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A276205
a(0) = a(2) = a(3) = 0. For n>2 a(n) is the smallest nonnegative integer such that there is no arithmetic progression j,k,m,n (of length 4) such that a(j)+a(k)+a(m) = a(n).
4
0, 0, 0, 1, 0, 0, 2, 0, 0, 0, 1, 2, 1, 3, 0, 0, 0, 4, 0, 1, 2, 2, 3, 1, 4, 0, 0, 1, 0, 0, 0, 5, 3, 0, 7, 1, 0, 4, 2, 4, 2, 3, 5, 1, 1, 4, 1, 3, 0, 0, 0, 1, 0, 0, 2, 0, 0, 0, 9, 2, 8, 10, 0, 4, 0, 0, 0, 2, 1, 7, 13, 4, 12, 4, 6, 7, 4, 4, 2, 0, 10, 2, 2, 1, 3, 1, 0, 0, 0, 12, 0, 9, 1, 0, 5, 2, 1, 17, 0, 3, 5, 0, 1, 1, 0, 0, 8, 3, 0, 0, 0, 15, 12, 9, 10, 11, 1, 5
OFFSET
0,7
COMMENTS
This sequence, unlike A276204 (defined similarly) is seemingly irregular.
a(n) <= n/3. - Robert Israel, Aug 24 2016
The graph (and the definition) are reminiscent of A229037. - N. J. A. Sloane, Aug 29 2016
LINKS
EXAMPLE
For n = 6 we have that:
a(6)>0, because a(0)+a(2)+a(4)=0 and 0,2,4,6 is an arithmetic progression.
a(6)>1, because a(3)+a(4)+a(5)=1 and 3,4,5,6 is an arithmetic progression.
there is no such arithmetic progression j,k,m,6 that a(j)+a(k)+a(m)=2, so a(6) = 2.
MAPLE
for i from 0 to 2 do A[i]:= 0 od:
for n from 3 to 200 do
Forbid:= {seq(A[n-d]+A[n-2*d]+A[n-3*d], d=1..floor(n/3))};
A[n]:= min({$0..max(Forbid)+1} minus Forbid)
od:
seq(A[i], i=0..200); # Robert Israel, Aug 24 2016
CROSSREFS
Cf. A276204 (length 3), A276206 (length 5), A276207 (any length).
Cf. also A229037.
Sequence in context: A117408 A228360 A303138 * A244966 A079100 A296167
KEYWORD
nonn,look
AUTHOR
Michal Urbanski, Aug 24 2016
STATUS
approved