A276207 -id:A276207

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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).

+10
7

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,7

COMMENTS

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.

LINKS

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

EXAMPLE

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.

CROSSREFS

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

Cf. A033627, A229037.

KEYWORD

nonn,hear

AUTHOR

Michal Urbanski, Aug 24 2016

STATUS

approved

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).

+10
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

(list; graph; refs; listen; history; text; internal format)

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

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

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.

KEYWORD

nonn,look

AUTHOR

Michal Urbanski, Aug 24 2016

STATUS

approved

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).

+10
4

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, 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, 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

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,21

COMMENTS

The sequence has the same set of values as A051039 (4-Stohr sequence)

Conjecture 1:

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

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

1 There is no digit 4 in the quinary representation of n

Then a(n)=0

2 There is a digit 4 in the quinary representation of n

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).

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

Conjecture 2:

The sequence can be generated in a following way:

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

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.

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

LINKS

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

EXAMPLE

For n = 23 we have that:

a(23)>0, because a(3)+a(8)+a(13)+a(18)=0 and 3,8,13,18,23 is an arithmetic progression.

a(23)>1, because a(7)+a(11)+a(15)+a(19)=1 and 7,11,15,19,23 is an arithmetic progression.

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.

CROSSREFS

Cf. A276204 (length 3), A276205 (length 4), A276207 (any length).

Cf. A051039 (4-Stohr sequence).

KEYWORD

nonn

AUTHOR

Michal Urbanski, Aug 24 2016

STATUS

approved

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