Semigroups of Polyhedra with Prescribed Number of Lattice Points and the k-Frobenius Problem

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Problem: Reconstructed conjecture: for a positive integer triple $a=(a_1,a_2,a_3)$, usually with $\gcd(a_1,a_2,a_3)=1$, let

Let

with the common convention $g_i(a)=0$ if this set is empty. The conjecture says $g_i(a)$ is increasing in $i$ for $0\le i\le14$. This is the nontrivial “exactly $i$ representations” interpretation; the “at most $i$” interpretation would be monotone by inclusion and would not be a real conjecture.

Result: The conjecture is false.

Take

This is a primitive triple, since $\gcd(4,5,6)=1$, and it is not redundant: none of $4,5,6$ is a nonnegative integer combination of the other two.

Let $R(n)=R_{(4,5,6)}(n)$. A representation with $x_1>0$ gives one of $n-4$, while those with $x_1=0$ solve $5y+6z=n$. Thus

where $T(n)=\#\{(y,z)\in\mathbb Z_{\ge0}^2:5y+6z=n\}$.

Using this recurrence gives, for $n=4q+r$, $0\le r<4$, $0\le q\le11$:

So no $n\le47$ has exactly $10$ representations, while $R(43)=9$. Also $R(44),R(45),R(46),R(47)\ge11$. Since adding one copy of the generator $4$ injects representations of $n$ into representations of $n+4$, every $n\ge44$ has at least $11$ representations.

Hence there is no positive integer with exactly $10$ representations:

under the standard empty-set convention, while

Thus $g_9>g_{10}$, so the sequence is not even nondecreasing.

If $g_i$ is left undefined when no such integer exists, then $g_{10}(4,5,6)$ is not a number, so the stated “sequence of numbers” does not exist.

A natural repair would be to use $F_i(a)=\max\{n:R_a(n)\le i\}$, which is monotone trivially by inclusion, or to add an explicit existence hypothesis for all $g_i$. But the stated exact-$g_i$ conjecture is false.

Citation: No external citation is needed; the counterexample is elementary.