Extremal problems for affine cubes of integers - University of Manitoba

6.1 AP3-free partitioning
Adapting Behrend’s [1] proof (cf. [16], Theorem 6.6, lower bound) gives
a partition result.
Theorem 6.2 For sufficiently large n, there exists a partition [1, n] =
X1 ∪ . . . ∪ Xq, q < e 3√ ln n so that each Xi is AP3-free and |Xi| ≤ n/e ln 2√ ln n .
Proof: Set d = ⌊e √ ln n⌋ (not to be confused with the dimension d of the affine
cube as in the rest of this paper), k = ⌈ ln(n+1)
⌉ − 1 and for each x ∈ [1, n],
ln(2d)
write
k
x = xi(2d) i ,
i=0
where for each i, 0 ≤ xi ≤ 2d − 1. Note that due to our choice of k,
(2d) k < n + 1 ≤ (2d) k+1 holds. Put
For I ⊂ [0, k], set
and
Xn,d = {x ∈ [n] : 0 ≤ xi ≤ d − 1 for all i}.
yI =
d(2d)
i∈I
i
YI = yI + Xn,d = {yI + x : x ∈ Xn,d}.
This gives Y∅ = Xn,d and |YI| = |Xn,d| for every I ⊆ [0, k]. For x =
ki=0 xi(2d) i , let
I = I(x) = {i ∈ [0, k] : xi ≥ d};
then
x − yI(x) =
i∈I(x)
xi(2d) i +
is an element of Xn,d, and so [n] =
set
and for each I ⊆ [0, k], put
I⊆[0,k]
i∈I(x)
(xi − d)(2d) i
k
Xn,d,s = x ∈ Xn,d : x
i=0
2
i = s
YI,s = yI + Xn,d,s = {yI + x : x ∈ Xn,d,s}.
12
YI is a partition. For an integer s