6
votes
1answer
272 views
A sum-product estimate in Z/p^2Z
We are interested in a sum-product type estimate. Let $p$ be an odd prime, and let $A$ be the order $p-1$ subgroup of $(\mathbb{Z}/p^2\mathbb{Z})^\times$. That is, let $A = \langle …
1
vote
3answers
228 views
how to proof this Stirling related equation
here is what I need to proof, have no idea were to start. I know there is some connection with the Stirling theorem.
$$
\sum_{i=0}^{d}\binom{m}{i} \leq \left ( \frac{em}{d} \right …
1
vote
1answer
126 views
unique sums in a finite direct product of sets of integers
I am an algebraist, and I am wondering if there is a definition for the following:
Let $A_1$, $A_2$, $\ldots, A_n$ be sets of integers (or more generally, subsets of a group $G$). …
3
votes
1answer
222 views
Additive set with small sum set and large difference set
I have a question!
Can someone explain how (the intuition, method?) one can try to construct an additive set of cardinality $N$ with a small sum set (around $N$) and a very large …
4
votes
0answers
391 views
Cliques in the Paley graph and a problem of Sarkozy
The following question is motivated by pure curiosity; it is not a part
of any research project and I do not have any applications. The question
comes as an interpolation between t …
12
votes
1answer
389 views
The hypercube: $|A {\stackrel2+} E| \ge |A|$?
I have a good motivation to ask the question below, but since the post is
already a little long, and the problem looks rather natural and appealing
(well, to me, at least), I'd rat …
1
vote
1answer
138 views
Find an approximate expression of a sum of a product using the average of each item
Is it possible to find an approximate expression of $\frac{\sum_{i=1}^{n} k_i x_i}{\sum_{i=1}^{n} k_i}$ using $\langle k \rangle$, $\langle k^2 \rangle$, $\langle x \rangle$, and $ …
4
votes
1answer
365 views
Intersecting Hamming spheres: is $|A\stackrel k+E|\ge|A|$?
Since my original posting some ten days ago, I discovered an amazing
example which changed significantly my perception of the problem.
Accordingly, the whole post got re-written no …

