Median divisor of even perfect numbers












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I noticed that when divisors of even perfect numbers are listed in ascending order, the middle divisor (I guess the median), is always of the form $2^n$, some power of 2. If true is there a proof for this, or does it happen all the time? I only checked up to the 8th perfect number. Thank you and apologies for the possible silliness of the question.






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    $begingroup$


    I noticed that when divisors of even perfect numbers are listed in ascending order, the middle divisor (I guess the median), is always of the form $2^n$, some power of 2. If true is there a proof for this, or does it happen all the time? I only checked up to the 8th perfect number. Thank you and apologies for the possible silliness of the question.






    perfect-numbers







    share|cite|improve this question









    New contributor




    Soulis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      3












      3








      3





      $begingroup$


      I noticed that when divisors of even perfect numbers are listed in ascending order, the middle divisor (I guess the median), is always of the form $2^n$, some power of 2. If true is there a proof for this, or does it happen all the time? I only checked up to the 8th perfect number. Thank you and apologies for the possible silliness of the question.






      perfect-numbers







      share|cite|improve this question









      New contributor




      Soulis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      I noticed that when divisors of even perfect numbers are listed in ascending order, the middle divisor (I guess the median), is always of the form $2^n$, some power of 2. If true is there a proof for this, or does it happen all the time? I only checked up to the 8th perfect number. Thank you and apologies for the possible silliness of the question.






      perfect-numbers




      perfect-numbers






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      Soulis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      edited 3 hours ago









      Parcly Taxel

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      asked 3 hours ago









      SoulisSoulis

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          2 Answers
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          $begingroup$

          We know that all even perfect numbers are of the form $(2^n-1)2^{n-1}$ where $2^n-1$ is prime, so there are $2n$ divisors. Ignoring the divisor that is the perfect number itself (so that the remaining $2n-1$ divisors sum to the perfect number), it is easy to see that those divisors have a simple ordering too:
          $$underbrace{1,2,dots,2^{n-1}}_{n},2^n-1,2(2^n-1),dots,2^{n-2}(2^n-1)$$
          The median is the $n$th divisor in this ordering, which we see is $2^{n-1}$. Therefore your claim is correct.






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            2












            $begingroup$

            Every even perfect number is of the form $n=2^{p-1}(2^p-1)$ with $2^p-1$ prime.



            Therefore, the factors of $n$ are



            $1, 2, 2^2, 2^3, ..., 2^{p-1}, 2^p-1, 2(2^p-1), 2^2(2^p-1), 2^3(2^p-1), ...2^{p-2}(2^p-1),$



            and the middle one is $2^{p-1}.$






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              2 Answers
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              $begingroup$

              We know that all even perfect numbers are of the form $(2^n-1)2^{n-1}$ where $2^n-1$ is prime, so there are $2n$ divisors. Ignoring the divisor that is the perfect number itself (so that the remaining $2n-1$ divisors sum to the perfect number), it is easy to see that those divisors have a simple ordering too:
              $$underbrace{1,2,dots,2^{n-1}}_{n},2^n-1,2(2^n-1),dots,2^{n-2}(2^n-1)$$
              The median is the $n$th divisor in this ordering, which we see is $2^{n-1}$. Therefore your claim is correct.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                We know that all even perfect numbers are of the form $(2^n-1)2^{n-1}$ where $2^n-1$ is prime, so there are $2n$ divisors. Ignoring the divisor that is the perfect number itself (so that the remaining $2n-1$ divisors sum to the perfect number), it is easy to see that those divisors have a simple ordering too:
                $$underbrace{1,2,dots,2^{n-1}}_{n},2^n-1,2(2^n-1),dots,2^{n-2}(2^n-1)$$
                The median is the $n$th divisor in this ordering, which we see is $2^{n-1}$. Therefore your claim is correct.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  We know that all even perfect numbers are of the form $(2^n-1)2^{n-1}$ where $2^n-1$ is prime, so there are $2n$ divisors. Ignoring the divisor that is the perfect number itself (so that the remaining $2n-1$ divisors sum to the perfect number), it is easy to see that those divisors have a simple ordering too:
                  $$underbrace{1,2,dots,2^{n-1}}_{n},2^n-1,2(2^n-1),dots,2^{n-2}(2^n-1)$$
                  The median is the $n$th divisor in this ordering, which we see is $2^{n-1}$. Therefore your claim is correct.






                  share|cite|improve this answer









                  $endgroup$



                  We know that all even perfect numbers are of the form $(2^n-1)2^{n-1}$ where $2^n-1$ is prime, so there are $2n$ divisors. Ignoring the divisor that is the perfect number itself (so that the remaining $2n-1$ divisors sum to the perfect number), it is easy to see that those divisors have a simple ordering too:
                  $$underbrace{1,2,dots,2^{n-1}}_{n},2^n-1,2(2^n-1),dots,2^{n-2}(2^n-1)$$
                  The median is the $n$th divisor in this ordering, which we see is $2^{n-1}$. Therefore your claim is correct.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  Parcly TaxelParcly Taxel

                  41.9k1372101




                  41.9k1372101























                      2












                      $begingroup$

                      Every even perfect number is of the form $n=2^{p-1}(2^p-1)$ with $2^p-1$ prime.



                      Therefore, the factors of $n$ are



                      $1, 2, 2^2, 2^3, ..., 2^{p-1}, 2^p-1, 2(2^p-1), 2^2(2^p-1), 2^3(2^p-1), ...2^{p-2}(2^p-1),$



                      and the middle one is $2^{p-1}.$






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        Every even perfect number is of the form $n=2^{p-1}(2^p-1)$ with $2^p-1$ prime.



                        Therefore, the factors of $n$ are



                        $1, 2, 2^2, 2^3, ..., 2^{p-1}, 2^p-1, 2(2^p-1), 2^2(2^p-1), 2^3(2^p-1), ...2^{p-2}(2^p-1),$



                        and the middle one is $2^{p-1}.$






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          Every even perfect number is of the form $n=2^{p-1}(2^p-1)$ with $2^p-1$ prime.



                          Therefore, the factors of $n$ are



                          $1, 2, 2^2, 2^3, ..., 2^{p-1}, 2^p-1, 2(2^p-1), 2^2(2^p-1), 2^3(2^p-1), ...2^{p-2}(2^p-1),$



                          and the middle one is $2^{p-1}.$






                          share|cite|improve this answer









                          $endgroup$



                          Every even perfect number is of the form $n=2^{p-1}(2^p-1)$ with $2^p-1$ prime.



                          Therefore, the factors of $n$ are



                          $1, 2, 2^2, 2^3, ..., 2^{p-1}, 2^p-1, 2(2^p-1), 2^2(2^p-1), 2^3(2^p-1), ...2^{p-2}(2^p-1),$



                          and the middle one is $2^{p-1}.$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 3 hours ago









                          J. W. TannerJ. W. Tanner

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                          2,0691117






















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