OFFSET
2,1
COMMENTS
Let f(n)=0 when n is even and f(n)=1 when n is odd.
If n is a square, we have a(n)=4+f(n).
If n is a nonsquare, we have a(n)=3-f(v(n)) where v(n) is the exponent of the largest power of 2 that divides n.
In particular, when n is an odd nonsquare, we have a(n)=3.
LINKS
Krishnendu Paul and Shameek Paul, Square-weighted zero-sum constants, arXiv:2202.13143 [math.NT], 2022-2024.
FORMULA
When n is a square, we have a(n)=4 if n is even, and a(n)=5 if n is odd.
Let v(n) be the exponent of the largest power of 2 which divides n. When n is not a square, we have a(n)=2 if v(n) is odd, and a(n)=3 if v(n) is even.
EXAMPLE
We will show that a(6)=2. Let A be the set of nonzero squares in Z_6. Then A={1,3,4}. By considering the sequence S=(1), we see that a(6) is at least 2. In order to show that a(6) is at most two, we need to show that every sequence in Z_6 of length 2 has an A-weighted zero-sum subsequence.
Let S=(x,y) be a sequence in Z_6. Suppose x is even. Then 3x=0 and so T=(x) is an A-weighted zero-sum subsequence of S. A similar argument holds if y is even. Hence, we may assume that both x and y are odd. As x+y is even, we see that 3(x+y)=0. Thus, it follows that S is an A-weighted zero-sum subsequence.
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Shameek Paul, Apr 05 2024
STATUS
approved