2 interesting problems

**PaulRS** · January 5th 2009, 02:53 PM

Let $g(n)$ be the greatest odd divisor of $n$ , show that $ \mathop {\lim }\limits_{n \to + \infty } \tfrac{1} {n} \cdot \sum\limits_{k = 1}^n {\tfrac{{g\left( k \right)}} {k}} $ exists and find it ( Bulgaria 1985)
Find all $ k \in \mathbb{Z}\text{ with }k \geq 2 $ such that $ n \not|g(k^n+1) $ , $ \forall n \in \mathbb{Z}/n > 1 $ (Olimpíada Rioplatense 2008) - $g(n)$ is defined as in 1 -

Have fun!

**NonCommAlg** · January 5th 2009, 04:39 PM

Originally Posted by PaulRS

Let $g(n)$ be the greatest odd divisor of $n$ , show that $ \mathop {\lim }\limits_{n \to + \infty } \tfrac{1} {n} \cdot \sum\limits_{k = 1}^n {\tfrac{{g\left( k \right)}} {k}} $ exists and find it ( Bulgaria 1985)

a simple observation shows that $\sum_{k=1}^n \frac{g(k)}{k} = \sum_{k=1}^m \frac{1}{2^{k-1}} \left \lfloor \frac{n+2^{k-1}}{2^k} \right \rfloor,$ where $m=\lfloor \log_2 n \rfloor + 1.$ the value of $m$ is not important. what we need is to see that $m \rightarrow \infty$ as $n \rightarrow \infty.$

now since $x-1 < \lfloor x \rfloor \leq x,$ the squeeze theorem gives us: $\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^n \frac{g(k)}{k}=\frac{2}{3}. \ \ \Box$

the second problem is left for other members to try!

**chiph588@** · January 5th 2009, 06:56 PM

Originally Posted by NonCommAlg

a simple observation shows that $\sum_{k=1}^n \frac{g(k)}{k} = \sum_{k=1}^m \frac{1}{2^{k-1}} \left \lfloor \frac{n+2^{k-1}}{2^k} \right \rfloor,$ where $m=\lfloor \log_2 n \rfloor + 1.$ the value of $m$ is not important. what we need is to see that $m \rightarrow \infty$ as $n \rightarrow \infty.$

now since $x-1 < \lfloor x \rfloor \leq x,$ the squeeze theorem gives us: $\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^n \frac{g(k)}{k}=\frac{2}{3}. \ \ \Box$

the second problem is left for other members to try!

How did you obtain $\sum_{k=1}^n \frac{g(k)}{k} = \sum_{k=1}^m \frac{1}{2^{k-1}} \left \lfloor \frac{n+2^{k-1}}{2^k} \right \rfloor,$ where $m=\lfloor \log_2 n \rfloor + 1$ ?

**PaulRS** · January 6th 2009, 01:34 AM

Originally Posted by chiph588@

How did you obtain $\sum_{k=1}^n \frac{g(k)}{k} = \sum_{k=1}^m \frac{1}{2^{k-1}} \left \lfloor \frac{n+2^{k-1}}{2^k} \right \rfloor,$ where $m=\lfloor \log_2 n \rfloor + 1$ ?

First note that: $ a + 2^{t } \equiv 0\left( {\bmod .2^{t+1} } \right) \Leftrightarrow a \equiv 2^{t} \left( {\bmod .2^{t+1} } \right) $ (sum $2^{t}$ on both sides)

The idea is that $ a = 2^t \cdot s{\text{ with }}s \ne \dot 2 $ iff $a\equiv{2^{t}}(\bmod.2^{t+1})$

Indeed, to prove this:
$ a = 2^t \cdot \left( {2l + 1} \right) \equiv 2^t \left( {\bmod .2^{t + 1} } \right) $ , conversly if $ a \equiv 2^t \left( {\bmod .2^{t + 1} } \right) $ then $ \left. {2^t } \right|{a } $ and $2 ^{t+1}\not|a$

Now since $ \left\lfloor {\tfrac{n} {a}} \right\rfloor - \left\lfloor {\tfrac{{n - 1}} {a}} \right\rfloor = \left\{ \begin{gathered} 1{\text{ if }}\left. a \right|n \hfill \\ 0{\text{ otherwise}} \hfill \\ \end{gathered} \right. $

It follows easily that: $ \sum\limits_{k = 1}^\infty {\tfrac{1} {{2^{k - 1} }} \cdot \left( {\left\lfloor {\tfrac{{n + 1 + 2^{k - 1} }} {{2^k }}} \right\rfloor - \left\lfloor {\tfrac{{n + 2^{k - 1} }} {{2^k }}} \right\rfloor } \right)} = \tfrac{{g\left( {n + 1} \right)}} {{n + 1}} $ (only one of the terms is not 0 )

My proof of (1) is based on the following observation: $ \tfrac{{g\left( n \right)}} {n} = 2 - \sum\limits_{\left. {2^s } \right|n{\text{ ; }}s \geqslant 0} {\tfrac{1} {{2^s }}} $ then by a simple counting argument ( $2^j$ appears as many times as multiples of $2^j$ there are in between 1 and n -both included-, that is $\left\lfloor {\tfrac{n} {{2^k }}} \right\rfloor$ ): $ \sum\limits_{k = 1}^n {\tfrac{{g\left( k \right)}} {k}} = 2 \cdot n - \sum\limits_{k = 0}^\infty {\tfrac{1} {{2^k }} \cdot \left\lfloor {\tfrac{n} {{2^k }}} \right\rfloor } $ ...

**chiph588@** · January 7th 2009, 06:13 PM

Sorry to be a bother, but I'm still a bit confused...

**chiph588@** · May 30th 2010, 05:21 PM

Originally Posted by PaulRS

Let $g(n)$ be the greatest odd divisor of $n$ , show that $ \mathop {\lim }\limits_{n \to + \infty } \tfrac{1} {n} \cdot \sum\limits_{k = 1}^n {\tfrac{{g\left( k \right)}} {k}} $ exists and find it ( Bulgaria 1985)
Find all $ k \in \mathbb{Z}\text{ with }k \geq 2 $ such that $ n \not|g(k^n+1) $ , $ \forall n \in \mathbb{Z}/n > 1 $ (Olimpíada Rioplatense 2008) - $g(n)$ is defined as in 1 -

Have fun!

Any hints on 2?

**gmatt** · June 1st 2010, 09:01 AM

Originally Posted by PaulRS

Find all $ k \in \mathbb{Z}\text{ with }k \geq 2 $ such that $ n \not|g(k^n+1) $ , $ \forall n \in \mathbb{Z}/n > 1 $ (Olimpíada Rioplatense 2008) - $g(n)$ is defined as in 1 -

Correct me if I am wrong but the question is not clear as stated. Does the question mean to ask

$ r \not|g(k^r+1) $ , $ \forall r \in \mathbb{Z}/n > 1 $

or perhaps

$ r \not|g(k^n+1) $ , $ \forall r \in \mathbb{Z}/n > 1 $

or does it mean to ask

$ n\not|g(k^n+1) $ , $ \forall n $

the way that for all is used currently doesn't make sense to me, since the quantified variable $n$ is used to define the domain set $\mathbb{Z}/n$ .

**PaulRS** · November 13th 2010, 01:33 PM

The bar there ("/") means "such that". So what I mean is that that should hold for all integers n greater than 1.

Thread: 2 interesting problems

LinkBack

Thread Tools

Display

2 interesting problems

Similar Math Help Forum Discussions

Looks interesting

An interesting problem on C=A+B

Interesting one!

Interesting X

Interesting Problem

Search Tags