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Relationship between GCD, LCM and the Riemann Zeta function


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5












$begingroup$


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



$$
frac{1}{n}sum_{k = 1}^nsum_{i = 1}^{k} bigg(frac{gcd(k,i)}{text{lcm}(k,i)}bigg)^s approx zeta(s+1)
$$



or equivalently



$$
frac{1}{n}sum_{k = 1}^nsum_{i = 1}^{k} bigg(frac{gcd(k,i)^2}{ki}bigg)^s approx zeta(s+1)
$$



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











$endgroup$



















    5












    $begingroup$


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



    $$
    frac{1}{n}sum_{k = 1}^nsum_{i = 1}^{k} bigg(frac{gcd(k,i)}{text{lcm}(k,i)}bigg)^s approx zeta(s+1)
    $$



    or equivalently



    $$
    frac{1}{n}sum_{k = 1}^nsum_{i = 1}^{k} bigg(frac{gcd(k,i)^2}{ki}bigg)^s approx zeta(s+1)
    $$



    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











    $endgroup$















      5












      5








      5


      3



      $begingroup$


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



      $$
      frac{1}{n}sum_{k = 1}^nsum_{i = 1}^{k} bigg(frac{gcd(k,i)}{text{lcm}(k,i)}bigg)^s approx zeta(s+1)
      $$



      or equivalently



      $$
      frac{1}{n}sum_{k = 1}^nsum_{i = 1}^{k} bigg(frac{gcd(k,i)^2}{ki}bigg)^s approx zeta(s+1)
      $$



      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











      $endgroup$




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



      $$
      frac{1}{n}sum_{k = 1}^nsum_{i = 1}^{k} bigg(frac{gcd(k,i)}{text{lcm}(k,i)}bigg)^s approx zeta(s+1)
      $$



      or equivalently



      $$
      frac{1}{n}sum_{k = 1}^nsum_{i = 1}^{k} bigg(frac{gcd(k,i)^2}{ki}bigg)^s approx zeta(s+1)
      $$



      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




      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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          $begingroup$

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



          $$sum_{A,B, gcd(A,B) le G} frac{gcd(A,B)^s}{lcm(A,B)^s} = sum_{gle G} sum_{a,b, gcd(a,b)=1}frac{gcd(ag,bg)^s}{lcm(ag,bg)^s}$$



          $$= sum_{gle G} sum_{a,b, gcd(a,b)=1}frac{g^s}{(abg)^s} = G sum_{a,b, gcd(a,b)=1}frac{1}{(ab)^s}$$
          $$ = G sum_d mu(d)sum_{u,v}frac{1}{(d^2uv)^s}= G (sum_d mu(d)d^{-2s})(sum_uu^{-s})(sum_v v^{-s}) = G frac{zeta(s)^2}{zeta(2s)}$$



          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
            1






            active

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            active

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            active

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            6












            $begingroup$

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



            $$sum_{A,B, gcd(A,B) le G} frac{gcd(A,B)^s}{lcm(A,B)^s} = sum_{gle G} sum_{a,b, gcd(a,b)=1}frac{gcd(ag,bg)^s}{lcm(ag,bg)^s}$$



            $$= sum_{gle G} sum_{a,b, gcd(a,b)=1}frac{g^s}{(abg)^s} = G sum_{a,b, gcd(a,b)=1}frac{1}{(ab)^s}$$
            $$ = G sum_d mu(d)sum_{u,v}frac{1}{(d^2uv)^s}= G (sum_d mu(d)d^{-2s})(sum_uu^{-s})(sum_v v^{-s}) = G frac{zeta(s)^2}{zeta(2s)}$$



            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



              $$sum_{A,B, gcd(A,B) le G} frac{gcd(A,B)^s}{lcm(A,B)^s} = sum_{gle G} sum_{a,b, gcd(a,b)=1}frac{gcd(ag,bg)^s}{lcm(ag,bg)^s}$$



              $$= sum_{gle G} sum_{a,b, gcd(a,b)=1}frac{g^s}{(abg)^s} = G sum_{a,b, gcd(a,b)=1}frac{1}{(ab)^s}$$
              $$ = G sum_d mu(d)sum_{u,v}frac{1}{(d^2uv)^s}= G (sum_d mu(d)d^{-2s})(sum_uu^{-s})(sum_v v^{-s}) = G frac{zeta(s)^2}{zeta(2s)}$$



              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



                $$sum_{A,B, gcd(A,B) le G} frac{gcd(A,B)^s}{lcm(A,B)^s} = sum_{gle G} sum_{a,b, gcd(a,b)=1}frac{gcd(ag,bg)^s}{lcm(ag,bg)^s}$$



                $$= sum_{gle G} sum_{a,b, gcd(a,b)=1}frac{g^s}{(abg)^s} = G sum_{a,b, gcd(a,b)=1}frac{1}{(ab)^s}$$
                $$ = G sum_d mu(d)sum_{u,v}frac{1}{(d^2uv)^s}= G (sum_d mu(d)d^{-2s})(sum_uu^{-s})(sum_v v^{-s}) = G frac{zeta(s)^2}{zeta(2s)}$$



                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



                $$sum_{A,B, gcd(A,B) le G} frac{gcd(A,B)^s}{lcm(A,B)^s} = sum_{gle G} sum_{a,b, gcd(a,b)=1}frac{gcd(ag,bg)^s}{lcm(ag,bg)^s}$$



                $$= sum_{gle G} sum_{a,b, gcd(a,b)=1}frac{g^s}{(abg)^s} = G sum_{a,b, gcd(a,b)=1}frac{1}{(ab)^s}$$
                $$ = G sum_d mu(d)sum_{u,v}frac{1}{(d^2uv)^s}= G (sum_d mu(d)d^{-2s})(sum_uu^{-s})(sum_v v^{-s}) = G frac{zeta(s)^2}{zeta(2s)}$$



                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




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