If $a_1=a_2=a_3=1$ then if $n>3$ show that $2^gcd(a_n-2,a_n-3)+a_n-1=2n-5.$

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Question: I write $a_1=a_2=a_3=1$ and set $a_n=2^gcd(a_n-2,a_n-3)+a_n-1.$ For $n>3$ is it true and how
can I show that $a_n=2n-5$ ? In particular I would like to show that that the sequence $a_n_n=1^infty$ is the sequence $1,1,1,3,5,7,9,11,13,15,ldots$




For example if $n=7$ then $a_7=2^gcd(a_5,a_4)+a_6=2^gcd(3,5)+2=2^1+7=9.$ And $2*7-5=14-5=9.$




sequences-and-series algebra-precalculus elementary-number-theory




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    Seems like the critical point is just that $gcd(a_n,a_n-1)=1$ for all $n$. Good place to start.
    – lulu
    2 days ago














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Question: I write $a_1=a_2=a_3=1$ and set $a_n=2^gcd(a_n-2,a_n-3)+a_n-1.$ For $n>3$ is it true and how
can I show that $a_n=2n-5$ ? In particular I would like to show that that the sequence $a_n_n=1^infty$ is the sequence $1,1,1,3,5,7,9,11,13,15,ldots$




For example if $n=7$ then $a_7=2^gcd(a_5,a_4)+a_6=2^gcd(3,5)+2=2^1+7=9.$ And $2*7-5=14-5=9.$




sequences-and-series algebra-precalculus elementary-number-theory




share|cite|improve this question















  • 2




    Seems like the critical point is just that $gcd(a_n,a_n-1)=1$ for all $n$. Good place to start.
    – lulu
    2 days ago












up vote
0
down vote

favorite









up vote
0
down vote

favorite












Question: I write $a_1=a_2=a_3=1$ and set $a_n=2^gcd(a_n-2,a_n-3)+a_n-1.$ For $n>3$ is it true and how
can I show that $a_n=2n-5$ ? In particular I would like to show that that the sequence $a_n_n=1^infty$ is the sequence $1,1,1,3,5,7,9,11,13,15,ldots$




For example if $n=7$ then $a_7=2^gcd(a_5,a_4)+a_6=2^gcd(3,5)+2=2^1+7=9.$ And $2*7-5=14-5=9.$




sequences-and-series algebra-precalculus elementary-number-theory




share|cite|improve this question












Question: I write $a_1=a_2=a_3=1$ and set $a_n=2^gcd(a_n-2,a_n-3)+a_n-1.$ For $n>3$ is it true and how
can I show that $a_n=2n-5$ ? In particular I would like to show that that the sequence $a_n_n=1^infty$ is the sequence $1,1,1,3,5,7,9,11,13,15,ldots$




For example if $n=7$ then $a_7=2^gcd(a_5,a_4)+a_6=2^gcd(3,5)+2=2^1+7=9.$ And $2*7-5=14-5=9.$





sequences-and-series algebra-precalculus elementary-number-theory





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asked 2 days ago









Anthony Hernandez

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  • 2




    Seems like the critical point is just that $gcd(a_n,a_n-1)=1$ for all $n$. Good place to start.
    – lulu
    2 days ago












  • 2




    Seems like the critical point is just that $gcd(a_n,a_n-1)=1$ for all $n$. Good place to start.
    – lulu
    2 days ago







2




2




Seems like the critical point is just that $gcd(a_n,a_n-1)=1$ for all $n$. Good place to start.
– lulu
2 days ago




Seems like the critical point is just that $gcd(a_n,a_n-1)=1$ for all $n$. Good place to start.
– lulu
2 days ago










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@lulu per you comment (sort of) I have the following as a possible solution:




By definition $2^gcd(a_n-2,a_n-3)+a_n-1=a_n$ and so $2^gcd(a_n-2,a_n-3)=a_n-a_n-1 $ for every $n$. I can take "logs" on the LHS and RHS of the equality to get $gcd(a_n-2,a_n-3)log2=log(a_n-a_n-1)$ consequently $$gcd(a_n-2,a_n-3)=log(a_n-a_n-1)above 1.5pt log2$$ If $log(a_n-a_n-1)above 1.5pt log2=2x$ for some $xin mathbbN$ then $log(a_n-a_n-1)=2xlog2$ and so $a_n-a_n-1=4^x$ which is a contradiction. So I must have $log(a_n-a_n-1)above 1.5pt log2=2x+1$ in which case $a_n-a_n-1=2^2x+1$ which is a contradiction for $x>1$. So $x=0$ and $log(a_n-a_n-1)above 1.5pt log2=1.$ By equality $gcd(a_n-2,a_n-3)=1.$ So I have shown that $a_n=2+a_n-1.$ I know that $a_4=3$ and so I have shown that $a_n_n=1^infty$ is the sequence $1,1,1,3,5,7,ldots$ Now by recursion I have that



beginalign
a_n&=2+a_n-1\
&=2+2+a_n-2\
&=2+2+2+a_n-3\
&=underbrace2+2+2+ldots+2_text$n-4$ times+1+1+1\
&=2(n-4)+3\
&=2n-5
endalign







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    @lulu per you comment (sort of) I have the following as a possible solution:




