmetic progression.
%e A276206 a(23)>1, because a(7)+a(11)+a(15)+a(19)=1 and 7,11,15,19,23 is an arithmetic progression.
%e A276206 There is no such arithmetic progression i,j,k,m,23 that a(i)+a(j)+a(k)+a(m)=2, so a(23) = 2.
%Y A276206 Cf. A276204 (length 3), A276205 (length 4), A276207 (any length).
%Y A276206 Cf. A051039 (4-Stohr sequence).
%K A276206 nonn
%O A276206 0,21
%A A276206 _Michal Urbanski_, Aug 24 2016