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Math Help - 2 interesting problems

  1. #1
    Super Member PaulRS's Avatar
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    2 interesting problems

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


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  2. #2
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    Quote Originally Posted by PaulRS View Post
    1. Let g(n) be the greatest odd divisor of n, show that <br />
\mathop {\lim }\limits_{n \to + \infty } \tfrac{1}<br />
{n} \cdot \sum\limits_{k = 1}^n {\tfrac{{g\left( k \right)}}<br />
{k}} <br />
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!
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by NonCommAlg View Post
    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?
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  4. #4
    Super Member PaulRS's Avatar
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    Quote Originally Posted by chiph588@ View Post
    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: <br />
a + 2^{t }  \equiv 0\left( {\bmod .2^{t+1} } \right) \Leftrightarrow a  \equiv 2^{t} \left( {\bmod .2^{t+1} } \right)<br />
(sum 2^{t} on both sides)

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

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

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

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


    My proof of (1) is based on the following observation: <br />
\tfrac{{g\left( n \right)}}<br />
{n} = 2 - \sum\limits_{\left. {2^s } \right|n{\text{ ; }}s \geqslant 0} {\tfrac{1}<br />
{{2^s }}} <br />
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}<br />
{{2^k }}} \right\rfloor  ): <br />
\sum\limits_{k = 1}^n {\tfrac{{g\left( k \right)}}<br />
{k}}  = 2 \cdot n - \sum\limits_{k = 0}^\infty  {\tfrac{1}<br />
{{2^k }} \cdot \left\lfloor {\tfrac{n}<br />
{{2^k }}} \right\rfloor } <br />
...
    Last edited by PaulRS; January 6th 2009 at 04:04 AM.
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Sorry to be a bother, but I'm still a bit confused...
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by PaulRS View Post
    1. Let g(n) be the greatest odd divisor of n, show that <br />
\mathop {\lim }\limits_{n \to  + \infty } \tfrac{1}<br />
{n} \cdot \sum\limits_{k = 1}^n {\tfrac{{g\left( k \right)}}<br />
{k}} <br />
exists and find it ( Bulgaria 1985)
    2. Find all <br />
k \in \mathbb{Z}\text{ with }k \geq 2<br />
such that <br />
n \not|g(k^n+1)<br />
, <br />
\forall n \in \mathbb{Z}/n > 1<br />
(Olimpíada Rioplatense 2008) - g(n) is defined as in 1 -


    Have fun!
    Any hints on 2?
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  7. #7
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    Quote Originally Posted by PaulRS View Post
    Find all <br />
 k \in \mathbb{Z}\text{ with }k \geq 2<br />
such that <br />
n \not|g(k^n+1)<br />
, <br />
\forall n \in \mathbb{Z}/n > 1<br />
(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

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

    or perhaps

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

    or does it mean to ask

    <br />
    n\not|g(k^n+1)<br />
  , <br />
  \forall n<br />

    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.
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  8. #8
    Super Member PaulRS's Avatar
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    The bar there ("/") means "such that". So what I mean is that that should hold for all integers n greater than 1.
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