Monomial-count conjecture for symmetrized Mordell–Tornheim zeta values

Exact expansion and support of the symmetrized Mordell--Tornheim values

Let

$$\overline{\zeta}_n =\sum_{\substack{a_1,\ldots,a_n\in\mathbb Z\setminus\{0\}\\ a_1+\cdots+a_n=0}} \frac{1}{|a_1\cdots a_n|} \qquad(n\geq1).$$

We prove the stronger coefficient-level statement

$$\boxed{\displaystyle \overline{\zeta}_n =n! \sum_{\substack{m_2,\ldots,m_n\geq0\\ 2m_2+3m_3+\cdots+nm_n=n}} \prod_{j=2}^{n} \frac{1}{m_j!} \left(\frac{2^j-2}{j}\,\zeta(j)\right)^{m_j}.}$$

Each displayed monomial has a strictly positive integer coefficient, and the monomials are indexed bijectively by the partitions of $n$ having no part equal to $1$. Consequently, their number is exactly

$$p(n)-p(n-1).$$

Here monomials are understood in the source's formal Bell-polynomial expansion, not modulo algebraic relations among numerical zeta values.

Proof

Dobrowolski's Theorem 2.2, equivalently equation (4.9) of the cited paper, establishes the identity

$$\overline{\zeta}_n =B_n\left( 0, (2^2-2)\Gamma(2)\zeta(2), \ldots, (2^n-2)\Gamma(n)\zeta(n) \right),$$

where $B_n$ is the complete exponential Bell polynomial. We use this existing theorem as the starting point; the remaining task is to determine every coefficient and prove that none vanishes.

Introduce independent formal indeterminates $Z_2,Z_3,\ldots$ with weighted degrees

$$\operatorname{wt}(Z_j)=j.$$

Define $P_0=1$ and, for $n\geq1$, define

$$P_n(Z_2,\ldots,Z_n) =B_n\left( 0, (2^2-2)(2-1)!Z_2, \ldots, (2^n-2)(n-1)!Z_n \right).$$

The defining exponential generating function of the complete Bell polynomials gives

$$\sum_{n=0}^{\infty}P_n\frac{x^n}{n!} =\exp\left( \sum_{j=2}^{\infty} \frac{2^j-2}{j}\,Z_jx^j \right).$$

Expanding the exponential as a product of formal power series yields

$$\begin{aligned} \sum_{n=0}^{\infty}P_n\frac{x^n}{n!} &= \prod_{j=2}^{\infty} \exp\left(\frac{2^j-2}{j}Z_jx^j\right)\\ &= \prod_{j=2}^{\infty} \sum_{m_j=0}^{\infty} \frac{1}{m_j!} \left(\frac{2^j-2}{j}Z_jx^j\right)^{m_j}. \end{aligned}$$

Therefore, for every multiplicity vector satisfying $\sum_{j=2}^n jm_j=n$, the coefficient of its associated monomial is

$$\left[\prod_{j=2}^{n}Z_j^{m_j}\right]P_n =\frac{n!}{\prod_{j=2}^{n}m_j!} \prod_{j=2}^{n} \left(\frac{2^j-2}{j}\right)^{m_j}>0.$$

In fact, this coefficient is an integer: equivalently, it factors as

$$\left[\prod_{j=2}^{n}Z_j^{m_j}\right]P_n = \frac{n!}{\prod_{j=2}^{n}(j!)^{m_j}m_j!} \prod_{j=2}^{n} \big((2^j-2)(j-1)!\big)^{m_j}.$$

The first factor counts the set partitions of an $n$-element set having exactly $m_j$ blocks of size $j$ for every $j$; it is therefore a positive integer. Every factor in the second product is a positive integer as well. Thus every allowed formal monomial occurs exactly once and has a strictly positive integer coefficient.

Its weighted degree is

$$\operatorname{wt}\left(\prod_{j=2}^n Z_j^{m_j}\right) =\sum_{j=2}^n jm_j=n,$$

so $P_n$ is weighted-homogeneous. The multiplicity vector corresponds bijectively to the integer partition

$$n=\underbrace{2+\cdots+2}_{m_2\text{ times}} +\underbrace{3+\cdots+3}_{m_3\text{ times}} +\cdots +\underbrace{n+\cdots+n}_{m_n\text{ times}}.$$

Among the $p(n)$ partitions of $n$, those containing a part $1$ correspond bijectively to the $p(n-1)$ partitions of $n-1$ by deleting one part $1$. Hence the number containing no part $1$, and therefore the exact formal monomial count, is

$$\#\operatorname{supp}(P_n)=p(n)-p(n-1).$$

More precisely, if $p_k(s)$ denotes the number of partitions of $s$ into exactly $k$ positive parts, the number of monomials containing exactly $k$ zeta factors, counted with multiplicity, is

$$\#\left\{ (m_2,\ldots,m_n): \sum_{j=2}^n jm_j=n,\quad \sum_{j=2}^n m_j=k \right\} =p_k(n-k),$$

because subtracting $1$ from each of the $k$ parts gives a partition of $n-k$ into exactly $k$ positive parts.

For $n=1$, the defining sum is empty and $\overline{\zeta}_1=P_1=0$. Since $p(1)=p(0)=1$, the claimed number of monomials is correctly $0$.

Finally, evaluating $Z_j=\zeta(j)$ recovers the boxed identity. This evaluation does not assert that the numerical zeta monomials are algebraically independent. For example,

$$P_4(Z_2,Z_3,Z_4)=12Z_2^2+84Z_4, \qquad \zeta(4)=\frac25\zeta(2)^2.$$

Although the two evaluated terms can subsequently be combined numerically, the source itself lists both as distinct monomials, exactly as required by its conjecture.

This proves Conjecture A.2 of Przemysław Dobrowolski, Evaluation of the symmetrized Mordell--Tornheim zeta function, https://arxiv.org/abs/2603.20550. The separate Conjectures A.1 and A.3 are not addressed.