Let \( \mathbb{F}_2^n \) be the finite field of cardinality \( 2^n \). For all large \( n \), any subset \( A \subset \mathbb{F}_2^n \times \mathbb{F}_2^n \) of cardinality
\[
|A| \gtrsim \frac{4^n \log \log n}{\log n},
\]
must contain three points \( \{(x, y), (x + d, y), (x, y + d)\} \) for \( x, y, d \in \mathbb{F}_2^n \) and \( d \neq 0 \). Our argument is an elaboration of an argument of Shkredov [14], building upon the finite field analog of Ben Green [10]. The interest in our result is in the exponent on \( \log n \), which is larger than has been obtained previously.

We present analytical properties of a sequence of integers related to the evaluation of a rational integral. We also discuss an algorithm for the evaluation of the 2-adic valuation of these integers that has a combinatorial interpretation.

It is proposed that finding the recursion relation and generating function for the (colored) Motzkin numbers of higher rank introduced recently is an interesting problem.

Let \( \mathbb{F}_2^n \) be the finite field of cardinality \( 2^n \). For all large \( n \), any subset \( A \subset \mathbb{F}_2^n \times \mathbb{F}_2^n \) of cardinality \[|A| \gtrsim \frac{4^n \log \log n}{\log n}, \] must contain three points \( \{(x, y), (x + d, y), (x, y + d)\} \) for \( x, y, d \in \mathbb{F}_2^n \) and \( d \neq 0 \). Our argument is an elaboration of an argument of Shkredov [14], building upon the finite field analog of Ben Green [10]. The interest in our result is in the exponent on \( \log n \), which is larger than has been obtained previously.

Let \( S \) be a finite set of positive integers with largest element \( m \). Let us randomly select a composition \( a \) of the integer \( n \) with parts in \( S \), and let \( m(a) \) be the multiplicity of \( m \) as a part of \( a \). Let \( 0 \leq r < q \) be integers, with \( q \geq 2 \), and let \( p_{n,r} \) be the probability that \( m(a) \) is congruent to \( r \mod q \). We show that if \( S \) satisfies a certain simple condition, then \( \lim_{n \to \infty} p_{n,r} = 1/q \). In fact, we show that an obvious necessary condition on \( S \) turns out to be sufficient.