A276206 -id:A276206 - cp's OEIS frontend

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).

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, 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, 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

Offset: 0

Michal Urbanski, Aug 24 2016

The sequence has the same set of values as A033627.
The sequence has a kind of a "triple rhythm", compare the distribution of zeros to the Cantor set.
Conjecture 1:
One can calculate a(n) in a following, non-recursive way, using the ternary representation of n.
Let n>=0 be an integer. We consider two cases:
1. There is no digit 2 in the ternary representation of n. Then a(n) = 0.
2. There is a digit 2 in the ternary representation of n.
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).
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.
Conjecture 2:
The sequence can be generated in a following way:
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.
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.
The assumption about primality is essential, for complex lengths of the arithmetic progression the sequence is irregular, see A276205.

			For n = 6 we have that:
a(6)>0, because a(0)+a(3)=0 and 0,3,6 is an arithmetic progression.
a(6)>1, because a(4)+a(5)=0 and 4,5,6 is an arithmetic progression.
There is no such arithmetic progression k,m,6 that a(k)+a(m) = 2, so a(6) = 2.

Michal Urbanski, Table of n, a(n) for n = 0..199999

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

Cf. A033627, A229037.

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).

Original entry on oeis.org

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

Michal Urbanski, Aug 24 2016

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

			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.

Michal Urbanski, Table of n, a(n) for n = 0..49999

Cf. A276204 (length 3), A276206 (length 5), A276207 (any length).

Cf. also A229037.

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

A276207 a(0) = a(1) = 0. For n>1 a(n) is the smallest nonnegative integer such that there is no arithmetic progression m_1,m_2,...,m_i,n (2<=i

Original entry on oeis.org

0, 0, 1, 0, 0, 2, 4, 1, 1, 0, 0, 5, 0, 0, 3, 7, 2, 3, 2, 4, 1, 1, 12, 1, 5, 1, 1, 0, 0, 13, 0, 0, 10, 9, 6, 7, 0, 0, 18, 0, 0, 15, 4, 14, 7, 6, 8, 2, 6, 3, 16, 3, 3, 2, 3, 7, 1, 10, 25, 8, 5, 1, 1, 1, 4, 14, 27, 4, 1, 1, 10, 2, 2, 6, 1, 26, 8, 1, 19, 1, 1, 0, 0, 13, 0, 0, 7, 24, 2, 19, 0, 0, 34, 0, 0, 29, 32, 32, 5, 15, 21, 14, 15, 6, 6, 24, 13, 39, 0, 0, 24, 0

Offset: 0

Michal Urbanski, Aug 24 2016

nonn

The distribution of zeros is the same as in A276204.

			For n = 5 we have that:
a(5)>0, because a(3)+a(4)=0 and 3,4,5 is an arithmetic progression
a(5)>1, because a(2)+a(3)+a(4)=1 and 2,3,4,5 is an arithmetic progression
there is no such arithmetic progression m_1,m_2,...,m_i,5 that a(m_1)+a(m_2)+...+a(m_i)=2, so a(5) = 2.

Michal Urbanski, Table of n, a(n) for n = 0..49999

Cf. A276204 (length 3), A276205 (length 4), A276206 (length 5).

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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).

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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).

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A276207 a(0) = a(1) = 0. For n>1 a(n) is the smallest nonnegative integer such that there is no arithmetic progression m_1,m_2,...,m_i,n (2<=i

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