Mixing
Prime Number Theorem Related functions Divisibility Relation Counter Sought
Clash Royale CLAN TAG#URR8PPP
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In looking at the two functions defined:
$$psi_0(x)=ln( operatornamelcm(1,2,3,...,lfloor x rfloor))$$
$$psi_1(x)=sum _j=1^ lfloor x rfloor sum _i=0^ Bigllfloor frac ln left( x right) ln left( p_j
right) Bigrrfloor +1ln left( p_j^i right)
$$
(where $p_n$ is the $n^th$ prime and $lcm$ denontes the lowest common multiple of the arguments enclosed)
I am interested in finding the minimum value (if it exists) of $n in mathbb N$ that satisfies:
$$ Bigglfracleft(lfloor ln left( psi_0(n)+psi_1(n)right)
rfloor +1 right) !lfloor sqrt n rfloor!Biggr ne 0$$
where $x$ denotes the fractional part of $x$.
some values evaluated:
$$fracleft(lfloor ln left( psi_0(8)+psi_1(8)right)
rfloor +1 right) !lfloor sqrt 8 rfloor!=12$$
$$fracleft(lfloor ln left( psi_0(12)+psi_1(12)right)
rfloor +1 right) !lfloor sqrt 12 rfloor!=20$$
$$fracleft(lfloor ln left( psi_0(20)+psi_1(20)right)
rfloor +1 right) !lfloor sqrt 20 rfloor!=5$$
The value is less than 1 at $n=36$ indicating this to be the immeadiate border of the region of $mathbb N$ for which the inequality Carl mentioned begins to be true (inductively reasoning).
Beyond $0<n<32$ I am not as yet able to produce a result, float approximations continue to imply that the value is 0 up to $n=40$, however really what is needed here is someone with more experience in number theory to assess the situation and give an opinion as to whether it is worth pursuing or not.
Closely related to the relations in a previous questions I posted here and here
I will try my best to follow along with Carl's answer, he has skipped a few steps that are probably what he may consider too obvious to show, but so far:
Because:
$$psi_0(x)=ln(operatornamelcm(1,2,3,...,x-1,lfloor xrfloor))=alpha,ln left( 2 right) +beta,ln left( 3 right)... +sum _j=
1^ lfloor x rfloor ln left( p_j right)
$$ for some $$alpha, beta,... in mathbb N$$
And similarly:
$$psi_1(x)=sum _j=1^ lfloor x rfloor sum _i=0^ Bigllfloor frac ln left( x right) ln left( p_j
right) Bigrrfloor +1ln left( p_j^i right) =frac12sum _j=1^ lfloor x rfloor ln left( p_j
right) left( Bigllfloor frac ln left( x right)
ln left( p_j right) Bigrrfloor +2 right) left( Bigllfloor frac ln left( x right) ln left( p_j right)
Bigrrfloor +1 right)$$
$$=alpha',ln left( 2 right) +beta',ln left( 3 right)... +sum _j=
1^ lfloor x rfloor ln left( p_j right) $$
for some $$alpha', beta',... in mathbb N$$
The asymptotic relation I think I originally started from, which think is actually false as I originally stated, but again, just curious about the division relation on the naturals I really am new to asymptotics:
$$sum _j=1^ lfloor x rfloor left( Bigllfloor
frac ln left( x right) ln left( p_j right)
Bigrrfloor +1 right) ln left( p_j right)+psi_0(x) sim x
$$
elementary-number-theory prime-numbers
asked Jul 15 at 19:59
Adam
13112
 |Â
show 6 more comments
up vote
2
down vote
favorite
In looking at the two functions defined:
$$psi_0(x)=ln( operatornamelcm(1,2,3,...,lfloor x rfloor))$$
$$psi_1(x)=sum _j=1^ lfloor x rfloor sum _i=0^ Bigllfloor frac ln left( x right) ln left( p_j
right) Bigrrfloor +1ln left( p_j^i right)
$$
(where $p_n$ is the $n^th$ prime and $lcm$ denontes the lowest common multiple of the arguments enclosed)
I am interested in finding the minimum value (if it exists) of $n in mathbb N$ that satisfies:
$$ Bigglfracleft(lfloor ln left( psi_0(n)+psi_1(n)right)
rfloor +1 right) !lfloor sqrt n rfloor!Biggr ne 0$$
where $x$ denotes the fractional part of $x$.
some values evaluated:
$$fracleft(lfloor ln left( psi_0(8)+psi_1(8)right)
rfloor +1 right) !lfloor sqrt 8 rfloor!=12$$
$$fracleft(lfloor ln left( psi_0(12)+psi_1(12)right)
rfloor +1 right) !lfloor sqrt 12 rfloor!=20$$
$$fracleft(lfloor ln left( psi_0(20)+psi_1(20)right)
rfloor +1 right) !lfloor sqrt 20 rfloor!=5$$
The value is less than 1 at $n=36$ indicating this to be the immeadiate border of the region of $mathbb N$ for which the inequality Carl mentioned begins to be true (inductively reasoning).
Beyond $0<n<32$ I am not as yet able to produce a result, float approximations continue to imply that the value is 0 up to $n=40$, however really what is needed here is someone with more experience in number theory to assess the situation and give an opinion as to whether it is worth pursuing or not.
