Some 3-adic congruences for binomial sums (Q2254818)

Let \(\nu_3(n)\) denote the largest integer \(k\), such \(3^k\) divides \(n\). The paper verifies the following conjectures of \textit{Z.-W. Sun} [Acta Arith. 148, No. 1, 63--76 (2011; Zbl 1305.11014)]: for an integer \(m=3\ell +1\), (i) for every positive integer \(n\), \[ \nu_3\Bigl(\frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{m^k}\binom{2k}{k}\Bigl)\geq \min \{\nu_3(n), \nu_3(m-1)-1\}; \] (i') for every positive integer \(a\geq \nu_3(m-1)\), \[ \frac{1}{3^a} \sum_{k=0}^{3^a-1} \frac{1}{m^k}\binom{2k}{k} \equiv \frac{m-1}{3} \bmod 3^{\nu_3(m-1)}; \] (ii) for every positive integer \(n\), \[ \nu_3\Bigl(\frac{1}{n}\sum_{k=0}^{n-1} \binom{n-1}{k} \binom{2k}{k}\Bigl) \geq \min \{\nu_3(n), \nu_3(m-1)\}-1; \] (ii') for every positive integer \(a\geq \nu_3(m-1)\), \[ \frac{1}{3^a} \sum_{k=0}^{3^a-1} \frac{(-1)^k}{m^k}\binom{3^a-1}{k}\binom{2k}{k} \equiv -\frac{m-1}{3} \bmod 3^{\nu_3(m-1)}; \] (iii) for every integer \(a\geq 2\), \[ \frac{1}{3^a} \sum_{k=0}^{3^a-1} (-1)^k \binom{3^a-1}{k}\binom{2k}{k} \equiv -3^{a-1} \bmod 3^a. \]