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).

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

Offset: 0

Author

Michal Urbanski, Aug 24 2016

Keywords

• nonn

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.

Examples

			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.
		

Links

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

Crossrefs

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

Cf. A051039 (4-Stohr sequence).