Closely related to the relations in a previous questions I posted here and here
I will try my best to follow along with Carl's answer, he has skipped a few steps that are probably what he may consider too obvious to show, but so far:
Because:
$$psi_0(x)=ln(operatornamelcm(1,2,3,...,x-1,lfloor xrfloor))=alpha,ln left( 2 right) +beta,ln left( 3 right)... +sum _j=
1^ lfloor x rfloor ln left( p_j right)
$$ for some $$alpha, beta,... in mathbb N$$
And similarly:
$$psi_1(x)=sum _j=1^ lfloor x rfloor sum _i=0^ Bigllfloor frac ln left( x right) ln left( p_j
right) Bigrrfloor +1ln left( p_j^i right) =frac12sum _j=1^ lfloor x rfloor ln left( p_j
right) left( Bigllfloor frac ln left( x right)
ln left( p_j right) Bigrrfloor +2 right) left( Bigllfloor frac ln left( x right) ln left( p_j right)
Bigrrfloor +1 right)$$
$$=alpha',ln left( 2 right) +beta',ln left( 3 right)... +sum _j=
1^ lfloor x rfloor ln left( p_j right) $$
for some $$alpha', beta',... in mathbb N$$
The asymptotic relation I think I originally started from, which think is actually false as I originally stated, but again, just curious about the division relation on the naturals I really am new to asymptotics:
$$sum _j=1^ lfloor x rfloor left( Bigllfloor
frac ln left( x right) ln left( p_j right)
Bigrrfloor +1 right) ln left( p_j right)+psi_0(x) sim x
$$
elementary-number-theory prime-numbers
asked Jul 15 at 19:59
Adam
13112
What is the source of this problem? Also, if I'm not mistaken, you're just asking for when $lfloor ln(psi_0(n)+psi_1(n))rfloor < lfloor sqrtnrfloor$, right?
– Carl Schildkraut
Jul 15 at 22:02
That's correct yes. It arose from when I was studying the Chebyshev functions and asymptotic relations in prime number theory, after referred to do so after posting another question here.
– Adam
Jul 15 at 22:09
If you can show me how to establish the bounds for that inequality, it would also be helpful I think, it was my impression they both increased without bound (denominator and numerator) and
– Adam
Jul 15 at 22:11
Well not entirely sir, as I said it arose from when I was studying the Chebyshev functions, so it was a variation of those in trying to reproduce in a manner that I could compute numerical data from. I suppose if I changed any part it may not hold the same divisibility relation
– Adam
Jul 15 at 22:18
I can't prove it, but it looks like $psi_1(n)sim psi_0(n)sim n$ (the second one follows from the prime number theorem but I can't prove the first yet).
– Carl Schildkraut
Jul 15 at 22:21
 |Â
show 6 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In looking at the two functions defined:
$$psi_0(x)=ln( operatornamelcm(1,2,3,...,lfloor x rfloor))$$
$$psi_1(x)=sum _j=1^ lfloor x rfloor sum _i=0^ Bigllfloor frac ln left( x right) ln left( p_j
right) Bigrrfloor +1ln left( p_j^i right)
$$
(where $p_n$ is the $n^th$ prime and $lcm$ denontes the lowest common multiple of the arguments enclosed)
I am interested in finding the minimum value (if it exists) of $n in mathbb N$ that satisfies:
$$ Bigglfracleft(lfloor ln left( psi_0(n)+psi_1(n)right)
rfloor +1 right) !lfloor sqrt n rfloor!Biggr ne 0$$
where $x$ denotes the fractional part of $x$.
some values evaluated:
$$fracleft(lfloor ln left( psi_0(8)+psi_1(8)right)
rfloor +1 right) !lfloor sqrt 8 rfloor!=12$$
$$fracleft(lfloor ln left( psi_0(12)+psi_1(12)right)
rfloor +1 right) !lfloor sqrt 12 rfloor!=20$$
$$fracleft(lfloor ln left( psi_0(20)+psi_1(20)right)
rfloor +1 right) !lfloor sqrt 20 rfloor!=5$$
The value is less than 1 at $n=36$ indicating this to be the immeadiate border of the region of $mathbb N$ for which the inequality Carl mentioned begins to be true (inductively reasoning).
Beyond $0<n<32$ I am not as yet able to produce a result, float approximations continue to imply that the value is 0 up to $n=40$, however really what is needed here is someone with more experience in number theory to assess the situation and give an opinion as to whether it is worth pursuing or not.
Closely related to the relations in a previous questions I posted here and here
I will try my best to follow along with Carl's answer, he has skipped a few steps that are probably what he may consider too obvious to show, but so far:
Because:
$$psi_0(x)=ln(operatornamelcm(1,2,3,...,x-1,lfloor xrfloor))=alpha,ln left( 2 right) +beta,ln left( 3 right)... +sum _j=
1^ lfloor x rfloor ln left( p_j right)
$$ for some $$alpha, beta,... in mathbb N$$
And similarly:
$$psi_1(x)=sum _j=1^ lfloor x rfloor sum _i=0^ Bigllfloor frac ln left( x right) ln left( p_j
right) Bigrrfloor +1ln left( p_j^i right) =frac12sum _j=1^ lfloor x rfloor ln left( p_j
right) left( Bigllfloor frac ln left( x right)
ln left( p_j right) Bigrrfloor +2 right) left( Bigllfloor frac ln left( x right) ln left( p_j right)
Bigrrfloor +1 right)$$
$$=alpha',ln left( 2 right) +beta',ln left( 3 right)... +sum _j=
1^ lfloor x rfloor ln left( p_j right) $$
for some $$alpha', beta',... in mathbb N$$
The asymptotic relation I think I originally started from, which think is actually false as I originally stated, but again, just curious about the division relation on the naturals I really am new to asymptotics:
$$sum _j=1^ lfloor x rfloor left( Bigllfloor
frac ln left( x right) ln left( p_j right)
Bigrrfloor +1 right) ln left( p_j right)+psi_0(x) sim x
$$
elementary-number-theory prime-numbers
asked Jul 15 at 19:59
Adam
13112
In looking at the two functions defined:
$$psi_0(x)=ln( operatornamelcm(1,2,3,...,lfloor x rfloor))$$
$$psi_1(x)=sum _j=1^ lfloor x rfloor sum _i=0^ Bigllfloor frac ln left( x right) ln left( p_j
right) Bigrrfloor +1ln left( p_j^i right)
$$
(where $p_n$ is the $n^th$ prime and $lcm$ denontes the lowest common multiple of the arguments enclosed)
I am interested in finding the minimum value (if it exists) of $n in mathbb N$ that satisfies:
$$ Bigglfracleft(lfloor ln left( psi_0(n)+psi_1(n)right)
rfloor +1 right) !lfloor sqrt n rfloor!Biggr ne 0$$
where $x$ denotes the fractional part of $x$.
