In a perfect number $2^{p−1} times (2^p − 1)$, the ratio of $p$ to the digits in its perfect number...












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I was reading about Mersenne primes and perfect numbers, and how the expression $2^{p−1} times (2^p − 1)$, where $p$ is any prime number, can be used to generate perfect numbers when $2^{p−1}$ is a Mersenne prime. I also found a claim on Wikipedia that "The ratio ($p$ / digits) approaches $log(10) / log(4) = 1.6609640474ldots$" but I cannot find the proof that shows this to be true. Why is it $log(10)/log(4)$?






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    $begingroup$


    I was reading about Mersenne primes and perfect numbers, and how the expression $2^{p−1} times (2^p − 1)$, where $p$ is any prime number, can be used to generate perfect numbers when $2^{p−1}$ is a Mersenne prime. I also found a claim on Wikipedia that "The ratio ($p$ / digits) approaches $log(10) / log(4) = 1.6609640474ldots$" but I cannot find the proof that shows this to be true. Why is it $log(10)/log(4)$?






    prime-numbers perfect-numbers







    share|cite|improve this question











    $endgroup$















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      2








      2


      3



      $begingroup$


      I was reading about Mersenne primes and perfect numbers, and how the expression $2^{p−1} times (2^p − 1)$, where $p$ is any prime number, can be used to generate perfect numbers when $2^{p−1}$ is a Mersenne prime. I also found a claim on Wikipedia that "The ratio ($p$ / digits) approaches $log(10) / log(4) = 1.6609640474ldots$" but I cannot find the proof that shows this to be true. Why is it $log(10)/log(4)$?






      prime-numbers perfect-numbers







      share|cite|improve this question











      $endgroup$




      I was reading about Mersenne primes and perfect numbers, and how the expression $2^{p−1} times (2^p − 1)$, where $p$ is any prime number, can be used to generate perfect numbers when $2^{p−1}$ is a Mersenne prime. I also found a claim on Wikipedia that "The ratio ($p$ / digits) approaches $log(10) / log(4) = 1.6609640474ldots$" but I cannot find the proof that shows this to be true. Why is it $log(10)/log(4)$?






      prime-numbers perfect-numbers




      prime-numbers perfect-numbers






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      edited Dec 7 '18 at 3:48









      zipirovich

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      asked Dec 7 '18 at 3:39









      Sara Sara

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          $begingroup$

          Because the number of digits in an integer $N$ is approximately $log_{10}N$. The precise formula is $lfloorlog_{10}Nrfloor+1$, but when $N$ is large this rounding is insignificant.



          For $N=2^{p−1}times(2^p−1)$, and assuming $p$ (and therefore, $N$) is large, this gives:
          $$begin{aligned}
          log_{10}N&=log_{10}left(2^{p−1}times(2^p−1)right)=\
          &=log_{10}left(2^{p−1}right)+log_{10}left(2^p−1right)approx\
          &approxlog_{10}left(2^{p−1}right)+log_{10}left(2^pright)=\
          &=(p-1)log_{10}2+plog_{10}2=\
          &=(2p-1)log_{10}2approx\
          &approx2plog_{10}2=\
          &=plog_{10}4,
          end{aligned}$$
          and then the change of base formula finishes the proof.






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            $begingroup$

            Because the number of digits in an integer $N$ is approximately $log_{10}N$. The precise formula is $lfloorlog_{10}Nrfloor+1$, but when $N$ is large this rounding is insignificant.



            For $N=2^{p−1}times(2^p−1)$, and assuming $p$ (and therefore, $N$) is large, this gives:
            $$begin{aligned}
            log_{10}N&=log_{10}left(2^{p−1}times(2^p−1)right)=\
            &=log_{10}left(2^{p−1}right)+log_{10}left(2^p−1right)approx\
            &approxlog_{10}left(2^{p−1}right)+log_{10}left(2^pright)=\
            &=(p-1)log_{10}2+plog_{10}2=\
            &=(2p-1)log_{10}2approx\
            &approx2plog_{10}2=\
            &=plog_{10}4,
            end{aligned}$$
            and then the change of base formula finishes the proof.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              Because the number of digits in an integer $N$ is approximately $log_{10}N$. The precise formula is $lfloorlog_{10}Nrfloor+1$, but when $N$ is large this rounding is insignificant.



              For $N=2^{p−1}times(2^p−1)$, and assuming $p$ (and therefore, $N$) is large, this gives:
              $$begin{aligned}
              log_{10}N&=log_{10}left(2^{p−1}times(2^p−1)right)=\
              &=log_{10}left(2^{p−1}right)+log_{10}left(2^p−1right)approx\
              &approxlog_{10}left(2^{p−1}right)+log_{10}left(2^pright)=\
              &=(p-1)log_{10}2+plog_{10}2=\
              &=(2p-1)log_{10}2approx\
              &approx2plog_{10}2=\
              &=plog_{10}4,
              end{aligned}$$
              and then the change of base formula finishes the proof.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                Because the number of digits in an integer $N$ is approximately $log_{10}N$. The precise formula is $lfloorlog_{10}Nrfloor+1$, but when $N$ is large this rounding is insignificant.



                For $N=2^{p−1}times(2^p−1)$, and assuming $p$ (and therefore, $N$) is large, this gives:
                $$begin{aligned}
                log_{10}N&=log_{10}left(2^{p−1}times(2^p−1)right)=\
                &=log_{10}left(2^{p−1}right)+log_{10}left(2^p−1right)approx\
                &approxlog_{10}left(2^{p−1}right)+log_{10}left(2^pright)=\
                &=(p-1)log_{10}2+plog_{10}2=\
                &=(2p-1)log_{10}2approx\
                &approx2plog_{10}2=\
                &=plog_{10}4,
                end{aligned}$$
                and then the change of base formula finishes the proof.






                share|cite|improve this answer









                $endgroup$



                Because the number of digits in an integer $N$ is approximately $log_{10}N$. The precise formula is $lfloorlog_{10}Nrfloor+1$, but when $N$ is large this rounding is insignificant.



                For $N=2^{p−1}times(2^p−1)$, and assuming $p$ (and therefore, $N$) is large, this gives:
                $$begin{aligned}
                log_{10}N&=log_{10}left(2^{p−1}times(2^p−1)right)=\
                &=log_{10}left(2^{p−1}right)+log_{10}left(2^p−1right)approx\
                &approxlog_{10}left(2^{p−1}right)+log_{10}left(2^pright)=\
                &=(p-1)log_{10}2+plog_{10}2=\
                &=(2p-1)log_{10}2approx\
                &approx2plog_{10}2=\
                &=plog_{10}4,
                end{aligned}$$
                and then the change of base formula finishes the proof.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 7 '18 at 3:57









                zipirovichzipirovich

                11.2k11631




                11.2k11631






























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