Is this sequence always periodic?












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$begingroup$


The sequence $a_1,a_2,dots$ contains only positive reals. Suppose that for any $i$ positive integer
$$n-(a_i+a_{i+1})+dfrac{a_ia_{i+1}-a_{i+2}+1}{n}=1$$
where $n$ is a fixed positive real.



Is it always true that the sequence $a_i$ is periodic? This problem is from Rookie Team Contest MOP 2001. I need your help!






linear-algebra sequences-and-series







share|cite|improve this question









$endgroup$












  • $begingroup$
    Try $n=1$, so then $a_{i+2} = 1-(a_i-a_{i+1}) + a_ia_{i+1}$ and it looks like if initialized properly the $a_i$ sequence will always increase.
    $endgroup$
    – Michael
    Oct 22 '18 at 22:16










  • $begingroup$
    As Michael commented, for $n=1$, we have $a_{i+2}=(a_{i+1}-1)(a_i-1)$. So, we see that $a_i$ is increasing for $a_1=a_2=3$, for example.
    $endgroup$
    – mathlove
    Oct 23 '18 at 5:43










  • $begingroup$
    But what if $nneq 1$. I mean in the problem we don’t know $n$, we only know that it is a fixed positive real and satisfy the equation!
    $endgroup$
    – Yeah
    Oct 23 '18 at 7:26
















0












$begingroup$


The sequence $a_1,a_2,dots$ contains only positive reals. Suppose that for any $i$ positive integer
$$n-(a_i+a_{i+1})+dfrac{a_ia_{i+1}-a_{i+2}+1}{n}=1$$
where $n$ is a fixed positive real.



Is it always true that the sequence $a_i$ is periodic? This problem is from Rookie Team Contest MOP 2001. I need your help!






linear-algebra sequences-and-series







share|cite|improve this question









$endgroup$












  • $begingroup$
    Try $n=1$, so then $a_{i+2} = 1-(a_i-a_{i+1}) + a_ia_{i+1}$ and it looks like if initialized properly the $a_i$ sequence will always increase.
    $endgroup$
    – Michael
    Oct 22 '18 at 22:16










  • $begingroup$
    As Michael commented, for $n=1$, we have $a_{i+2}=(a_{i+1}-1)(a_i-1)$. So, we see that $a_i$ is increasing for $a_1=a_2=3$, for example.
    $endgroup$
    – mathlove
    Oct 23 '18 at 5:43










  • $begingroup$
    But what if $nneq 1$. I mean in the problem we don’t know $n$, we only know that it is a fixed positive real and satisfy the equation!
    $endgroup$
    – Yeah
    Oct 23 '18 at 7:26














0












0








0





$begingroup$


The sequence $a_1,a_2,dots$ contains only positive reals. Suppose that for any $i$ positive integer
$$n-(a_i+a_{i+1})+dfrac{a_ia_{i+1}-a_{i+2}+1}{n}=1$$
where $n$ is a fixed positive real.



Is it always true that the sequence $a_i$ is periodic? This problem is from Rookie Team Contest MOP 2001. I need your help!






linear-algebra sequences-and-series







share|cite|improve this question









$endgroup$




The sequence $a_1,a_2,dots$ contains only positive reals. Suppose that for any $i$ positive integer
$$n-(a_i+a_{i+1})+dfrac{a_ia_{i+1}-a_{i+2}+1}{n}=1$$
where $n$ is a fixed positive real.