some values evaluated:
$$fracleft(lfloor ln left( psi_0(8)+psi_1(8)right)
rfloor +1 right) !lfloor sqrt 8 rfloor!=12$$
$$fracleft(lfloor ln left( psi_0(12)+psi_1(12)right)
rfloor +1 right) !lfloor sqrt 12 rfloor!=20$$
$$fracleft(lfloor ln left( psi_0(20)+psi_1(20)right)
rfloor +1 right) !lfloor sqrt 20 rfloor!=5$$
The value is less than 1 at $n=36$ indicating this to be the immeadiate border of the region of $mathbb N$ for which the inequality Carl mentioned begins to be true (inductively reasoning).
Beyond $0<n<32$ I am not as yet able to produce a result, float approximations continue to imply that the value is 0 up to $n=40$, however really what is needed here is someone with more experience in number theory to assess the situation and give an opinion as to whether it is worth pursuing or not.
Closely related to the relations in a previous questions I posted here and here
I will try my best to follow along with Carl's answer, he has skipped a few steps that are probably what he may consider too obvious to show, but so far:
Because:
$$psi_0(x)=ln(operatornamelcm(1,2,3,...,x-1,lfloor xrfloor))=alpha,ln left( 2 right) +beta,ln left( 3 right)... +sum _j=
1^ lfloor x rfloor ln left( p_j right)
$$ for some $$alpha, beta,... in mathbb N$$
And similarly:
$$psi_1(x)=sum _j=1^ lfloor x rfloor sum _i=0^ Bigllfloor frac ln left( x right) ln left( p_j
right) Bigrrfloor +1ln left( p_j^i right) =frac12sum _j=1^ lfloor x rfloor ln left( p_j
right) left( Bigllfloor frac ln left( x right)
ln left( p_j right) Bigrrfloor +2 right) left( Bigllfloor frac ln left( x right) ln left( p_j right)
Bigrrfloor +1 right)$$
$$=alpha',ln left( 2 right) +beta',ln left( 3 right)... +sum _j=
1^ lfloor x rfloor ln left( p_j right) $$
for some $$alpha', beta',... in mathbb N$$
The asymptotic relation I think I originally started from, which think is actually false as I originally stated, but again, just curious about the division relation on the naturals I really am new to asymptotics:
$$sum _j=1^ lfloor x rfloor left( Bigllfloor
frac ln left( x right) ln left( p_j right)
Bigrrfloor +1 right) ln left( p_j right)+psi_0(x) sim x
$$
elementary-number-theory prime-numbers
asked Jul 15 at 19:59
Adam
13112
edited Jul 16 at 4:03
asked Jul 15 at 19:59
Adam
13112
asked Jul 15 at 19:59
Adam
13112
asked Jul 15 at 19:59
Adam
13112
13112
What is the source of this problem? Also, if I'm not mistaken, you're just asking for when $lfloor ln(psi_0(n)+psi_1(n))rfloor < lfloor sqrtnrfloor$, right?
– Carl Schildkraut
Jul 15 at 22:02
That's correct yes. It arose from when I was studying the Chebyshev functions and asymptotic relations in prime number theory, after referred to do so after posting another question here.
– Adam
Jul 15 at 22:09
If you can show me how to establish the bounds for that inequality, it would also be helpful I think, it was my impression they both increased without bound (denominator and numerator) and
– Adam
Jul 15 at 22:11
Well not entirely sir, as I said it arose from when I was studying the Chebyshev functions, so it was a variation of those in trying to reproduce in a manner that I could compute numerical data from. I suppose if I changed any part it may not hold the same divisibility relation
– Adam
Jul 15 at 22:18
I can't prove it, but it looks like $psi_1(n)sim psi_0(n)sim n$ (the second one follows from the prime number theorem but I can't prove the first yet).
– Carl Schildkraut
Jul 15 at 22:21
 |Â
show 6 more comments
What is the source of this problem? Also, if I'm not mistaken, you're just asking for when $lfloor ln(psi_0(n)+psi_1(n))rfloor < lfloor sqrtnrfloor$, right?
– Carl Schildkraut
Jul 15 at 22:02
That's correct yes. It arose from when I was studying the Chebyshev functions and asymptotic relations in prime number theory, after referred to do so after posting another question here.
