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Congruences concerning Legendre polynomials II | Zhi-Hong Sun
; | Date: |
17 Dec 2010 | Abstract: | Let $p>3$ be a prime, and let $m$ be an integer with $p
mid m$. In the paper
we solve some conjectures of Z.W. Sun concerning $sum_{k=0}^{p-1} (1/{m^k})
inom{2k}k^3$ modulo ${p^2}$. In particular, we show that
$sum_{k=0}^{{p-1/}2}inom{2k}k^3e 0mod {p^2}$ for $pe 3,5,6mod 7$. Let
$P_n(x)$ be the Legendre polynomials. In the paper we also show that $P_{[p/4]}
(t)e -ig({-6}/pig)sum_{x=0}^{p-1}ig({x^3-{3(3t+5)}2x-9t-7}/pig)mod
p$ and determine $P_{{p-1}/2}(sqrt 2), P_{{p-1/}2}({3sqrt 2/}4),
P_{{p-1}/2}(sqrt {-3}),P_{{p-1}/2}({sqrt 3}/2), P_{{p-1}/2}(sqrt {-63}),
P_{{p-1}/2}({3sqrt 7}/8)mod p$, where $t$ is a rational $p-$integer, $[x]$ is
the greatest integer not exceeding $x$ and $(a/p)$ is the Legendre symbol. As
consequences we determine $P_{[p/4]}(t)mod p$ in the cases $t=-5/3,-7/9,
65/63$ and confirm several conjectures of Z.W. Sun on
$sum_{k=0}^{p-1}inom{4k}{2k}inom{2k}km^{-k}mod p$ in the cases
$m=48,63,72,128$. We also determine $sum_{k=0}^{[p/4]}{(4k)!}/{m^kcdot
k!^4}mod p$ for $m=-144,-1024,-3969,648$. | Source: | arXiv, 1012.3898 | Services: | Forum | Review | PDF | Favorites |
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