All Questions
5 questions
13
votes
1
answer
584
views
A congruence for a product of binomial coefficients?
For every prime $p\geq 5$ one seems to have the congruence
$$(-1)^{(p-1)/2}\prod_{k=0}^{p-1}{p-1\choose k}\equiv 1-p+\frac{3}{2}p^2-\frac{7}{6}p^3\pmod{p^4}\ .$$
(I have checked all primes up to $5000$...
12
votes
1
answer
838
views
Numbers $k$ with $\{\binom nk:\ n\in\mathbb N\}$ dense in $\mathbb Z_p$ for any prime $p\le k$
Let $k$ be a positive integer and let $p$ be a prime. In my 2011 PAMS paper joint with my former student W. Zhang [Proc. Amer. Math. Soc. 139(2011), 1569-1577], we studied when $$S(k)=\left\{\binom nk:...
7
votes
0
answers
213
views
Does Morley's congruence characterize primes greater than $3$?
In 1895 Morley showed that $$\binom{p-1}{(p-1)/2}\equiv(-1)^{\frac{p-1}2}4^{p-1}\pmod{p^3}$$
for any prime $p>3$.
In 2009, I formulated the following conjecture concerning the converse of Morley's ...
4
votes
3
answers
544
views
On a theorem of Hensel about congruence of binomial coefficient
In the paper Binomial coefficients modulo prime powers, Andrew Granville stated the following theorem:
Let $n, m$ and $r=n-m$ be three given positive integer and $p^k$ is the exact power of $p$ ...
22
votes
1
answer
1k
views
Does $\binom{2n}{n} \equiv 2 \pmod p$ ever hold?
Well, the title does not tell the whole story; the complete question is:
Are there any primes of the form $p=2n(n-1)+1$, with integer $n\ge 1$, such that
$$ \binom{2n}{n} \equiv 2\pmod p ? $$
...