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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). 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).

Sequence in context: A309509 A216953 A326786 * A334222 A124752 A293730

Adjacent sequences:  A276203 A276204 A276205 * A276207 A276208 A276209

KEYWORD

nonn

AUTHOR

Michal Urbanski, Aug 24 2016

STATUS

approved

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Last modified September 18 16:42 EDT 2020. Contains 337170 sequences. (Running on oeis4.)