Let \( a_n \) be a sequence defined recursively by \( a_0 = a_1 = \dots = a_5 = 1 \) and
\[
a_n = \frac{a_{n-1} a_{n-5} + a_{n-2} a_{n-4} + a_{n-3}^2}{a_{n-6}}
\]
for \( n \geq 6 \). Prove or disprove that every \( a_n \) is an integer.
A real sequence \( x_1, x_2, x_3, \cdots \) satisfies the relation \( x_{n+2} = x_{n+1} + x_n \) for \( n = 1, 2, 3, \cdots \). If a number \( r \) satisfies \( x_i = x_j = r \) for some \( i \) and \( j \) \( (i \neq j) \), we say that \( r \) is a repeated number in this sequence. Prove that there can be more than \( 2013 \) repeated numbers in such a sequence, but it is impossible to have infinitely many repeated numbers.
Let \( N > 1000 \) be an integer. Define a sequence \( A_n \) by
\[
A_0 = 1, \, A_1 = 0, \, A_{2k+1} = \frac{2k}{2k+1} A_{2k} + \frac{1}{2k+1} A_{2k-1}, \, A_{2k} = \frac{2k-1}{2k} \frac{A_{2k-1}}{N} + \frac{1}{2k} A_{2k-2}.
\]
Show that the following inequality holds for any integer \( k \) with \( 1 \leq k \leq (1/2) N^{1/3} \).
\[
A_{2k-2} \leq \frac{1}{\sqrt{(2k-2)!}}.
\]
Let \(a_0=3\) and \(a_{n}=a_{n-1}+\sqrt{a_{n-1}^2+3}\) for all \(n\ge 1\). Determine \[\lim_{n\to\infty}\frac{a_n}{2^n}.\]
Evaluate the sum \[ \sum_{k=0}^{[n/2]} (-4)^{n-k} \binom{n-k}{k} ,\] where [x] denotes the greatest integer less than or equal to x.
Let \(a_1\le a_2\le \cdots \le a_k\) and \(b_1\le b_2\le \cdots \le b_l\) be sequences of positive integers at most M. Prove that if \[ \sum_{i=1}^{k} a_i^n = \sum_{j=1}^l b_j^n\] for all \(1\le n\le M\), then \(k=l\) and \(a_i=b_i\) for all \(1\le i\le k\).
Let \(a_0=a\) and \(a_{n+1}=a_n (a_n^2-3)\). Find all real values \(a\) such that the sequence \(\{a_n\}\) converges.
Let \(a_1<\cdots\) be a sequence of positive integers such that \(\log a_1, \log a_2,\log a_3,\cdots\) are linearly independent over the rational field \(\mathbb Q\). Prove that \(\lim_{k\to \infty} a_k/k=\infty\).
Let \(a_1=\sqrt{1+2}\),
\(a_2=\sqrt{1+2\sqrt{1+3}}\),
\(a_3=\sqrt{1+2\sqrt{1+3\sqrt{1+4}}}\), …,
\(a_n=\sqrt{1+2\sqrt{1+3\sqrt {\cdots \sqrt{\sqrt{\sqrt{\cdots\sqrt{1+n\sqrt{1+(n+1)}}}}}}}}\), … .
Prove that \(\displaystyle\lim_{n\to \infty} \frac{a_{n+1}-a_{n}}{a_n-a_{n-1}}=\frac12\).