Second Chebyshev function and Unique prime factorization
$begingroup$
the identity for the second Chebyshev function:
$$sum _{j=1}^{pi left( x right) } Biggllfloor {frac {ln
left( x right) }{ln left( p_{{j}} right) }} Biggrrfloor ln
left( p_{{j}} right)=ln(operatorname{lcm}(1,2,3,...,lfloor x rfloor))quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(0)
$$
Any natural number $N$ has a unique prime factorization product:
$$N=p_{1,N}^{v_{1,N}}p_{2,N}^{v_{2,N}}p_{3,N}^{v_{3,N}}cdotcdotcdot p_{omega(N),N}^{v_{omega(N),N}}quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(1)$$
Allows us to make the following assertion about the natural logarithm of any such natural:
$$ln left( N right) =sum _{j=1}^{omega left( N right) }v_{{j,N}}
ln left( p_{{j,N}} right)
quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(1')$$
My question is as to whether or not stating the above is sufficient justification for stating the following to be true:
$pi(x)=omega(operatorname{lcm}(1,2,3,...,lfloor xrfloor))quad quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(a)$
the largest multiplicity any factor of $operatorname{lcm}(1,2,3,...,lfloor xrfloor)$ may have is $ lfloorfrac{ln left( x right) }{2}rfloor quadquadquadquadquadquadquad,,,(b)$
Both (a) and (b) I concluded from (0) & ('1), I just feel as if I am missing something that is necessary in a direct proof of these statements.
number-theory
asked Aug 7 '18 at 19:42
AdamAdam
54114
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add a comment |
$begingroup$
the identity for the second Chebyshev function:
$$sum _{j=1}^{pi left( x right) } Biggllfloor {frac {ln
left( x right) }{ln left( p_{{j}} right) }} Biggrrfloor ln
left( p_{{j}} right)=ln(operatorname{lcm}(1,2,3,...,lfloor x rfloor))quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(0)
$$
Any natural number $N$ has a unique prime factorization product:
$$N=p_{1,N}^{v_{1,N}}p_{2,N}^{v_{2,N}}p_{3,N}^{v_{3,N}}cdotcdotcdot p_{omega(N),N}^{v_{omega(N),N}}quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(1)$$
Allows us to make the following assertion about the natural logarithm of any such natural:
$$ln left( N right) =sum _{j=1}^{omega left( N right) }v_{{j,N}}
ln left( p_{{j,N}} right)
quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(1')$$
My question is as to whether or not stating the above is sufficient justification for stating the following to be true:
$pi(x)=omega(operatorname{lcm}(1,2,3,...,lfloor xrfloor))quad quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(a)$
the largest multiplicity any factor of $operatorname{lcm}(1,2,3,...,lfloor xrfloor)$ may have is $ lfloorfrac{ln left( x right) }{2}rfloor quadquadquadquadquadquadquad,,,(b)$
Both (a) and (b) I concluded from (0) & ('1), I just feel as if I am missing something that is necessary in a direct proof of these statements.
number-theory
asked Aug 7 '18 at 19:42
AdamAdam
54114
$endgroup$
add a comment |
$begingroup$
the identity for the second Chebyshev function:
$$sum _{j=1}^{pi left( x right) } Biggllfloor {frac {ln
left( x right) }{ln left( p_{{j}} right) }} Biggrrfloor ln
left( p_{{j}} right)=ln(operatorname{lcm}(1,2,3,...,lfloor x rfloor))quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(0)
$$
Any natural number $N$ has a unique prime factorization product:
$$N=p_{1,N}^{v_{1,N}}p_{2,N}^{v_{2,N}}p_{3,N}^{v_{3,N}}cdotcdotcdot p_{omega(N),N}^{v_{omega(N),N}}quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(1)$$
Allows us to make the following assertion about the natural logarithm of any such natural:
$$ln left( N right) =sum _{j=1}^{omega left( N right) }v_{{j,N}}
ln left( p_{{j,N}} right)
quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(1')$$
My question is as to whether or not stating the above is sufficient justification for stating the following to be true:
$pi(x)=omega(operatorname{lcm}(1,2,3,...,lfloor xrfloor))quad quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(a)$
the largest multiplicity any factor of $operatorname{lcm}(1,2,3,...,lfloor xrfloor)$ may have is $ lfloorfrac{ln left( x right) }{2}rfloor quadquadquadquadquadquadquad,,,(b)$
Both (a) and (b) I concluded from (0) & ('1), I just feel as if I am missing something that is necessary in a direct proof of these statements.
