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

Less than 1 year old · traced to 2026

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.

References

Primary source

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

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed progress

A reader-supplied calculation claims the statement fails at the prime three, but this has not been independently checked, while the source paper records the assertions only as conjectures.

The problem proposes three congruences linking Domb numbers with two Lucas sequences. Sun’s 2026 paper records the exact assertions as Conjecture 3.4, not as proved results.

July 2026 conjecture record

The paper gives no proof or disproof of these three Lucas-sequence congruences; its confirmed Domb-number result concerns a different sum.

Posted attempt

A reader-supplied calculation claims that part (i) fails at $p=3$ : $(\frac{3}{3})=0$ , $(\frac{3}{13})=1$ , and using $D(0)=1$ , $D(1)=4$ , $D(2)=28$ , $u_0=0$ , $u_1=1$ , $u_2=11$ gives $\sum_{k=0}^{2}D(k)u_k=312\equiv6\pmod 9$ . The attempt is not independently verified; it says nothing about $p>3$ or part (ii).

Current status (as of August 2026): The three congruences have no verified proof, while an unverified calculation claims to disprove part (i) at $p=3$ ; the cases with $p>3$ and part (ii) remain open.

Sources

arxiv.org

CounterexampleThis solution needs a summarySee full solutionHide full solution

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.