Geometric Moments and Rational Lambert Values: Proofs and Further Results

Testing for additional cyclotomic factors.

Independently of the moment question, let $v_{{\mathrm{\Phi }}_d}$ denote the multiplicity of the cyclotomic factor ${\mathrm{\Phi }}_d$ in a nonzero polynomial in $\mathbb{Q}[p]$, and set $v_{{\mathrm{\Phi }}_d}(0)=\mathrm{\infty }$. For a $2\times 2$ matrix with entry valuations $0,2,2,5$, the two determinant products have valuations $5$ and $4$. The lower one is unique, so the determinant has valuation $4$. A tie is different: all four entries of $\left(\begin{array}{cc}1& 1\\ 1& 1+{\mathrm{\Phi }}_d\end{array}\right)$ have valuation zero, but its determinant is ${\mathrm{\Phi }}_d$. At general rank, we must find the least total valuation and then test whether the terms attaining it cancel. For $N,d\ge 1$, define

The last expression is a minimum-cost assignment: row $i$ is matched to column $\sigma (i)$ at cost $t_{i+\sigma (i),d}$. The valuation of a polynomial in $Y$ means the minimum valuation of its coefficients in $\mathbb{Q}[p]$. Each determinant term is divisible by ${\mathrm{\Phi }}_d^{e_{N,d}}$. Set

where the bar is reduction in $(\mathbb{Q}[p]/({\mathrm{\Phi }}_d))[Y]$. Then $v_{{\mathrm{\Phi }}_d}{\mathcal{H}}_N=e_{N,d}$ exactly when ${\mathrm{\Gamma }}_{N,d}\ne 0$. This isolates the cancellation that a genuine additional factor would require. The assignment bound records the smallest valuation available among the determinant terms. Excess divisibility is possible only if their leading residues cancel. To test that cancellation by one determinant, we distribute the minimum valuation between rows and columns. The integer costs are finite, since every ${\alpha }_m$ is nonzero. Minimum-cost assignment duality gives integer row and column potentials with

Here is a direct construction, so their existence is not an additional hypothesis. Choose a minimum-cost permutation $\sigma$. On the column indices put a directed edge from $\sigma (i)$ to $j$ of length $t_{i+j,d}-t_{i+\sigma (i),d}$. A negative directed cycle would improve $\sigma$ by reassigning the corresponding rows along that cycle; there is therefore no such cycle. Add a new vertex with an edge of length zero to every column. The shortest-path distances $v_j$ are finite integers and satisfy

Putting $u_i=t_{i+\sigma (i),d}-v_{\sigma (i)}$ gives the inequalities above, with equality on the chosen assignment and hence equality of totals.

Divide row $i$ by ${\mathrm{\Phi }}_d^{u_i}$ and column $j$ by ${\mathrm{\Phi }}_d^{v_j}$ over $\mathbb{Q}(p)$. The inequalities ensure that every resulting entry is in $\mathbb{Q}[p,Y]$, even if some potentials are negative. Reduce these entries modulo ${\mathrm{\Phi }}_d$. Only those with $u_i+v_j=t_{i+j,d}$ survive. The determinant of the reduced matrix is ${\mathrm{\Gamma }}_{N,d}$, because every non-minimal permutation vanishes on reduction. Thus the valuation exceeds $e_{N,d}$ exactly when the reduced matrix is singular over $(\mathbb{Q}[p]/({\mathrm{\Phi }}_d))(Y)$. A kernel relation valid at just one value of $Y$ is insufficient: the determinant must vanish as a polynomial in $Y$. Conversely, one nonzero value ${\mathrm{\Gamma }}_{N,d}(Y_0)$ proves equality with the assignment bound. Neither assignment duality nor the existence of the potentials forces the reduced determinant to vanish.

An exact calculation outside Lean gives the following monic contents. Here ${\mathrm{cont}}_Y$ means the monic gcd in $\mathbb{Q}[p]$ of the coefficients in $Y$:

These factorisations attain the assignment bound for every $d$ at these five ranks. The residue certificates verify equality for $d\le 8$; the displayed complete factorisations contain no other cyclotomic factors, so the nonnegative assignment bound is zero at all remaining $d$. The power of $p$ is not cyclotomic content.

For the table, an exact program computes the full polynomials ${\mathcal{H}}_N(Y_0;p)$ at $Y_0=0,\dots ,N$ by fraction-free elimination, checking every polynomial division. Their monic gcd in $\mathbb{Q}[p]$ is ${\mathrm{cont}}_Y{\mathcal{H}}_N$. A divisor of all coefficients divides every evaluated polynomial; conversely, the $N+1$ values recover those coefficients by rational interpolation in $Y$. This proves the reverse divisibility over $\mathbb{Q}[p]$, where the interpolation denominators are units. The calculation includes every factor, not just a prescribed list of cyclotomic candidates.

The same program produces a nonzero residue witness for every pair

There are $76$ such pairs. For each, the certificate gives an optimal assignment, integer dual potentials, an integer $Y_0\in \{0,\dots ,N\}$ and the nonzero residue ${\mathrm{\Gamma }}_{N,d}(Y_0)$. The signed minimum-cost subset recurrence and the determinant of the reduced matrix give the same residue. A separate check computes the full polynomial ${\mathcal{H}}_N(Y_0;p)$ and verifies directly that division by ${\mathrm{\Phi }}_d^{e_{N,d}}$ leaves exactly this nonzero residue. At $N=2,d=1$, for example, the entry valuations are $\left(\begin{array}{cc}0& 2\\ 2& 5\end{array}\right)$, $e_{2,1}=4$, and ${\mathrm{\Gamma }}_{2,1}(0)=-9$.

The source coefficients, determinant-value polynomials and witnesses are computed by that program, which also repeats the direct determinant checks; its recorded result lists every witness. At ranks six to eight these certificates cover only the stated cyclotomic-index window; complete coefficient contents are computed only through rank five.

These computations distinguish systematic common factors from cancellation beyond entry valuations, a distinction also important in fraction-free matrix decompositions [40]. In particular the gcd argument at the points $0,\dots ,N$ is over $\mathbb{Q}[p]$: the rational Vandermonde inverse is not generally integral, so it must not be used to assert the corresponding coefficient gcd in $\mathbb{Z}[p]$.

Proposition 4 of Krattenthaler–Rochev–Väänänen–Zudilin [13] obtains cyclotomic factors through root-of-unity annihilation for a different first-order tail recurrence. No corresponding recurrence has been established for this moving-diagonal pencil. The finite witnesses prove equality with the assignment bound at these $76$ pairs. They do not exclude excess at larger ranks or, at ranks six to eight, cyclotomic indices outside the stated window. Whether minimum-valuation residues ever cancel systematically, and whether $s_m$ is a Stieltjes sequence, remain separate questions. The remainder measure answers neither.