Dirichlet sum involving coprimes to p#












2












$begingroup$


This exercise seemed straightforward but I have not managed to do the following proof. Let $p#$ be the product of primes not exceeding p. Let $c(n)$ be the nth coprime to $p#$ (mod 2,3,...,p). Let $X =prod_{p_ileq p}phi(p_i).$ Of course there are X coprimes for given p, beginning in numerical order with 1.



To show: In the limit as p gets large, the sum $$sum_{nleq X}frac{c(n)}{n}sim p#. $$



Some evidence--



begin{array} {|r|r|}
hline
p & sum c(n)/n &p#~~~~& sum c(n)n^{-1}/p# \
hline
5 &24.89 &30& 0.829\
hline
7 &201.05 &210&0.957 \
hline
11&2295.08&2310& 0.9935\
hline
13&30008.69 &30030&0.99929 \
hline
17& 510480.82&510510&0.999943\
hline
end{array}



Putting $C(x)=sum_{nleq x}frac{c(n)}{n}$ we can use Abel summation to put



$$sum_{nleq x}frac{c(n)}{n}= frac{1}{x}C(x)+int_1^xfrac{C(t)}{t^2}dt$$



and if $x=X,$ I think it's easy to show the first term on the right is $frac{p#}{2}.$ Assuming that's true, is there enough information to argue that the integral has the same value? If not, is there a better way?






number-theory prime-numbers dirichlet-series







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    This exercise seemed straightforward but I have not managed to do the following proof. Let $p#$ be the product of primes not exceeding p. Let $c(n)$ be the nth coprime to $p#$ (mod 2,3,...,p). Let $X =prod_{p_ileq p}phi(p_i).$ Of course there are X coprimes for given p, beginning in numerical order with 1.



    To show: In the limit as p gets large, the sum $$sum_{nleq X}frac{c(n)}{n}sim p#. $$



    Some evidence--



    begin{array} {|r|r|}
    hline
    p & sum c(n)/n &p#~~~~& sum c(n)n^{-1}/p# \
    hline
    5 &24.89 &30& 0.829\
    hline
    7 &201.05 &210&0.957 \
    hline
    11&2295.08&2310& 0.9935\
    hline
    13&30008.69 &30030&0.99929 \
    hline
    17& 510480.82&510510&0.999943\
    hline
    end{array}



    Putting $C(x)=sum_{nleq x}frac{c(n)}{n}$ we can use Abel summation to put



    $$sum_{nleq x}frac{c(n)}{n}= frac{1}{x}C(x)+int_1^xfrac{C(t)}{t^2}dt$$



    and if $x=X,$ I think it's easy to show the first term on the right is $frac{p#}{2}.$ Assuming that's true, is there enough information to argue that the integral has the same value? If not, is there a better way?






    number-theory prime-numbers dirichlet-series







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      This exercise seemed straightforward but I have not managed to do the following proof. Let $p#$ be the product of primes not exceeding p. Let $c(n)$ be the nth coprime to $p#$ (mod 2,3,...,p). Let $X =prod_{p_ileq p}phi(p_i).$ Of course there are X coprimes for given p, beginning in numerical order with 1.



      To show: In the limit as p gets large, the sum $$sum_{nleq X}frac{c(n)}{n}sim p#. $$



      Some evidence--



      begin{array} {|r|r|}
      hline
      p & sum c(n)/n &p#~~~~& sum c(n)n^{-1}/p# \
      hline
      5 &24.89 &30& 0.829\
      hline
      7 &201.05 &210&0.957 \
      hline
      11&2295.08&2310& 0.9935\
      hline
      13&30008.69 &30030&0.99929 \
      hline
      17& 510480.82&510510&0.999943\
      hline
      end{array}



      Putting $C(x)=sum_{nleq x}frac{c(n)}{n}$ we can use Abel summation to put



      $$sum_{nleq x}frac{c(n)}{n}= frac{1}{x}C(x)+int_1^xfrac{C(t)}{t^2}dt$$



      and if $x=X,$ I think it's easy to show the first term on the right is $frac{p#}{2}.$ Assuming that's true, is there enough information to argue that the integral has the same value? If not, is there a better way?






      number-theory prime-numbers dirichlet-series







      share|cite|improve this question











      $endgroup$




      This exercise seemed straightforward but I have not managed to do the following proof. Let $p#$ be the product of primes not exceeding p. Let $c(n)$ be the nth coprime to $p#$ (mod 2,3,...,p). Let $X =prod_{p_ileq p}phi(p_i).$ Of course there are X coprimes for given p, beginning in numerical order with 1.