– Adam
Jul 15 at 22:09
If you can show me how to establish the bounds for that inequality, it would also be helpful I think, it was my impression they both increased without bound (denominator and numerator) and
– Adam
Jul 15 at 22:11
Well not entirely sir, as I said it arose from when I was studying the Chebyshev functions, so it was a variation of those in trying to reproduce in a manner that I could compute numerical data from. I suppose if I changed any part it may not hold the same divisibility relation
– Adam
Jul 15 at 22:18
I can't prove it, but it looks like $psi_1(n)sim psi_0(n)sim n$ (the second one follows from the prime number theorem but I can't prove the first yet).
– Carl Schildkraut
Jul 15 at 22:21
What is the source of this problem? Also, if I'm not mistaken, you're just asking for when $lfloor ln(psi_0(n)+psi_1(n))rfloor < lfloor sqrtnrfloor$, right?
– Carl Schildkraut
Jul 15 at 22:02
What is the source of this problem? Also, if I'm not mistaken, you're just asking for when $lfloor ln(psi_0(n)+psi_1(n))rfloor < lfloor sqrtnrfloor$, right?
– Carl Schildkraut
Jul 15 at 22:02
That's correct yes. It arose from when I was studying the Chebyshev functions and asymptotic relations in prime number theory, after referred to do so after posting another question here.
– Adam
Jul 15 at 22:09
That's correct yes. It arose from when I was studying the Chebyshev functions and asymptotic relations in prime number theory, after referred to do so after posting another question here.
– Adam
Jul 15 at 22:09
If you can show me how to establish the bounds for that inequality, it would also be helpful I think, it was my impression they both increased without bound (denominator and numerator) and
– Adam
Jul 15 at 22:11
If you can show me how to establish the bounds for that inequality, it would also be helpful I think, it was my impression they both increased without bound (denominator and numerator) and
– Adam
Jul 15 at 22:11
Well not entirely sir, as I said it arose from when I was studying the Chebyshev functions, so it was a variation of those in trying to reproduce in a manner that I could compute numerical data from. I suppose if I changed any part it may not hold the same divisibility relation
– Adam
Jul 15 at 22:18
Well not entirely sir, as I said it arose from when I was studying the Chebyshev functions, so it was a variation of those in trying to reproduce in a manner that I could compute numerical data from. I suppose if I changed any part it may not hold the same divisibility relation
– Adam
Jul 15 at 22:18
I can't prove it, but it looks like $psi_1(n)sim psi_0(n)sim n$ (the second one follows from the prime number theorem but I can't prove the first yet).
– Carl Schildkraut
Jul 15 at 22:21
I can't prove it, but it looks like $psi_1(n)sim psi_0(n)sim n$ (the second one follows from the prime number theorem but I can't prove the first yet).
– Carl Schildkraut
Jul 15 at 22:21
 |Â
show 6 more comments
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
As
$$psi_0(x)=sum_p^ileq xln(p),$$
we have
beginalign
psi_0(x)+psi_1(x)
&=sum_pleq xleft[lfloorlog_p(x)rfloorln(p)+sum_i=0^lfloorlog_p(x)rfloor+1iln(p)right]\
&=frac12sum_pleq xln(p)left[lfloorlog_p(x)rfloor^2+5lfloorlog_p(x)rfloor+2right]\
&=vartheta(x)+3psi_0(x)+frac12left[sum_pleq sqrtx ln(p)left(lfloor log_p(x)rfloor^2-lfloor log_p(x)rfloorright)right]\
&=vartheta(x)+3psi_0(x)+frac12sum_pleq sqrtx ln(p)lfloor log_p(x)rfloorleft(lfloor log_p(x)rfloor-1right)\
&leq vartheta(x)+3psi_0(x)+fraclfloor log_2(x)rfloor-12sum_pleq sqrtx ln(p)lfloor log_p(x)rfloor\
&= vartheta(x)+3psi_0(x)+fraclfloor log_2(x)rfloor-12psi_0(sqrtx)\
&leq vartheta(x)+3psi_0(x)+log(x)psi_0(sqrtx).
endalign
As $psi_0(x)sim x$ by the Prime Number Theorem, this is asymptotic to
$$4x+csqrtxlog(x),$$
so
$$log(psi_0(x)+psi_1(x))sim log(4x) << sqrtx.$$
As your condition is
$$lfloor ln(psi_0(n)+psi_1(n))rfloor < lfloor sqrtnrfloor,$$
this is true for all sufficiently large $n$.
We claim that, for all $n>100$,
$$ln(psi_0(n)+psi_1(n)) + 1 < sqrtn.$$
Indeed, we already know
$$psi_0(n)+psi_1(n) leq vartheta(x)+3psi_0(x)+log(x)psi_0(sqrtx);$$
using bounds from here we have that
$$psi_0(x)+psi_1(x)leq 1.000028x+3cdot 1.03883x+1.03883left(log(x)sqrtxright);$$
as
$$frac12log(x)=log(sqrtx)sqrtxleq x,$$
we then have
$$psi_0(x)+psi_1(x)leq (4.116518)x+(2cdot 1.03883)x < 7x.$$
So, we only need to prove that
$$ln(7x)+1<sqrtx Leftrightarrow 0<sqrtx-1-ln(7x)$$
for $x>100$. At $x=100$, this is true. Its derivative is
$$frac12sqrtx-fracln(7)x=frac12xleft(sqrtx-2ln(7)right),$$
which is $>0$ if $x>4ln(7)^2$, which is true for all $x>100$. This finishes the proof.