number-theory
asked Aug 7 '18 at 19:42
AdamAdam
54114
$endgroup$
the identity for the second Chebyshev function:
$$sum _{j=1}^{pi left( x right) } Biggllfloor {frac {ln
left( x right) }{ln left( p_{{j}} right) }} Biggrrfloor ln
left( p_{{j}} right)=ln(operatorname{lcm}(1,2,3,...,lfloor x rfloor))quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(0)
$$
Any natural number $N$ has a unique prime factorization product:
$$N=p_{1,N}^{v_{1,N}}p_{2,N}^{v_{2,N}}p_{3,N}^{v_{3,N}}cdotcdotcdot p_{omega(N),N}^{v_{omega(N),N}}quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(1)$$
Allows us to make the following assertion about the natural logarithm of any such natural:
$$ln left( N right) =sum _{j=1}^{omega left( N right) }v_{{j,N}}
ln left( p_{{j,N}} right)
quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(1')$$
My question is as to whether or not stating the above is sufficient justification for stating the following to be true:
$pi(x)=omega(operatorname{lcm}(1,2,3,...,lfloor xrfloor))quad quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquad(a)$
the largest multiplicity any factor of $operatorname{lcm}(1,2,3,...,lfloor xrfloor)$ may have is $ lfloorfrac{ln left( x right) }{2}rfloor quadquadquadquadquadquadquad,,,(b)$
Both (a) and (b) I concluded from (0) & ('1), I just feel as if I am missing something that is necessary in a direct proof of these statements.
number-theory
number-theory
asked Aug 7 '18 at 19:42
AdamAdam
54114
asked Aug 7 '18 at 19:42
AdamAdam
54114
edited Aug 15 '18 at 16:50
Adam
asked Aug 7 '18 at 19:42
AdamAdam
54114
asked Aug 7 '18 at 19:42
AdamAdam
54114
asked Aug 7 '18 at 19:42
AdamAdam
54114
54114
add a comment |
add a comment |
1 Answer
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$begingroup$
Hint: If you are not sure whether the stated justification is sufficient or not, this is an indication that you should add some more information.
In the first case you could argue:
- The arithmetical function $pi(x)$ counts the number of primes less or equal to $x$. Since this is the same as the index of the greatest prime of the numbers $1,2,dots,x$, the claim
begin{align*}
pi(x)=omega(mathrm{lcm}(1,2,3,...,lfloor xrfloor))
end{align*}
follows.
The second case is not valid:
We have as counterexample for instance $x=5$. We obtain
begin{align*}
mathrm{lcm}(1,2,3,4,5)&=mathrm{lcm}(1,2,3,2^2,5)tag{1}\
&=2^2cdot3cdot 5
end{align*}
We see the multiplicity of $2$ in (1) is 2, but $leftlfloorfrac{ln(5)}{2}rightrfloor=0$.
answered Dec 1 '18 at 22:59
Markus ScheuerMarkus Scheuer
60.4k455144
$endgroup$
$begingroup$
Much appreciated Markus I really am not sure now why I thought $(b)$ was true to be honest
$endgroup$
– Adam
Dec 3 '18 at 3:12
1
$begingroup$
@Adam: You're welcome.
$endgroup$
– Markus Scheuer
Dec 3 '18 at 5:38
add a comment |
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$begingroup$
Hint: If you are not sure whether the stated justification is sufficient or not, this is an indication that you should add some more information.
In the first case you could argue:
- The arithmetical function $pi(x)$ counts the number of primes less or equal to $x$. Since this is the same as the index of the greatest prime of the numbers $1,2,dots,x$, the claim
begin{align*}
pi(x)=omega(mathrm{lcm}(1,2,3,...,lfloor xrfloor))
end{align*}
follows.
The second case is not valid:
We have as counterexample for instance $x=5$. We obtain
begin{align*}
mathrm{lcm}(1,2,3,4,5)&=mathrm{lcm}(1,2,3,2^2,5)tag{1}\
&=2^2cdot3cdot 5
end{align*}
We see the multiplicity of $2$ in (1) is 2, but $leftlfloorfrac{ln(5)}{2}rightrfloor=0$.
answered Dec 1 '18 at 22:59
Markus ScheuerMarkus Scheuer
60.4k455144
$endgroup$
$begingroup$
Much appreciated Markus I really am not sure now why I thought $(b)$ was true to be honest
$endgroup$
– Adam
Dec 3 '18 at 3:12
1
$begingroup$
@Adam: You're welcome.
$endgroup$
– Markus Scheuer
Dec 3 '18 at 5:38
add a comment |
$begingroup$
Hint: If you are not sure whether the stated justification is sufficient or not, this is an indication that you should add some more information.