      To show: In the limit as p gets large, the sum $$sum_{nleq X}frac{c(n)}{n}sim p#. $$



      Some evidence--



      begin{array} {|r|r|}
      hline
      p & sum c(n)/n &p#~~~~& sum c(n)n^{-1}/p# \
      hline
      5 &24.89 &30& 0.829\
      hline
      7 &201.05 &210&0.957 \
      hline
      11&2295.08&2310& 0.9935\
      hline
      13&30008.69 &30030&0.99929 \
      hline
      17& 510480.82&510510&0.999943\
      hline
      end{array}



      Putting $C(x)=sum_{nleq x}frac{c(n)}{n}$ we can use Abel summation to put



      $$sum_{nleq x}frac{c(n)}{n}= frac{1}{x}C(x)+int_1^xfrac{C(t)}{t^2}dt$$



      and if $x=X,$ I think it's easy to show the first term on the right is $frac{p#}{2}.$ Assuming that's true, is there enough information to argue that the integral has the same value? If not, is there a better way?






      number-theory prime-numbers dirichlet-series




      number-theory prime-numbers dirichlet-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 2 at 7:58







      daniel

















      asked Feb 1 at 9:17









      danieldaniel

      6,26922258




      6,26922258






















          1 Answer
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          $begingroup$

          Let $# k = prod_{p le k} p = e^{sum_{p le k} log p} = e^{k+o(k)}$ (by the PNT) then



          $$sum_{n le N, gcd(n,# k) = 1} 1 = sum_{d | # k} mu(d) lfloor frac{N}d rfloor = sum_{d | # k} (mu(d)frac{N}d + O(1)) = N prod_{p le k} (1-p^{-1}) +O(2^{pi(k)})$$



          With $c_k(m)$ the $m$-th integer coprime to $# k$ then $m = sum_{n le N, gcd(n,# k) = 1} 1$ and $gcd(N,# k) = 1$ implies $c_k(m) = N$ so $$c_k(N prod_{p le k} (1-p^{-1}) +O(2^{pi(k)})) = N+O(2^{pi(k)}) , qquad c_k(m) = frac{m}{prod_{p le k} (1-p^{-1})}+O(2^{1+pi(k)})$$



          Whence $$sum_{m le varphi(# k)} frac{c_k(m)}{m}=sum_{m le varphi(# k)}frac{1}{prod_{p le k} (1-p^{-1})}+frac{O(2^{1+pi(k)})}{m} \=frac{varphi(# k)}{varphi(#k)/#k}+O(2^{1+pi(k)} log varphi(# k))= # k+O(2^{1+pi(k)} k) $$



          where all the $O$ constants are $1$






          share|cite|improve this answer











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          • $begingroup$
            This is a magisterial answer. Working through it was a really helpful exercise for me, thanks.
            $endgroup$
            – daniel
            Feb 2 at 9:28












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          1 Answer
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          1 Answer
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          active

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          active

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          2












          $begingroup$

          Let $# k = prod_{p le k} p = e^{sum_{p le k} log p} = e^{k+o(k)}$ (by the PNT) then



          $$sum_{n le N, gcd(n,# k) = 1} 1 = sum_{d | # k} mu(d) lfloor frac{N}d rfloor = sum_{d | # k} (mu(d)frac{N}d + O(1)) = N prod_{p le k} (1-p^{-1}) +O(2^{pi(k)})$$



          With $c_k(m)$ the $m$-th integer coprime to $# k$ then $m = sum_{n le N, gcd(n,# k) = 1} 1$ and $gcd(N,# k) = 1$ implies $c_k(m) = N$ so $$c_k(N prod_{p le k} (1-p^{-1}) +O(2^{pi(k)})) = N+O(2^{pi(k)}) , qquad c_k(m) = frac{m}{prod_{p le k} (1-p^{-1})}+O(2^{1+pi(k)})$$



          Whence $$sum_{m le varphi(# k)} frac{c_k(m)}{m}=sum_{m le varphi(# k)}frac{1}{prod_{p le k} (1-p^{-1})}+frac{O(2^{1+pi(k)})}{m} \=frac{varphi(# k)}{varphi(#k)/#k}+O(2^{1+pi(k)} log varphi(# k))= # k+O(2^{1+pi(k)} k) $$



          where all the $O$ constants are $1$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            This is a magisterial answer. Working through it was a really helpful exercise for me, thanks.
            $endgroup$
            – daniel
            Feb 2 at 9:28
















          2












          $begingroup$

          Let $# k = prod_{p le k} p = e^{sum_{p le k} log p} = e^{k+o(k)}$ (by the PNT) then



          $$sum_{n le N, gcd(n,# k) = 1} 1 = sum_{d | # k} mu(d) lfloor frac{N}d rfloor = sum_{d | # k} (mu(d)frac{N}d + O(1)) = N prod_{p le k} (1-p^{-1}) +O(2^{pi(k)})$$