I have numerically verified that your condition is true for all $ngeq 25$ but no $nleq 24$ using the following code:
def psi_sum(n):
prod=1
for p in sympy.ntheory.primerange(1,n+1):
n2=n
k=-1
while n2>0:
n2//=p
k+=1
v=(k**2+5*k+2)//2
prod*=(p**v)
return prod
f=lambda n:int(log(log(psi_sum(n))))
g=lambda n:int(sqrt(n))
print([n for n in range(2,101) if f(n)<g(n)])
answered Jul 15 at 23:10
Carl Schildkraut
8,28711238
Wow! It may take me some time to see how you were able to change things to a single sum
– Adam
Jul 15 at 23:13
Sorry can you please teach me how you are able to assert it to be true for all $n$ greater than or equal to 25, I have no troubles with the first 25,it's confidence beyond I can't do!
– Adam
Jul 15 at 23:22
I don't know it exactly, but I've checked it numerically through $100$. Later today I'll have more time to use explicit bounds on this - I should be able to prove it for all $n>100$ analytically.
– Carl Schildkraut
Jul 15 at 23:34
I'm sorry sir don't you mean $n$ up to 100, rather than beyond 100?
– Adam
Jul 15 at 23:43
@Adam This should be a complete answer now; see my latest edits.
– Carl Schildkraut
Jul 15 at 23:50
 |Â
show 1 more comment
up vote
0
down vote
$n=36$ is the answer to my original question
$$ Bigglfracleft(lfloor ln left( psi_0(36)+psi_1(36)right)
rfloor +1 right) !lfloor sqrt 36 rfloor!Biggr =frac16$$
answered Jul 16 at 4:04
Adam
13112
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
As
$$psi_0(x)=sum_p^ileq xln(p),$$
we have
beginalign
psi_0(x)+psi_1(x)
&=sum_pleq xleft[lfloorlog_p(x)rfloorln(p)+sum_i=0^lfloorlog_p(x)rfloor+1iln(p)right]\
&=frac12sum_pleq xln(p)left[lfloorlog_p(x)rfloor^2+5lfloorlog_p(x)rfloor+2right]\
&=vartheta(x)+3psi_0(x)+frac12left[sum_pleq sqrtx ln(p)left(lfloor log_p(x)rfloor^2-lfloor log_p(x)rfloorright)right]\
&=vartheta(x)+3psi_0(x)+frac12sum_pleq sqrtx ln(p)lfloor log_p(x)rfloorleft(lfloor log_p(x)rfloor-1right)\
&leq vartheta(x)+3psi_0(x)+fraclfloor log_2(x)rfloor-12sum_pleq sqrtx ln(p)lfloor log_p(x)rfloor\
&= vartheta(x)+3psi_0(x)+fraclfloor log_2(x)rfloor-12psi_0(sqrtx)\
&leq vartheta(x)+3psi_0(x)+log(x)psi_0(sqrtx).
endalign
As $psi_0(x)sim x$ by the Prime Number Theorem, this is asymptotic to
$$4x+csqrtxlog(x),$$
so
$$log(psi_0(x)+psi_1(x))sim log(4x) << sqrtx.$$
As your condition is
$$lfloor ln(psi_0(n)+psi_1(n))rfloor < lfloor sqrtnrfloor,$$
this is true for all sufficiently large $n$.
We claim that, for all $n>100$,
$$ln(psi_0(n)+psi_1(n)) + 1 < sqrtn.$$
Indeed, we already know
$$psi_0(n)+psi_1(n) leq vartheta(x)+3psi_0(x)+log(x)psi_0(sqrtx);$$
using bounds from here we have that
$$psi_0(x)+psi_1(x)leq 1.000028x+3cdot 1.03883x+1.03883left(log(x)sqrtxright);$$
as
$$frac12log(x)=log(sqrtx)sqrtxleq x,$$
we then have
$$psi_0(x)+psi_1(x)leq (4.116518)x+(2cdot 1.03883)x < 7x.$$
So, we only need to prove that
$$ln(7x)+1<sqrtx Leftrightarrow 0<sqrtx-1-ln(7x)$$
for $x>100$. At $x=100$, this is true. Its derivative is
$$frac12sqrtx-fracln(7)x=frac12xleft(sqrtx-2ln(7)right),$$
which is $>0$ if $x>4ln(7)^2$, which is true for all $x>100$. This finishes the proof.
I have numerically verified that your condition is true for all $ngeq 25$ but no $nleq 24$ using the following code:
def psi_sum(n):
prod=1
for p in sympy.ntheory.primerange(1,n+1):
n2=n
k=-1
while n2>0:
n2//=p
k+=1
v=(k**2+5*k+2)//2
prod*=(p**v)
return prod
f=lambda n:int(log(log(psi_sum(n))))
g=lambda n:int(sqrt(n))
print([n for n in range(2,101) if f(n)<g(n)])
answered Jul 15 at 23:10
Carl Schildkraut
8,28711238
Wow! It may take me some time to see how you were able to change things to a single sum
– Adam
Jul 15 at 23:13
Sorry can you please teach me how you are able to assert it to be true for all $n$ greater than or equal to 25, I have no troubles with the first 25,it's confidence beyond I can't do!
– Adam
Jul 15 at 23:22
I don't know it exactly, but I've checked it numerically through $100$. Later today I'll have more time to use explicit bounds on this - I should be able to prove it for all $n>100$ analytically.
– Carl Schildkraut
Jul 15 at 23:34
I'm sorry sir don't you mean $n$ up to 100, rather than beyond 100?
– Adam
Jul 15 at 23:43
@Adam This should be a complete answer now; see my latest edits.
