Let D(k) denote the Domb numbers, and let (u_n)_n≥ 0 and (v_n)_n≥ 0 be the Lucas sequences defined by u_0=0,quad u_1=1,quad u_n+1=11u_n-u_n-1quad(n=1,2,3,…), and v_0=2,quad v_1=11,quad v_n+1=11v_n-v_n-1quad(n=1,2,3,…). Congruences for Domb numbers and Lucas sequences. For every odd prime p: (i) if one of the Legendre symbols (fracp3) and (fracp13) is 1, then ∑_k=0^p-1D(k)u_k equiv 0 pmodp^2. Moreover, if (fracp3)=(fracp13)=1, then ∑_k=0^p-1D(k)u_k equiv 0 pmodp^3. (ii) If the Jacobi symbol (fracp39) is -1, then ∑_k=0^p-1D(k)v_k equiv 0 pmodp^2. These are further conjectural supercongruences relating Domb numbers to Lucas sequences; the supplied text gives no evidence of a proof or disproof.

Congruences for Domb numbers and Lucas sequences

From papers

Let $D(k)$ denote the Domb numbers, and let $(u_n)_{n\geq 0}$ and $(v_n)_{n\geq 0}$ be the Lucas sequences defined by

u_0=0,\quad u_1=1,\quad u_{n+1}=11u_n-u_{n-1}\quad(n=1,2,3,\ldots),

and

v_0=2,\quad v_1=11,\quad v_{n+1}=11v_n-v_{n-1}\quad(n=1,2,3,\ldots).

Congruences for Domb numbers and Lucas sequences. For every odd prime $p$ : (i) if one of the Legendre symbols $(\frac{p}{3})$ and $(\frac{p}{13})$ is $1$ , then

\sum_{k=0}^{p-1}D(k)u_k\equiv 0\pmod{p^2}.

Moreover, if $(\frac{p}{3})=(\frac{p}{13})=1$ , then

\sum_{k=0}^{p-1}D(k)u_k\equiv 0\pmod{p^3}.

(ii) If the Jacobi symbol $(\frac{p}{39})$ is $-1$ , then

\sum_{k=0}^{p-1}D(k)v_k\equiv 0\pmod{p^2}.

These are further conjectural supercongruences relating Domb numbers to Lucas sequences; the supplied text gives no evidence of a proof or disproof.

Progress summary

Open

The conjecture is recorded in a recent paper, but no proof, disproof, or independent verification has been found.

The problem asserts three divisibility properties linking Domb numbers with two Lucas sequences. A related conjecture was reportedly made by the author in 2019, but the supplied sources identify no established resolution.

July 2026 arXiv restatement

An arXiv paper records the exact claims as Conjecture 3.4, including the stronger modulus $p^3$ case. It labels them conjectures rather than theorems and supplies no proof, disproof, or verification; the other retrieved papers concern different congruences.

Current status (as of August 2026): The three congruences remain conjectural, with no publicly verified proof or counterexample found in the retrieved sources.

Sources

arxiv.org

Sources & referencesView supporting material

Primary source

Zhi-Wei Sun, “A new kind of numbers and related congruences”, arXiv:2607.07638 (2026).

1 week ago

Assertion (i), as stated for every odd prime, is false at $p=3$ .

Indeed,

\left(\frac{3}{3}\right)=0, \qquad \left(\frac{3}{13}\right)=1,

since $4^2\equiv3\pmod{13}$ . Thus the hypothesis that at least one of these two Legendre symbols equals $1$ is satisfied.

The first three Domb numbers are

D(0)=1,\qquad D(1)=4,\qquad D(2)=28.

For the specified Lucas sequence,

u_0=0,\qquad u_1=1,\qquad u_2=11u_1-u_0=11.

Consequently

\sum_{k=0}^{p-1}D(k)u_k =D(0)u_0+D(1)u_1+D(2)u_2 =0+4+28\cdot11 =312 \equiv6\not\equiv0\pmod9.

Hence the proposed congruence modulo $p^2$ fails for the odd prime $p=3$ , disproving the universal statement. Restricting the first assertion to $p>3$ would exclude this counterexample; no conclusion about that modified assertion or the other proposed congruences follows here.

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Shivam Patel · 1 day ago