THE LARGEST PROJECTIVE CUBE-FREE SUBSETS OF Z_2^n

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Problem: Reconstructed statement: in $G=\mathbb Z/2^n\mathbb Z$, let

A set is centred if it fills layers $L_1,L_2,\dots$ in order, with at most one partially filled layer. For $A\subseteq G$, define the number of $d$-cubes by

This matches the paper’s ordered Schur-triple convention for $d=2$. The conjecture asserts that for $n\ge k\ge2$ and $|A|=M$, centred sets minimize $Q_{2^k}(A)$.

There is an off-by-one ambiguity in the paper’s “analog of Samotij” wording, but the counterexample below also refutes the $2^{k-1}$-version by taking $k=3$.

Result: The conjecture is false.

Take $n=4$, $k=2$, $M=13$, so $G=\mathbb Z/16\mathbb Z$ and we count $4$-cubes. The centred $13$-sets are

and

They have equal cube counts by $x\mapsto -x$.

Now let

Then $|A|=13$, but $A$ is not centred since it contains $8\in L_4$ while omitting elements of $L_2$.

For $F\subseteq G$, write $N(F)=Q_4(G\setminus F)$. A direct valuation enumeration in $\mathbb Z/16\mathbb Z$ gives:

where $e_i=v_2(x_i)$, $m$ is the coordinate permutation factor, and all omitted valuation patterns contribute $0$. Each $t_F$ counts fixed-order choices with all nonempty subset sums outside $F$.

Thus

while

Hence a non-centred set of size $13$ has strictly fewer $4$-cubes than every centred set of size $13$. This disproves the conjecture.

Audit: the parameters satisfy $4=n\ge k=2\ge2$, the size is $M=13$, the cube-counting convention is exactly the ordered-generator convention extending Schur triples, and no extra hypotheses are used.

Citation: Conjecture and terminology: Jason Long and Adam Zsolt Wagner, “The largest projective cube-free subsets of $\mathbb Z_{2^n}$,” arXiv:1810.01225, Conjecture 1.9. The counterexample above is not cited from the literature.