The Domb numbers are given by $$D_n=\sum_{k=0}^n\binom{n}{k}^2\binom{2k}k\binom{2(n-k)}{n-k}\ \ \ (n=0,1,2,\ldots).$$ Such numbers have combinatorial interpretation, see, e.g., http://oeis.org/A002895.

I have the following three conjectures on congruences involving the Domb numbers.

Conjecture 1 (2019). For any odd prime $p$, we have $$\sum_{k=1}^{p-1}\frac{D_k}k\equiv\left(\frac p3\right)\frac 25pB_{p-2}\left(\frac13\right)\pmod{p^2},$$ where $(-)$ is the Legendre symbol and $B_{p-2}(x)$ is the Bernoulli polynomial of degree $p-2$.

Conjecture 2 (2020). For any prime $p>5$, we have $$\sum_{k=1}^{p-1}\frac1k\left(D_k-\frac{4D_{k-1}}{64^{k-1}}\right)\equiv-\frac{16}3p^2B_{p-3}\pmod{p^3},$$ where $B_0,B_1,\ldots$ are the Bernoulli numbers.

Conjecture 3 (2013). For any prime $p>3$, we have $$\det[D_{i+j}]_{0\le i,j\le p-1}\equiv\begin{cases}(\frac{-1}p)(4x^2-2p)\pmod{p^2}&\mbox{if}\ p=x^2+3y^2\ (x,y\in\mathbb Z),\\0\pmod{p^2}&\mbox{if}\ p\equiv 2\pmod 3.\end{cases}$$

Conjectures 1 and 3 appeared in Conjecture 79 of my published paper Open Conjectures on Congruences. I have not made Conjecture 2 public before, I can prove the congruence in Conjecture 2 modulo $p$.

QUESTION. Any ideas towards solving these conjectures?

  • $\begingroup$ how could you prove the conjecture 2 modulo $p$? $\endgroup$ – vidyarthi Aug 8 at 10:39
  • $\begingroup$ Because $D_k\equiv 64^kD_{p-1-k}\pmod p.$ $\endgroup$ – Zhi-Wei Sun Aug 10 at 12:33

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