    By definition $2^gcd(a_n-2,a_n-3)+a_n-1=a_n$ and so $2^gcd(a_n-2,a_n-3)=a_n-a_n-1 $ for every $n$. I can take "logs" on the LHS and RHS of the equality to get $gcd(a_n-2,a_n-3)log2=log(a_n-a_n-1)$ consequently $$gcd(a_n-2,a_n-3)=log(a_n-a_n-1)above 1.5pt log2$$ If $log(a_n-a_n-1)above 1.5pt log2=2x$ for some $xin mathbbN$ then $log(a_n-a_n-1)=2xlog2$ and so $a_n-a_n-1=4^x$ which is a contradiction. So I must have $log(a_n-a_n-1)above 1.5pt log2=2x+1$ in which case $a_n-a_n-1=2^2x+1$ which is a contradiction for $x>1$. So $x=0$ and $log(a_n-a_n-1)above 1.5pt log2=1.$ By equality $gcd(a_n-2,a_n-3)=1.$ So I have shown that $a_n=2+a_n-1.$ I know that $a_4=3$ and so I have shown that $a_n_n=1^infty$ is the sequence $1,1,1,3,5,7,ldots$ Now by recursion I have that



    beginalign
    a_n&=2+a_n-1\
    &=2+2+a_n-2\
    &=2+2+2+a_n-3\
    &=underbrace2+2+2+ldots+2_text$n-4$ times+1+1+1\
    &=2(n-4)+3\
    &=2n-5
    endalign







    share|cite|improve this answer



























      up vote
      0
      down vote













      @lulu per you comment (sort of) I have the following as a possible solution:




      By definition $2^gcd(a_n-2,a_n-3)+a_n-1=a_n$ and so $2^gcd(a_n-2,a_n-3)=a_n-a_n-1 $ for every $n$. I can take "logs" on the LHS and RHS of the equality to get $gcd(a_n-2,a_n-3)log2=log(a_n-a_n-1)$ consequently $$gcd(a_n-2,a_n-3)=log(a_n-a_n-1)above 1.5pt log2$$ If $log(a_n-a_n-1)above 1.5pt log2=2x$ for some $xin mathbbN$ then $log(a_n-a_n-1)=2xlog2$ and so $a_n-a_n-1=4^x$ which is a contradiction. So I must have $log(a_n-a_n-1)above 1.5pt log2=2x+1$ in which case $a_n-a_n-1=2^2x+1$ which is a contradiction for $x>1$. So $x=0$ and $log(a_n-a_n-1)above 1.5pt log2=1.$ By equality $gcd(a_n-2,a_n-3)=1.$ So I have shown that $a_n=2+a_n-1.$ I know that $a_4=3$ and so I have shown that $a_n_n=1^infty$ is the sequence $1,1,1,3,5,7,ldots$ Now by recursion I have that



      beginalign
      a_n&=2+a_n-1\
      &=2+2+a_n-2\
      &=2+2+2+a_n-3\
      &=underbrace2+2+2+ldots+2_text$n-4$ times+1+1+1\
      &=2(n-4)+3\
      &=2n-5
      endalign







      share|cite|improve this answer

























        up vote
        0
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        down vote









        @lulu per you comment (sort of) I have the following as a possible solution:




        By definition $2^gcd(a_n-2,a_n-3)+a_n-1=a_n$ and so $2^gcd(a_n-2,a_n-3)=a_n-a_n-1 $ for every $n$. I can take "logs" on the LHS and RHS of the equality to get $gcd(a_n-2,a_n-3)log2=log(a_n-a_n-1)$ consequently $$gcd(a_n-2,a_n-3)=log(a_n-a_n-1)above 1.5pt log2$$ If $log(a_n-a_n-1)above 1.5pt log2=2x$ for some $xin mathbbN$ then $log(a_n-a_n-1)=2xlog2$ and so $a_n-a_n-1=4^x$ which is a contradiction. So I must have $log(a_n-a_n-1)above 1.5pt log2=2x+1$ in which case $a_n-a_n-1=2^2x+1$ which is a contradiction for $x>1$. So $x=0$ and $log(a_n-a_n-1)above 1.5pt log2=1.$ By equality $gcd(a_n-2,a_n-3)=1.$ So I have shown that $a_n=2+a_n-1.$ I know that $a_4=3$ and so I have shown that $a_n_n=1^infty$ is the sequence $1,1,1,3,5,7,ldots$ Now by recursion I have that



        beginalign
        a_n&=2+a_n-1\
        &=2+2+a_n-2\
        &=2+2+2+a_n-3\
        &=underbrace2+2+2+ldots+2_text$n-4$ times+1+1+1\
        &=2(n-4)+3\
        &=2n-5
        endalign







        share|cite|improve this answer















        @lulu per you comment (sort of) I have the following as a possible solution:




        By definition $2^gcd(a_n-2,a_n-3)+a_n-1=a_n$ and so $2^gcd(a_n-2,a_n-3)=a_n-a_n-1 $ for every $n$. I can take "logs" on the LHS and RHS of the equality to get $gcd(a_n-2,a_n-3)log2=log(a_n-a_n-1)$ consequently $$gcd(a_n-2,a_n-3)=log(a_n-a_n-1)above 1.5pt log2$$ If $log(a_n-a_n-1)above 1.5pt log2=2x$ for some $xin mathbbN$ then $log(a_n-a_n-1)=2xlog2$ and so $a_n-a_n-1=4^x$ which is a contradiction. So I must have $log(a_n-a_n-1)above 1.5pt log2=2x+1$ in which case $a_n-a_n-1=2^2x+1$ which is a contradiction for $x>1$. So $x=0$ and $log(a_n-a_n-1)above 1.5pt log2=1.$ By equality $gcd(a_n-2,a_n-3)=1.$ So I have shown that $a_n=2+a_n-1.$ I know that $a_4=3$ and so I have shown that $a_n_n=1^infty$ is the sequence $1,1,1,3,5,7,ldots$ Now by recursion I have that



        beginalign
        a_n&=2+a_n-1\
        &=2+2+a_n-2\
        &=2+2+2+a_n-3\
        &=underbrace2+2+2+ldots+2_text$n-4$ times+1+1+1\
        &=2(n-4)+3\
        &=2n-5
        endalign








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