Is it always true that the sequence $a_i$ is periodic? This problem is from Rookie Team Contest MOP 2001. I need your help!






linear-algebra sequences-and-series




linear-algebra sequences-and-series






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Oct 22 '18 at 21:39









YeahYeah

826




826












  • $begingroup$
    Try $n=1$, so then $a_{i+2} = 1-(a_i-a_{i+1}) + a_ia_{i+1}$ and it looks like if initialized properly the $a_i$ sequence will always increase.
    $endgroup$
    – Michael
    Oct 22 '18 at 22:16










  • $begingroup$
    As Michael commented, for $n=1$, we have $a_{i+2}=(a_{i+1}-1)(a_i-1)$. So, we see that $a_i$ is increasing for $a_1=a_2=3$, for example.
    $endgroup$
    – mathlove
    Oct 23 '18 at 5:43










  • $begingroup$
    But what if $nneq 1$. I mean in the problem we don’t know $n$, we only know that it is a fixed positive real and satisfy the equation!
    $endgroup$
    – Yeah
    Oct 23 '18 at 7:26


















  • $begingroup$
    Try $n=1$, so then $a_{i+2} = 1-(a_i-a_{i+1}) + a_ia_{i+1}$ and it looks like if initialized properly the $a_i$ sequence will always increase.
    $endgroup$
    – Michael
    Oct 22 '18 at 22:16










  • $begingroup$
    As Michael commented, for $n=1$, we have $a_{i+2}=(a_{i+1}-1)(a_i-1)$. So, we see that $a_i$ is increasing for $a_1=a_2=3$, for example.
    $endgroup$
    – mathlove
    Oct 23 '18 at 5:43










  • $begingroup$
    But what if $nneq 1$. I mean in the problem we don’t know $n$, we only know that it is a fixed positive real and satisfy the equation!
    $endgroup$
    – Yeah
    Oct 23 '18 at 7:26
















$begingroup$
Try $n=1$, so then $a_{i+2} = 1-(a_i-a_{i+1}) + a_ia_{i+1}$ and it looks like if initialized properly the $a_i$ sequence will always increase.
$endgroup$
– Michael
Oct 22 '18 at 22:16




$begingroup$
Try $n=1$, so then $a_{i+2} = 1-(a_i-a_{i+1}) + a_ia_{i+1}$ and it looks like if initialized properly the $a_i$ sequence will always increase.
$endgroup$
– Michael
Oct 22 '18 at 22:16












$begingroup$
As Michael commented, for $n=1$, we have $a_{i+2}=(a_{i+1}-1)(a_i-1)$. So, we see that $a_i$ is increasing for $a_1=a_2=3$, for example.
$endgroup$
– mathlove
Oct 23 '18 at 5:43




$begingroup$
As Michael commented, for $n=1$, we have $a_{i+2}=(a_{i+1}-1)(a_i-1)$. So, we see that $a_i$ is increasing for $a_1=a_2=3$, for example.
$endgroup$
– mathlove
Oct 23 '18 at 5:43












$begingroup$
But what if $nneq 1$. I mean in the problem we don’t know $n$, we only know that it is a fixed positive real and satisfy the equation!
$endgroup$
– Yeah
Oct 23 '18 at 7:26




$begingroup$
But what if $nneq 1$. I mean in the problem we don’t know $n$, we only know that it is a fixed positive real and satisfy the equation!
$endgroup$
– Yeah
Oct 23 '18 at 7:26










1 Answer
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$begingroup$

By multiplying both sides in $n$ we have $$n^2-n(a_i+a_{i+1})+a_ia_{i+1}-a_{i+2}+1=n$$therefore $$a_{i+2}=(a_i-n)(a_{i+1}-n)-n+1$$defining $b_i=a_i-n$ we have $$b_{i+2}=b_ib_{i+1}-2n+1$$therefore $b_i$ is periodic if and only if $a_i$ is periodic. Now choose $n={1over 2}$ therefore $$b_{i+2}=b_{i+1}b_i$$ where for $b_1=2$ and $b_2=2$ we have$${b_3=2^2\b_4=2^3\b_5=2^5\b_6=2^8\b_7=2^{11}\vdots}$$therefore $$Large b_i=2^{f_i}$$ where $f_i$ is the famous Fibonacci sequence. Since $f_i$ grows unbounded, so do $b_i$ and $a_i$ which implies on non-periodicity of $a_i$.






