A fixed, uncertified candidate against maximum-sum orbit matching

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This is not a counterexample to Erdős 963. Nor is it yet a counterexample to the maximum-sum complement-orbit assertion: the essential maximum-dimension claim below remains unproved.

The fixed configuration

Let

$$\begin{aligned} B&=\{1463,1477,1505,1533,1603,1687\},\\ E&=B\cup\{9233\},\qquad c=35,\\ X&=\{1000000001,1000000017,1000000053,1000000061\},\\ D&=E\cup X, \end{aligned}$$

and

$$C=\{1,9,11,17,19,25,27,35,43,51,59,61,71,79,87,95,131\}, \qquad O=D\cup C.$$

All entries are positive odd, $|D|=11$, and $|O|=28$.

The construction uses

$$F=\{9,11,15,19,29,41\},\quad B=7(200+F),\quad \sum B=9268,\quad9233=\sum B-c.$$

Dissociation of $F$ was proved in the minimum-sum note.

Proved: $D$ is dissociated

A ternary relation on $200+F$ first forces its coefficient sum to be zero, since $\sum F=124<200$, and then reduces to a relation on $F$. Thus $B$ is dissociated. A representation of $9233$ on $B$ would give

$$35=\sum_{b\in B}(1-\varepsilon_b)b,$$

impossible because every summand is nonnegative and $\min B>35$. Hence $E$ is dissociated.

Write $X=A+\{0,16,52,60\}$ with $A=1000000001$. In a relation on $D$, the coefficient sum on $X$ must vanish:

$$\sum E+0+16+52+60=18629<A.$$

Every element of $E$ is divisible by seven. The four offsets have distinct residues modulo seven, and their six pair sums

$$16,52,60,68,76,112$$

also have distinct residues modulo seven. A nonzero balanced ternary combination of four offsets would compare either two singletons or two disjoint pairs, so cannot vanish modulo seven. Thus all $X$ coefficients are zero. Dissociation of $E$ eliminates the rest.

Proved: the seventeen colors occupy exactly sixteen orbits

The relation

$$c=\sum B-9233 \tag{1}$$

uses every coordinate of $E$. The six offset differences are $16,52,60,36,44,8$, and their balanced two-versus-two differences, up to sign, are $96,24,8$. Consequently $C$ consists exactly of the distinct positive absolute values of

$$c,\quad c\pm16,\ c\pm52,\ c\pm60,\ c\pm36,\ c\pm44,\ c\pm8, \quad c\pm96,\ c\pm24.$$

These give the specified central square, full-support, and four-dimensional face representations on $D$.

There are no representations with different outside signs. To check this, let $a\in C$ have any ternary representation on $D$. The coefficient sum on $X$ is again zero, since $a\le131$ and the remaining terms are far smaller than $A$. The offset contribution $\tau$ has $|\tau|\le96$. Thus its $E$ contribution satisfies

$$|a-\tau|\le227.$$

It is odd, since $\tau$ is even.

If the coefficient of $9233$ is zero, write the $B$ contribution as $7(200s+t)$, where $|t|\le124$. Its absolute value at most $227$ forces $s=0$. Since the elements of $F$ are odd, $s=0$ makes $t$ even, so the contribution is even, a contradiction.

If the coefficient of $9233$ is $-1$, the contribution is

$$35-\sum_{b\in B}(1-\varepsilon_b)b.$$

Its absolute value at most $227$ forces every $\varepsilon_b=1$, because $\min B=1463$. Similarly, coefficient $+1$ forces every $\varepsilon_b=-1$. Thus every representation uses precisely the outside sign pattern in (1), or its negative.

In the folded cube of $\Sigma(D)$, fixing those outside signs up to reversal leaves one four-coordinate cube $Q$ of sixteen complement orbits. Every color in $C$ has its entire neighborhood in $Q$. The representation (1) of $c$ has all four $X$ coordinates free and supplies every orbit of $Q$. Therefore

$$|N(C)|=16<17=|C|. \tag{2}$$

This is Hall failure for the particular dissociated $D$.

The unproved claim and precise next task

It is not proved that $d(O)=11$. The only dimension experiment consisted of 1000 seeded greedy trials; the largest set found had size eleven. That is not an upper bound.

The exact unresolved finite task is to produce either a dissociated twelve-subset of $O$, or an exact relation-cover certificate for every twelve-subset. No such certificate was obtained.

If $d(O)=11$, then $D$ is automatically the unique maximum-sum maximum subset: its entries are the eleven largest elements of $O$. In that event (2) would refute the maximum-sum orbit-matching route. If a twelve-subset is found, this candidate is rejected. Neither outcome would resolve the original cardinality bound by itself.

Origin of the central configuration

For four variable coordinates $x_1,\ldots,x_4$, set

$$u=x_1+x_2-x_3-x_4,\quad v=x_1-x_2+x_3-x_4,\quad w=x_1-x_2-x_3+x_4.$$

The central colors are absolute values of

$$c,\quad c\pm u,\ c\pm v,\ c\pm w,\quad c\pm (u\pm v)/2,\quad c\pm (u\pm w)/2,\quad c\pm(v\pm w)/2.$$

After multiplication by two these are an affine image of $\{0,\pm2e_i,\pm e_i\pm e_j\}$, the $C_3$ root configuration with its origin. This identity alone supplies no augmentation.

In particular, the six axial forms are not a universal replacement. Put

$$B_0=3F,\quad E_0=B_0\cup\{367\},\quad c_0=5.$$

Then $E_0$ is dissociated and $c_0=\sum B_0-367$ is its unique ternary representation. Reduction modulo three forces the coefficient of $367$ to be $-1$, and positivity then forces all other coefficients to be $+1$. Dissociation follows because a representation of $367$ on $B_0$ would give a nonnegative combination of its elements equal to $5<\min B_0$.

Nevertheless both $3c_0$ and $6c_0$ remain represented after every single deletion from $E_0$. For $3c_0$, scale by three the four robust identities for $5$ in the minimum-sum note. For $6c_0$, use

$$10=19-9=29-19=41+9-11-29$$

scaled by three. Each family of supports has empty intersection. Since the six axial forms sum to $6c_0$, these identities block that proposed six-form augmentation even after one outside deletion. Higher-multiple relations need not provide an additional representation of the center itself.