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Relationship between GCD, LCM and the Riemann Zeta function

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8

4

$begingroup$

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 $zeta(s+1)$ ?

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

share|cite|improve this question

edited Jul 14 at 20:25

Nilotpal Kanti Sinha

asked Jul 14 at 19:22

Nilotpal Kanti SinhaNilotpal Kanti Sinha

5,60422 gold badges1616 silver badges4242 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What precisely do you mean by "asymptotic"? As $ntoinfty$? As $stoinfty$? Both?
  $endgroup$
  – Greg Martin
  Jul 15 at 8:11
  

  
  
  
  

add a comment | 

8

4

$begingroup$

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 $zeta(s+1)$ ?

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

share|cite|improve this question

edited Jul 14 at 20:25

Nilotpal Kanti Sinha

asked Jul 14 at 19:22

Nilotpal Kanti SinhaNilotpal Kanti Sinha

5,60422 gold badges1616 silver badges4242 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What precisely do you mean by "asymptotic"? As $ntoinfty$? As $stoinfty$? Both?
  $endgroup$
  – Greg Martin
  Jul 15 at 8:11
  

  
  
  
  

add a comment | 

$begingroup$

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 $zeta(s+1)$ ?

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

share|cite|improve this question

edited Jul 14 at 20:25

Nilotpal Kanti Sinha

asked Jul 14 at 19:22

Nilotpal Kanti SinhaNilotpal Kanti Sinha

5,60422 gold badges1616 silver badges4242 bronze badges

$endgroup$

share|cite|improve this question

edited Jul 14 at 20:25

Nilotpal Kanti Sinha

asked Jul 14 at 19:22

Nilotpal Kanti SinhaNilotpal Kanti Sinha

5,60422 gold badges1616 silver badges4242 bronze badges

share|cite|improve this question

edited Jul 14 at 20:25

Nilotpal Kanti Sinha

asked Jul 14 at 19:22

Nilotpal Kanti SinhaNilotpal Kanti Sinha

5,60422 gold badges1616 silver badges4242 bronze badges

asked Jul 14 at 19:22

Nilotpal Kanti SinhaNilotpal Kanti Sinha

5,60422 gold badges1616 silver badges4242 bronze badges

asked Jul 14 at 19:22

Nilotpal Kanti SinhaNilotpal Kanti Sinha

5,60422 gold badges1616 silver badges4242 bronze badges

1 Answer
1

active

oldest

votes

8

$begingroup$

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

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

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

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

share|cite|improve this answer

edited Jul 15 at 8:08

Greg Martin

38.1k22 gold badges3636 silver badges6767 bronze badges

answered Jul 14 at 20:32

reunsreuns

24k22 gold badges1414 silver badges6262 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This does not answer the question asked: the first expression is not the same function as in the OP..
  $endgroup$
  – Greg Martin
  Jul 15 at 8:09
  
  

  
  
  
  

add a comment | 

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8

$begingroup$

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

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

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

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

share|cite|improve this answer

edited Jul 15 at 8:08

Greg Martin

38.1k22 gold badges3636 silver badges6767 bronze badges

answered Jul 14 at 20:32

reunsreuns

24k22 gold badges1414 silver badges6262 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This does not answer the question asked: the first expression is not the same function as in the OP..
  $endgroup$
  – Greg Martin
  Jul 15 at 8:09
  
  

  
  
  
  

add a comment | 

8

$begingroup$

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

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

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

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

share|cite|improve this answer

edited Jul 15 at 8:08

Greg Martin

38.1k22 gold badges3636 silver badges6767 bronze badges

answered Jul 14 at 20:32

reunsreuns

24k22 gold badges1414 silver badges6262 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This does not answer the question asked: the first expression is not the same function as in the OP..
  $endgroup$
  – Greg Martin
  Jul 15 at 8:09
  
  

  
  
  
  

add a comment | 

$begingroup$

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

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

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

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

share|cite|improve this answer

edited Jul 15 at 8:08

Greg Martin

38.1k22 gold badges3636 silver badges6767 bronze badges

answered Jul 14 at 20:32

reunsreuns

24k22 gold badges1414 silver badges6262 bronze badges

$endgroup$

share|cite|improve this answer

edited Jul 15 at 8:08

Greg Martin

38.1k22 gold badges3636 silver badges6767 bronze badges

answered Jul 14 at 20:32

reunsreuns

24k22 gold badges1414 silver badges6262 bronze badges

edited Jul 15 at 8:08

Greg Martin

38.1k22 gold badges3636 silver badges6767 bronze badges

edited Jul 15 at 8:08

Greg Martin

38.1k22 gold badges3636 silver badges6767 bronze badges

answered Jul 14 at 20:32

reunsreuns

24k22 gold badges1414 silver badges6262 bronze badges

answered Jul 14 at 20:32

reunsreuns

24k22 gold badges1414 silver badges6262 bronze badges