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Relationship between GCD, LCM and the Riemann Zeta function


Values of the Riemann Zeta function and the Ramanujan Summation - How strong is the connection?Calculating the Zeroes of the Riemann-Zeta functionWhat is so interesting about the zeroes of the Riemann $zeta$ function?How to evaluate a zero of the Riemann zeta function?Is there any relationship between the Riemann z function and strange attractors?Relationship between Riemann Zeta function and Prime zeta functionQuestion on Integral Transform Related to Riemann Zeta Function $zeta(s)$Relation of prime indexed series with Riemann zeta functionRiemann zeta for real argument between 0 and 1 using Mellin, with short asymptotic expansionReference request on the Riemann Zeta function






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








5












$begingroup$


Let $zeta(s)$ be the Riemann zeta function. I observed that as for large $n$, as $s$ increased,



$$
frac1nsum_k = 1^nsum_i = 1^k bigg(fracgcd(k,i)textlcm(k,i)bigg)^s approx zeta(s+1)
$$



or equivalently



$$
frac1nsum_k = 1^nsum_i = 1^k bigg(fracgcd(k,i)^2kibigg)^s approx zeta(s+1)
$$



A few values of $s$, LHS and the RHS are given below



$$(3,1.221,1.202)$$
$$(4,1.084,1.0823)$$
$$(5,1.0372,1.0369)$$
$$(6,1.01737,1.01734)$$
$$(7,1.00835,1.00834)$$
$$(9,1.00494,1.00494)$$
$$(19,1.0000009539,1.0000009539)$$



Question: Is the LHS asymptotic to $zeta(s+1)$ ?










share|cite|improve this question











$endgroup$


















    5












    $begingroup$


    Let $zeta(s)$ be the Riemann zeta function. I observed that as for large $n$, as $s$ increased,



    $$
    frac1nsum_k = 1^nsum_i = 1^k bigg(fracgcd(k,i)textlcm(k,i)bigg)^s approx zeta(s+1)
    $$



    or equivalently



    $$
    frac1nsum_k = 1^nsum_i = 1^k bigg(fracgcd(k,i)^2kibigg)^s approx zeta(s+1)
    $$



    A few values of $s$, LHS and the RHS are given below



    $$(3,1.221,1.202)$$
    $$(4,1.084,1.0823)$$
    $$(5,1.0372,1.0369)$$
    $$(6,1.01737,1.01734)$$
    $$(7,1.00835,1.00834)$$
    $$(9,1.00494,1.00494)$$
    $$(19,1.0000009539,1.0000009539)$$



    Question: Is the LHS asymptotic to $zeta(s+1)$ ?










    share|cite|improve this question











    $endgroup$














      5












      5








      5


      3



      $begingroup$


      Let $zeta(s)$ be the Riemann zeta function. I observed that as for large $n$, as $s$ increased,



      $$
      frac1nsum_k = 1^nsum_i = 1^k bigg(fracgcd(k,i)textlcm(k,i)bigg)^s approx zeta(s+1)
      $$



      or equivalently



      $$
      frac1nsum_k = 1^nsum_i = 1^k bigg(fracgcd(k,i)^2kibigg)^s approx zeta(s+1)
      $$



      A few values of $s$, LHS and the RHS are given below



      $$(3,1.221,1.202)$$
      $$(4,1.084,1.0823)$$
      $$(5,1.0372,1.0369)$$
      $$(6,1.01737,1.01734)$$
      $$(7,1.00835,1.00834)$$
      $$(9,1.00494,1.00494)$$
      $$(19,1.0000009539,1.0000009539)$$



      Question: Is the LHS asymptotic to $zeta(s+1)$ ?










      share|cite|improve this question











      $endgroup$




      Let $zeta(s)$ be the Riemann zeta function. I observed that as for large $n$, as $s$ increased,



      $$
      frac1nsum_k = 1^nsum_i = 1^k bigg(fracgcd(k,i)textlcm(k,i)bigg)^s approx zeta(s+1)
      $$



      or equivalently



      $$
      frac1nsum_k = 1^nsum_i = 1^k bigg(fracgcd(k,i)^2kibigg)^s approx zeta(s+1)
      $$



      A few values of $s$, LHS and the RHS are given below



      $$(3,1.221,1.202)$$
      $$(4,1.084,1.0823)$$
      $$(5,1.0372,1.0369)$$
      $$(6,1.01737,1.01734)$$
      $$(7,1.00835,1.00834)$$
      $$(9,1.00494,1.00494)$$
      $$(19,1.0000009539,1.0000009539)$$



      Question: Is the LHS asymptotic to $zeta(s+1)$ ?







      number-theory elementary-number-theory summation prime-numbers analytic-number-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 8 hours ago