In the first case you could argue:
- The arithmetical function $pi(x)$ counts the number of primes less or equal to $x$. Since this is the same as the index of the greatest prime of the numbers $1,2,dots,x$, the claim
begin{align*}
pi(x)=omega(mathrm{lcm}(1,2,3,...,lfloor xrfloor))
end{align*}
follows.
The second case is not valid:
We have as counterexample for instance $x=5$. We obtain
begin{align*}
mathrm{lcm}(1,2,3,4,5)&=mathrm{lcm}(1,2,3,2^2,5)tag{1}\
&=2^2cdot3cdot 5
end{align*}
We see the multiplicity of $2$ in (1) is 2, but $leftlfloorfrac{ln(5)}{2}rightrfloor=0$.
answered Dec 1 '18 at 22:59
Markus ScheuerMarkus Scheuer
60.4k455144
$endgroup$
$begingroup$
Much appreciated Markus I really am not sure now why I thought $(b)$ was true to be honest
$endgroup$
– Adam
Dec 3 '18 at 3:12
1
$begingroup$
@Adam: You're welcome.
$endgroup$
– Markus Scheuer
Dec 3 '18 at 5:38
add a comment |
$begingroup$
Hint: If you are not sure whether the stated justification is sufficient or not, this is an indication that you should add some more information.
In the first case you could argue:
- The arithmetical function $pi(x)$ counts the number of primes less or equal to $x$. Since this is the same as the index of the greatest prime of the numbers $1,2,dots,x$, the claim
begin{align*}
pi(x)=omega(mathrm{lcm}(1,2,3,...,lfloor xrfloor))
end{align*}
follows.
The second case is not valid:
We have as counterexample for instance $x=5$. We obtain
begin{align*}
mathrm{lcm}(1,2,3,4,5)&=mathrm{lcm}(1,2,3,2^2,5)tag{1}\
&=2^2cdot3cdot 5
end{align*}
We see the multiplicity of $2$ in (1) is 2, but $leftlfloorfrac{ln(5)}{2}rightrfloor=0$.
answered Dec 1 '18 at 22:59
Markus ScheuerMarkus Scheuer
60.4k455144
$endgroup$
Hint: If you are not sure whether the stated justification is sufficient or not, this is an indication that you should add some more information.
In the first case you could argue:
- The arithmetical function $pi(x)$ counts the number of primes less or equal to $x$. Since this is the same as the index of the greatest prime of the numbers $1,2,dots,x$, the claim
begin{align*}
pi(x)=omega(mathrm{lcm}(1,2,3,...,lfloor xrfloor))
end{align*}
follows.
The second case is not valid:
We have as counterexample for instance $x=5$. We obtain
begin{align*}
mathrm{lcm}(1,2,3,4,5)&=mathrm{lcm}(1,2,3,2^2,5)tag{1}\
&=2^2cdot3cdot 5
end{align*}
We see the multiplicity of $2$ in (1) is 2, but $leftlfloorfrac{ln(5)}{2}rightrfloor=0$.
answered Dec 1 '18 at 22:59
Markus ScheuerMarkus Scheuer
60.4k455144
answered Dec 1 '18 at 22:59
Markus ScheuerMarkus Scheuer
60.4k455144
answered Dec 1 '18 at 22:59
Markus ScheuerMarkus Scheuer
60.4k455144
answered Dec 1 '18 at 22:59
Markus ScheuerMarkus Scheuer
60.4k455144
60.4k455144
$begingroup$
Much appreciated Markus I really am not sure now why I thought $(b)$ was true to be honest
$endgroup$
– Adam
Dec 3 '18 at 3:12
1
$begingroup$
@Adam: You're welcome.
$endgroup$
– Markus Scheuer
Dec 3 '18 at 5:38
add a comment |
$begingroup$
Much appreciated Markus I really am not sure now why I thought $(b)$ was true to be honest
$endgroup$
– Adam
Dec 3 '18 at 3:12
1
$begingroup$
@Adam: You're welcome.
$endgroup$
– Markus Scheuer
Dec 3 '18 at 5:38
$begingroup$
Much appreciated Markus I really am not sure now why I thought $(b)$ was true to be honest
$endgroup$
– Adam
Dec 3 '18 at 3:12
$begingroup$
Much appreciated Markus I really am not sure now why I thought $(b)$ was true to be honest
$endgroup$
– Adam
Dec 3 '18 at 3:12
1
1
$begingroup$
@Adam: You're welcome.
$endgroup$
– Markus Scheuer
Dec 3 '18 at 5:38
$begingroup$
@Adam: You're welcome.
$endgroup$
– Markus Scheuer
Dec 3 '18 at 5:38
add a comment |
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