Let P(x) have roots r_1,r_2,…,r_i, with respective multiplicities 1,m_2,…,m_i, where m_j≥ 1 for 2≤ j≤ i≤ k. Let a_1 be the coefficient of the principal root in the Binet expansion of q(1,t), and let Q_0(x) denote the polynomial used by the Zeroing Algorithm. Generalized principal-coefficient conjecture. The coefficient a_1 satisfies a_1=fracQ_0(r_1)∏_j=2^i(r_1-r_j)^m_j. This conjecture extends the corresponding result for ZLRRs whose roots all have multiplicity 1 to the case of arbitrary root multiplicities. Establishing it is more difficult because repeated roots make the Binet expansion of q(1,t) more complicated and prevent the Vandermonde-matrix argument used in the simple-root case.

The generalized principal-coefficient formula for Binet expansions

Let $P(x)$ have roots $r_1,r_2,\dots,r_i$ , with respective multiplicities $1,m_2,\dots,m_i$ , where $m_j\geq 1$ for $2\leq j\leq i\leq k$ . Let $a_1$ be the coefficient of the principal root in the Binet expansion of $q(1,t)$ , and let $Q_0(x)$ denote the polynomial used by the Zeroing Algorithm.

Generalized principal-coefficient conjecture. The coefficient $a_1$ satisfies

a_1=\frac{Q_0(r_1)}{\prod_{j=2}^{i}(r_1-r_j)^{m_j}}.

This conjecture extends the corresponding result for ZLRRs whose roots all have multiplicity $1$ to the case of arbitrary root multiplicities. Establishing it is more difficult because repeated roots make the Binet expansion of $q(1,t)$ more complicated and prevent the Vandermonde-matrix argument used in the simple-root case.

Progress summary

Solved

An unverified posted proof claims the formula works with repeated roots, but no independent verification or published follow-up has appeared.

The conjecture, stated as Conjecture 4.8 in 2020, extends the known simple-root formula to arbitrary multiplicities. The paper explicitly leaves the repeated-root case unproved.

Known results

The formula is established when all roots are simple; the paper says the Vandermonde-matrix argument does not extend directly to repeated roots.

Posted attempt

A posted argument claims a complete proof via the generating function $A(z)=B(z)/D(z)$ : extracting the simple pole at $z=1/r_1$ yields the stated coefficient, while repeated nonprincipal roots contribute only higher-order poles elsewhere. This proof has not been independently verified.

Current status (as of August 2026): The conjecture has an unverified complete-proof claim, but no corroborated proof, counterexample, or published progress; its mathematical status therefore remains unsettled.

Sources

arxiv.org

Sources & referencesView supporting material

Primary source

Thomas C. Martinez, Steven J. Miller, Clayton Mizgerd, Jack Murphy and Chenyang Sun, “Generalizing Zeckendorf's Theorem to Homogeneous Linear Recurrences, II”, arXiv:2009.07891 (2021).

1 week ago

Proof

Write $q_t=q(1,t)$ , and let

P(x)=x^k-c_1x^{k-1}-\cdots-c_k.

The coefficient recurrences in the Zeroing Algorithm give

q_t=c_1q_{t-1}+\cdots+c_kq_{t-k} \qquad (t\geq k),

while for $0\leq t<k$ the initial values satisfy

q_t-\sum_{s=1}^{t}c_s q_{t-s}=\beta_{t+1}.

Therefore the ordinary generating function $A(z)=\sum_{t\geq0}q_tz^t$ satisfies, as an identity of formal power series,

A(z)=\frac{B(z)}{D(z)},

where

B(z)=\beta_1+\beta_2z+\cdots+\beta_kz^{k-1} =z^{k-1}Q_0(1/z)

and

D(z)=1-c_1z-\cdots-c_kz^k=z^kP(1/z).

Because the roots of $P$ are $r_1,r_2,\ldots,r_i$ with respective multiplicities $1,m_2, \ldots,m_i$ , we have

D(z)=(1-r_1z)\prod_{j=2}^{i}(1-r_jz)^{m_j}.

In the Binet expansion, $r_1$ is simple, so its contribution to $q_t$ is $a_1r_1^t$ , whose generating function is

\frac{a_1}{1-r_1z}.

All contributions from the other roots have poles only at $z=1/r_j$ . Hence the usual simple-pole cover-up calculation gives

a_1 = \left.(1-r_1z)A(z)\right|_{z=1/r_1} = \frac{B(1/r_1)} {\prod_{j=2}^{i}(1-r_j/r_1)^{m_j}}.

Let

M=\sum_{j=2}^{i}m_j=k-1.

The reversal identity for $B$ gives

B(1/r_1)=\frac{Q_0(r_1)}{r_1^M},

and

\prod_{j=2}^{i}(1-r_j/r_1)^{m_j} = \frac{\prod_{j=2}^{i}(r_1-r_j)^{m_j}}{r_1^M}.

The powers of $r_1$ cancel, yielding

a_1= \frac{Q_0(r_1)} {\prod_{j=2}^{i}(r_1-r_j)^{m_j}}.

Thus Conjecture 4.8 holds. Repeated nonprincipal roots merely produce higher-order poles away from the simple principal pole and therefore do not affect this extraction.

Source: https://arxiv.org/abs/2009.07891

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