– Carl Schildkraut
Jul 15 at 23:50
 |Â
show 1 more comment
up vote
3
down vote
accepted
As
$$psi_0(x)=sum_p^ileq xln(p),$$
we have
beginalign
psi_0(x)+psi_1(x)
&=sum_pleq xleft[lfloorlog_p(x)rfloorln(p)+sum_i=0^lfloorlog_p(x)rfloor+1iln(p)right]\
&=frac12sum_pleq xln(p)left[lfloorlog_p(x)rfloor^2+5lfloorlog_p(x)rfloor+2right]\
&=vartheta(x)+3psi_0(x)+frac12left[sum_pleq sqrtx ln(p)left(lfloor log_p(x)rfloor^2-lfloor log_p(x)rfloorright)right]\
&=vartheta(x)+3psi_0(x)+frac12sum_pleq sqrtx ln(p)lfloor log_p(x)rfloorleft(lfloor log_p(x)rfloor-1right)\
&leq vartheta(x)+3psi_0(x)+fraclfloor log_2(x)rfloor-12sum_pleq sqrtx ln(p)lfloor log_p(x)rfloor\
&= vartheta(x)+3psi_0(x)+fraclfloor log_2(x)rfloor-12psi_0(sqrtx)\
&leq vartheta(x)+3psi_0(x)+log(x)psi_0(sqrtx).
endalign
As $psi_0(x)sim x$ by the Prime Number Theorem, this is asymptotic to
$$4x+csqrtxlog(x),$$
so
$$log(psi_0(x)+psi_1(x))sim log(4x) << sqrtx.$$
As your condition is
$$lfloor ln(psi_0(n)+psi_1(n))rfloor < lfloor sqrtnrfloor,$$
this is true for all sufficiently large $n$.
We claim that, for all $n>100$,
$$ln(psi_0(n)+psi_1(n)) + 1 < sqrtn.$$
Indeed, we already know
$$psi_0(n)+psi_1(n) leq vartheta(x)+3psi_0(x)+log(x)psi_0(sqrtx);$$
using bounds from here we have that
$$psi_0(x)+psi_1(x)leq 1.000028x+3cdot 1.03883x+1.03883left(log(x)sqrtxright);$$
as
$$frac12log(x)=log(sqrtx)sqrtxleq x,$$
we then have
$$psi_0(x)+psi_1(x)leq (4.116518)x+(2cdot 1.03883)x < 7x.$$
So, we only need to prove that
$$ln(7x)+1<sqrtx Leftrightarrow 0<sqrtx-1-ln(7x)$$
for $x>100$. At $x=100$, this is true. Its derivative is
$$frac12sqrtx-fracln(7)x=frac12xleft(sqrtx-2ln(7)right),$$
which is $>0$ if $x>4ln(7)^2$, which is true for all $x>100$. This finishes the proof.
I have numerically verified that your condition is true for all $ngeq 25$ but no $nleq 24$ using the following code:
def psi_sum(n):
prod=1
for p in sympy.ntheory.primerange(1,n+1):
n2=n
k=-1
while n2>0:
n2//=p
k+=1
v=(k**2+5*k+2)//2
prod*=(p**v)
return prod
f=lambda n:int(log(log(psi_sum(n))))
g=lambda n:int(sqrt(n))
print([n for n in range(2,101) if f(n)<g(n)])
answered Jul 15 at 23:10
Carl Schildkraut
8,28711238
Wow! It may take me some time to see how you were able to change things to a single sum
– Adam
Jul 15 at 23:13
Sorry can you please teach me how you are able to assert it to be true for all $n$ greater than or equal to 25, I have no troubles with the first 25,it's confidence beyond I can't do!
– Adam
Jul 15 at 23:22
I don't know it exactly, but I've checked it numerically through $100$. Later today I'll have more time to use explicit bounds on this - I should be able to prove it for all $n>100$ analytically.
– Carl Schildkraut
Jul 15 at 23:34
I'm sorry sir don't you mean $n$ up to 100, rather than beyond 100?
– Adam
Jul 15 at 23:43
@Adam This should be a complete answer now; see my latest edits.
– Carl Schildkraut
Jul 15 at 23:50
 |Â
show 1 more comment
up vote
3
down vote
accepted
up vote
3
down vote
accepted
As
$$psi_0(x)=sum_p^ileq xln(p),$$
we have
beginalign
psi_0(x)+psi_1(x)
&=sum_pleq xleft[lfloorlog_p(x)rfloorln(p)+sum_i=0^lfloorlog_p(x)rfloor+1iln(p)right]\
&=frac12sum_pleq xln(p)left[lfloorlog_p(x)rfloor^2+5lfloorlog_p(x)rfloor+2right]\
&=vartheta(x)+3psi_0(x)+frac12left[sum_pleq sqrtx ln(p)left(lfloor log_p(x)rfloor^2-lfloor log_p(x)rfloorright)right]\
&=vartheta(x)+3psi_0(x)+frac12sum_pleq sqrtx ln(p)lfloor log_p(x)rfloorleft(lfloor log_p(x)rfloor-1right)\
&leq vartheta(x)+3psi_0(x)+fraclfloor log_2(x)rfloor-12sum_pleq sqrtx ln(p)lfloor log_p(x)rfloor\
&= vartheta(x)+3psi_0(x)+fraclfloor log_2(x)rfloor-12psi_0(sqrtx)\
&leq vartheta(x)+3psi_0(x)+log(x)psi_0(sqrtx).
endalign
As $psi_0(x)sim x$ by the Prime Number Theorem, this is asymptotic to
$$4x+csqrtxlog(x),$$
so
$$log(psi_0(x)+psi_1(x))sim log(4x) << sqrtx.$$
As your condition is
$$lfloor ln(psi_0(n)+psi_1(n))rfloor < lfloor sqrtnrfloor,$$
this is true for all sufficiently large $n$.