      Nilotpal Kanti Sinha

















      asked 9 hours ago









      Nilotpal Kanti SinhaNilotpal Kanti Sinha

      5,3792 gold badges16 silver badges42 bronze badges




      5,3792 gold badges16 silver badges42 bronze badges




















          1 Answer
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          6












          $begingroup$

          With $(A,B) = (ga,gb), gcd(a,b)=1$ then



          $$sum_A,B, gcd(A,B) le G fracgcd(A,B)^slcm(A,B)^s = sum_gle G sum_a,b, gcd(a,b)=1fracgcd(ag,bg)^slcm(ag,bg)^s$$



          $$= sum_gle G sum_a,b, gcd(a,b)=1fracg^s(abg)^s = G sum_a,b, gcd(a,b)=1frac1(ab)^s$$
          $$ = G sum_d mu(d)sum_u,vfrac1(d^2uv)^s= G (sum_d mu(d)d^-2s)(sum_uu^-s)(sum_v v^-s) = G fraczeta(s)^2zeta(2s)$$



          As $s to infty$ it $to G zeta(s)^2 approx Gzeta(s+1)$






          share|cite|improve this answer









          $endgroup$















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            1 Answer
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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            6












            $begingroup$

            With $(A,B) = (ga,gb), gcd(a,b)=1$ then



            $$sum_A,B, gcd(A,B) le G fracgcd(A,B)^slcm(A,B)^s = sum_gle G sum_a,b, gcd(a,b)=1fracgcd(ag,bg)^slcm(ag,bg)^s$$



            $$= sum_gle G sum_a,b, gcd(a,b)=1fracg^s(abg)^s = G sum_a,b, gcd(a,b)=1frac1(ab)^s$$
            $$ = G sum_d mu(d)sum_u,vfrac1(d^2uv)^s= G (sum_d mu(d)d^-2s)(sum_uu^-s)(sum_v v^-s) = G fraczeta(s)^2zeta(2s)$$



            As $s to infty$ it $to G zeta(s)^2 approx Gzeta(s+1)$






            share|cite|improve this answer









            $endgroup$

















              6












              $begingroup$

              With $(A,B) = (ga,gb), gcd(a,b)=1$ then



              $$sum_A,B, gcd(A,B) le G fracgcd(A,B)^slcm(A,B)^s = sum_gle G sum_a,b, gcd(a,b)=1fracgcd(ag,bg)^slcm(ag,bg)^s$$



              $$= sum_gle G sum_a,b, gcd(a,b)=1fracg^s(abg)^s = G sum_a,b, gcd(a,b)=1frac1(ab)^s$$
              $$ = G sum_d mu(d)sum_u,vfrac1(d^2uv)^s= G (sum_d mu(d)d^-2s)(sum_uu^-s)(sum_v v^-s) = G fraczeta(s)^2zeta(2s)$$



              As $s to infty$ it $to G zeta(s)^2 approx Gzeta(s+1)$






              share|cite|improve this answer









              $endgroup$















                6












                6








                6





                $begingroup$

                With $(A,B) = (ga,gb), gcd(a,b)=1$ then



                $$sum_A,B, gcd(A,B) le G fracgcd(A,B)^slcm(A,B)^s = sum_gle G sum_a,b, gcd(a,b)=1fracgcd(ag,bg)^slcm(ag,bg)^s$$



                $$= sum_gle G sum_a,b, gcd(a,b)=1fracg^s(abg)^s = G sum_a,b, gcd(a,b)=1frac1(ab)^s$$
                $$ = G sum_d mu(d)sum_u,vfrac1(d^2uv)^s= G (sum_d mu(d)d^-2s)(sum_uu^-s)(sum_v v^-s) = G fraczeta(s)^2zeta(2s)$$



                As $s to infty$ it $to G zeta(s)^2 approx Gzeta(s+1)$






                share|cite|improve this answer









                $endgroup$



                With $(A,B) = (ga,gb), gcd(a,b)=1$ then



                $$sum_A,B, gcd(A,B) le G fracgcd(A,B)^slcm(A,B)^s = sum_gle G sum_a,b, gcd(a,b)=1fracgcd(ag,bg)^slcm(ag,bg)^s$$



                $$= sum_gle G sum_a,b, gcd(a,b)=1fracg^s(abg)^s = G sum_a,b, gcd(a,b)=1frac1(ab)^s$$
                $$ = G sum_d mu(d)sum_u,vfrac1(d^2uv)^s= G (sum_d mu(d)d^-2s)(sum_uu^-s)(sum_v v^-s) = G fraczeta(s)^2zeta(2s)$$



                As $s to infty$ it $to G zeta(s)^2 approx Gzeta(s+1)$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 8 hours ago









                reunsreuns

                23.6k2 gold badges14 silver badges61 bronze badges




                23.6k2 gold badges14 silver badges61 bronze badges



























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