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    $begingroup$

    By multiplying both sides in $n$ we have $$n^2-n(a_i+a_{i+1})+a_ia_{i+1}-a_{i+2}+1=n$$therefore $$a_{i+2}=(a_i-n)(a_{i+1}-n)-n+1$$defining $b_i=a_i-n$ we have $$b_{i+2}=b_ib_{i+1}-2n+1$$therefore $b_i$ is periodic if and only if $a_i$ is periodic. Now choose $n={1over 2}$ therefore $$b_{i+2}=b_{i+1}b_i$$ where for $b_1=2$ and $b_2=2$ we have$${b_3=2^2\b_4=2^3\b_5=2^5\b_6=2^8\b_7=2^{11}\vdots}$$therefore $$Large b_i=2^{f_i}$$ where $f_i$ is the famous Fibonacci sequence. Since $f_i$ grows unbounded, so do $b_i$ and $a_i$ which implies on non-periodicity of $a_i$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      By multiplying both sides in $n$ we have $$n^2-n(a_i+a_{i+1})+a_ia_{i+1}-a_{i+2}+1=n$$therefore $$a_{i+2}=(a_i-n)(a_{i+1}-n)-n+1$$defining $b_i=a_i-n$ we have $$b_{i+2}=b_ib_{i+1}-2n+1$$therefore $b_i$ is periodic if and only if $a_i$ is periodic. Now choose $n={1over 2}$ therefore $$b_{i+2}=b_{i+1}b_i$$ where for $b_1=2$ and $b_2=2$ we have$${b_3=2^2\b_4=2^3\b_5=2^5\b_6=2^8\b_7=2^{11}\vdots}$$therefore $$Large b_i=2^{f_i}$$ where $f_i$ is the famous Fibonacci sequence. Since $f_i$ grows unbounded, so do $b_i$ and $a_i$ which implies on non-periodicity of $a_i$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        By multiplying both sides in $n$ we have $$n^2-n(a_i+a_{i+1})+a_ia_{i+1}-a_{i+2}+1=n$$therefore $$a_{i+2}=(a_i-n)(a_{i+1}-n)-n+1$$defining $b_i=a_i-n$ we have $$b_{i+2}=b_ib_{i+1}-2n+1$$therefore $b_i$ is periodic if and only if $a_i$ is periodic. Now choose $n={1over 2}$ therefore $$b_{i+2}=b_{i+1}b_i$$ where for $b_1=2$ and $b_2=2$ we have$${b_3=2^2\b_4=2^3\b_5=2^5\b_6=2^8\b_7=2^{11}\vdots}$$therefore $$Large b_i=2^{f_i}$$ where $f_i$ is the famous Fibonacci sequence. Since $f_i$ grows unbounded, so do $b_i$ and $a_i$ which implies on non-periodicity of $a_i$.






        share|cite|improve this answer









        $endgroup$



        By multiplying both sides in $n$ we have $$n^2-n(a_i+a_{i+1})+a_ia_{i+1}-a_{i+2}+1=n$$therefore $$a_{i+2}=(a_i-n)(a_{i+1}-n)-n+1$$defining $b_i=a_i-n$ we have $$b_{i+2}=b_ib_{i+1}-2n+1$$therefore $b_i$ is periodic if and only if $a_i$ is periodic. Now choose $n={1over 2}$ therefore $$b_{i+2}=b_{i+1}b_i$$ where for $b_1=2$ and $b_2=2$ we have$${b_3=2^2\b_4=2^3\b_5=2^5\b_6=2^8\b_7=2^{11}\vdots}$$therefore $$Large b_i=2^{f_i}$$ where $f_i$ is the famous Fibonacci sequence. Since $f_i$ grows unbounded, so do $b_i$ and $a_i$ which implies on non-periodicity of $a_i$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 '18 at 17:27









        Mostafa AyazMostafa Ayaz

        14.8k3938




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