We claim that, for all $n>100$,
$$ln(psi_0(n)+psi_1(n)) + 1 < sqrtn.$$
Indeed, we already know
$$psi_0(n)+psi_1(n) leq vartheta(x)+3psi_0(x)+log(x)psi_0(sqrtx);$$
using bounds from here we have that
$$psi_0(x)+psi_1(x)leq 1.000028x+3cdot 1.03883x+1.03883left(log(x)sqrtxright);$$
as
$$frac12log(x)=log(sqrtx)sqrtxleq x,$$
we then have
$$psi_0(x)+psi_1(x)leq (4.116518)x+(2cdot 1.03883)x < 7x.$$
So, we only need to prove that
$$ln(7x)+1<sqrtx Leftrightarrow 0<sqrtx-1-ln(7x)$$
for $x>100$. At $x=100$, this is true. Its derivative is
$$frac12sqrtx-fracln(7)x=frac12xleft(sqrtx-2ln(7)right),$$
which is $>0$ if $x>4ln(7)^2$, which is true for all $x>100$. This finishes the proof.
I have numerically verified that your condition is true for all $ngeq 25$ but no $nleq 24$ using the following code:
def psi_sum(n):
prod=1
for p in sympy.ntheory.primerange(1,n+1):
n2=n
k=-1
while n2>0:
n2//=p
k+=1
v=(k**2+5*k+2)//2
prod*=(p**v)
return prod
f=lambda n:int(log(log(psi_sum(n))))
g=lambda n:int(sqrt(n))
print([n for n in range(2,101) if f(n)<g(n)])
answered Jul 15 at 23:10
Carl Schildkraut
8,28711238
As
$$psi_0(x)=sum_p^ileq xln(p),$$
we have
beginalign
psi_0(x)+psi_1(x)
&=sum_pleq xleft[lfloorlog_p(x)rfloorln(p)+sum_i=0^lfloorlog_p(x)rfloor+1iln(p)right]\
&=frac12sum_pleq xln(p)left[lfloorlog_p(x)rfloor^2+5lfloorlog_p(x)rfloor+2right]\
&=vartheta(x)+3psi_0(x)+frac12left[sum_pleq sqrtx ln(p)left(lfloor log_p(x)rfloor^2-lfloor log_p(x)rfloorright)right]\
&=vartheta(x)+3psi_0(x)+frac12sum_pleq sqrtx ln(p)lfloor log_p(x)rfloorleft(lfloor log_p(x)rfloor-1right)\
&leq vartheta(x)+3psi_0(x)+fraclfloor log_2(x)rfloor-12sum_pleq sqrtx ln(p)lfloor log_p(x)rfloor\
&= vartheta(x)+3psi_0(x)+fraclfloor log_2(x)rfloor-12psi_0(sqrtx)\
&leq vartheta(x)+3psi_0(x)+log(x)psi_0(sqrtx).
endalign
As $psi_0(x)sim x$ by the Prime Number Theorem, this is asymptotic to
$$4x+csqrtxlog(x),$$
so
$$log(psi_0(x)+psi_1(x))sim log(4x) << sqrtx.$$
As your condition is
$$lfloor ln(psi_0(n)+psi_1(n))rfloor < lfloor sqrtnrfloor,$$
this is true for all sufficiently large $n$.
We claim that, for all $n>100$,
$$ln(psi_0(n)+psi_1(n)) + 1 < sqrtn.$$
Indeed, we already know
$$psi_0(n)+psi_1(n) leq vartheta(x)+3psi_0(x)+log(x)psi_0(sqrtx);$$
using bounds from here we have that
$$psi_0(x)+psi_1(x)leq 1.000028x+3cdot 1.03883x+1.03883left(log(x)sqrtxright);$$
as
$$frac12log(x)=log(sqrtx)sqrtxleq x,$$
we then have
$$psi_0(x)+psi_1(x)leq (4.116518)x+(2cdot 1.03883)x < 7x.$$
So, we only need to prove that
$$ln(7x)+1<sqrtx Leftrightarrow 0<sqrtx-1-ln(7x)$$
for $x>100$. At $x=100$, this is true. Its derivative is
$$frac12sqrtx-fracln(7)x=frac12xleft(sqrtx-2ln(7)right),$$
which is $>0$ if $x>4ln(7)^2$, which is true for all $x>100$. This finishes the proof.
I have numerically verified that your condition is true for all $ngeq 25$ but no $nleq 24$ using the following code:
def psi_sum(n):
prod=1
for p in sympy.ntheory.primerange(1,n+1):
n2=n
k=-1
while n2>0:
n2//=p
k+=1
v=(k**2+5*k+2)//2
prod*=(p**v)
return prod
f=lambda n:int(log(log(psi_sum(n))))
g=lambda n:int(sqrt(n))
print([n for n in range(2,101) if f(n)<g(n)])
answered Jul 15 at 23:10
Carl Schildkraut
8,28711238
edited Jul 15 at 23:50
answered Jul 15 at 23:10
Carl Schildkraut
8,28711238
answered Jul 15 at 23:10
Carl Schildkraut
8,28711238
answered Jul 15 at 23:10
Carl Schildkraut
8,28711238
8,28711238
Wow! It may take me some time to see how you were able to change things to a single sum
– Adam
Jul 15 at 23:13
Sorry can you please teach me how you are able to assert it to be true for all $n$ greater than or equal to 25, I have no troubles with the first 25,it's confidence beyond I can't do!