          With $c_k(m)$ the $m$-th integer coprime to $# k$ then $m = sum_{n le N, gcd(n,# k) = 1} 1$ and $gcd(N,# k) = 1$ implies $c_k(m) = N$ so $$c_k(N prod_{p le k} (1-p^{-1}) +O(2^{pi(k)})) = N+O(2^{pi(k)}) , qquad c_k(m) = frac{m}{prod_{p le k} (1-p^{-1})}+O(2^{1+pi(k)})$$



          Whence $$sum_{m le varphi(# k)} frac{c_k(m)}{m}=sum_{m le varphi(# k)}frac{1}{prod_{p le k} (1-p^{-1})}+frac{O(2^{1+pi(k)})}{m} \=frac{varphi(# k)}{varphi(#k)/#k}+O(2^{1+pi(k)} log varphi(# k))= # k+O(2^{1+pi(k)} k) $$



          where all the $O$ constants are $1$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            This is a magisterial answer. Working through it was a really helpful exercise for me, thanks.
            $endgroup$
            – daniel
            Feb 2 at 9:28














          2












          2








          2





          $begingroup$

          Let $# k = prod_{p le k} p = e^{sum_{p le k} log p} = e^{k+o(k)}$ (by the PNT) then



          $$sum_{n le N, gcd(n,# k) = 1} 1 = sum_{d | # k} mu(d) lfloor frac{N}d rfloor = sum_{d | # k} (mu(d)frac{N}d + O(1)) = N prod_{p le k} (1-p^{-1}) +O(2^{pi(k)})$$



          With $c_k(m)$ the $m$-th integer coprime to $# k$ then $m = sum_{n le N, gcd(n,# k) = 1} 1$ and $gcd(N,# k) = 1$ implies $c_k(m) = N$ so $$c_k(N prod_{p le k} (1-p^{-1}) +O(2^{pi(k)})) = N+O(2^{pi(k)}) , qquad c_k(m) = frac{m}{prod_{p le k} (1-p^{-1})}+O(2^{1+pi(k)})$$



          Whence $$sum_{m le varphi(# k)} frac{c_k(m)}{m}=sum_{m le varphi(# k)}frac{1}{prod_{p le k} (1-p^{-1})}+frac{O(2^{1+pi(k)})}{m} \=frac{varphi(# k)}{varphi(#k)/#k}+O(2^{1+pi(k)} log varphi(# k))= # k+O(2^{1+pi(k)} k) $$



          where all the $O$ constants are $1$






          share|cite|improve this answer











          $endgroup$



          Let $# k = prod_{p le k} p = e^{sum_{p le k} log p} = e^{k+o(k)}$ (by the PNT) then



          $$sum_{n le N, gcd(n,# k) = 1} 1 = sum_{d | # k} mu(d) lfloor frac{N}d rfloor = sum_{d | # k} (mu(d)frac{N}d + O(1)) = N prod_{p le k} (1-p^{-1}) +O(2^{pi(k)})$$



          With $c_k(m)$ the $m$-th integer coprime to $# k$ then $m = sum_{n le N, gcd(n,# k) = 1} 1$ and $gcd(N,# k) = 1$ implies $c_k(m) = N$ so $$c_k(N prod_{p le k} (1-p^{-1}) +O(2^{pi(k)})) = N+O(2^{pi(k)}) , qquad c_k(m) = frac{m}{prod_{p le k} (1-p^{-1})}+O(2^{1+pi(k)})$$



          Whence $$sum_{m le varphi(# k)} frac{c_k(m)}{m}=sum_{m le varphi(# k)}frac{1}{prod_{p le k} (1-p^{-1})}+frac{O(2^{1+pi(k)})}{m} \=frac{varphi(# k)}{varphi(#k)/#k}+O(2^{1+pi(k)} log varphi(# k))= # k+O(2^{1+pi(k)} k) $$



          where all the $O$ constants are $1$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 1 at 19:47

























          answered Feb 1 at 18:44









          reunsreuns

          20.7k21353




          20.7k21353












          • $begingroup$
            This is a magisterial answer. Working through it was a really helpful exercise for me, thanks.
            $endgroup$
            – daniel
            Feb 2 at 9:28


















          • $begingroup$
            This is a magisterial answer. Working through it was a really helpful exercise for me, thanks.
            $endgroup$
            – daniel
            Feb 2 at 9:28
















          $begingroup$
          This is a magisterial answer. Working through it was a really helpful exercise for me, thanks.
          $endgroup$
          – daniel
          Feb 2 at 9:28




          $begingroup$
          This is a magisterial answer. Working through it was a really helpful exercise for me, thanks.
          $endgroup$
          – daniel
          Feb 2 at 9:28


















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