– Adam
Jul 15 at 23:22
I don't know it exactly, but I've checked it numerically through $100$. Later today I'll have more time to use explicit bounds on this - I should be able to prove it for all $n>100$ analytically.
– Carl Schildkraut
Jul 15 at 23:34
I'm sorry sir don't you mean $n$ up to 100, rather than beyond 100?
– Adam
Jul 15 at 23:43
@Adam This should be a complete answer now; see my latest edits.
– Carl Schildkraut
Jul 15 at 23:50
 |Â
show 1 more comment
Wow! It may take me some time to see how you were able to change things to a single sum
– Adam
Jul 15 at 23:13
Sorry can you please teach me how you are able to assert it to be true for all $n$ greater than or equal to 25, I have no troubles with the first 25,it's confidence beyond I can't do!
– Adam
Jul 15 at 23:22
I don't know it exactly, but I've checked it numerically through $100$. Later today I'll have more time to use explicit bounds on this - I should be able to prove it for all $n>100$ analytically.
– Carl Schildkraut
Jul 15 at 23:34
I'm sorry sir don't you mean $n$ up to 100, rather than beyond 100?
– Adam
Jul 15 at 23:43
@Adam This should be a complete answer now; see my latest edits.
– Carl Schildkraut
Jul 15 at 23:50
Wow! It may take me some time to see how you were able to change things to a single sum
– Adam
Jul 15 at 23:13
Wow! It may take me some time to see how you were able to change things to a single sum
– Adam
Jul 15 at 23:13
Sorry can you please teach me how you are able to assert it to be true for all $n$ greater than or equal to 25, I have no troubles with the first 25,it's confidence beyond I can't do!
– Adam
Jul 15 at 23:22
Sorry can you please teach me how you are able to assert it to be true for all $n$ greater than or equal to 25, I have no troubles with the first 25,it's confidence beyond I can't do!
– Adam
Jul 15 at 23:22
I don't know it exactly, but I've checked it numerically through $100$. Later today I'll have more time to use explicit bounds on this - I should be able to prove it for all $n>100$ analytically.
– Carl Schildkraut
Jul 15 at 23:34
I don't know it exactly, but I've checked it numerically through $100$. Later today I'll have more time to use explicit bounds on this - I should be able to prove it for all $n>100$ analytically.
– Carl Schildkraut
Jul 15 at 23:34
I'm sorry sir don't you mean $n$ up to 100, rather than beyond 100?
– Adam
Jul 15 at 23:43
I'm sorry sir don't you mean $n$ up to 100, rather than beyond 100?
– Adam
Jul 15 at 23:43
@Adam This should be a complete answer now; see my latest edits.
– Carl Schildkraut
Jul 15 at 23:50
@Adam This should be a complete answer now; see my latest edits.
– Carl Schildkraut
Jul 15 at 23:50
 |Â
show 1 more comment
up vote
0
down vote
$n=36$ is the answer to my original question
$$ Bigglfracleft(lfloor ln left( psi_0(36)+psi_1(36)right)
rfloor +1 right) !lfloor sqrt 36 rfloor!Biggr =frac16$$
answered Jul 16 at 4:04
Adam
13112
add a comment |Â
up vote
0
down vote
$n=36$ is the answer to my original question
$$ Bigglfracleft(lfloor ln left( psi_0(36)+psi_1(36)right)
rfloor +1 right) !lfloor sqrt 36 rfloor!Biggr =frac16$$
answered Jul 16 at 4:04
Adam
13112
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$n=36$ is the answer to my original question
$$ Bigglfracleft(lfloor ln left( psi_0(36)+psi_1(36)right)
rfloor +1 right) !lfloor sqrt 36 rfloor!Biggr =frac16$$
answered Jul 16 at 4:04
Adam
13112
$n=36$ is the answer to my original question
$$ Bigglfracleft(lfloor ln left( psi_0(36)+psi_1(36)right)
rfloor +1 right) !lfloor sqrt 36 rfloor!Biggr =frac16$$
answered Jul 16 at 4:04
Adam
13112
answered Jul 16 at 4:04
Adam
13112
answered Jul 16 at 4:04
Adam
13112
answered Jul 16 at 4:04
Adam
13112
13112
add a comment |Â
add a comment |Â
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What is the source of this problem? Also, if I'm not mistaken, you're just asking for when $lfloor ln(psi_0(n)+psi_1(n))rfloor < lfloor sqrtnrfloor$, right?
– Carl Schildkraut
Jul 15 at 22:02
That's correct yes. It arose from when I was studying the Chebyshev functions and asymptotic relations in prime number theory, after referred to do so after posting another question here.
– Adam
Jul 15 at 22:09
If you can show me how to establish the bounds for that inequality, it would also be helpful I think, it was my impression they both increased without bound (denominator and numerator) and
– Adam
Jul 15 at 22:11
Well not entirely sir, as I said it arose from when I was studying the Chebyshev functions, so it was a variation of those in trying to reproduce in a manner that I could compute numerical data from. I suppose if I changed any part it may not hold the same divisibility relation
– Adam
Jul 15 at 22:18
I can't prove it, but it looks like $psi_1(n)sim psi_0(n)sim n$ (the second one follows from the prime number theorem but I can't prove the first yet).
– Carl Schildkraut
Jul 15 at 22:21