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In looking at the two functions defined: $$\psi_{{0}}(x)=\ln( \operatorname{lcm}(1,2,3,...,\lfloor x \rfloor))$$

$$\psi_{{1}}(x)=\sum _{j=1}^{ \lfloor x \rfloor } \sum _{i=0}^{ \Bigl\lfloor {\frac {\ln \left( x \right) }{\ln \left( p_{{j}} \right) }} \Bigr\rfloor +1}\ln \left( {p_{{j}}}^{i} \right) $$ (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:

$$ {\Biggl\{\frac{\left(\lfloor \ln \left( \psi_{{0}}(n)+\psi_{{1}}(n)\right) \rfloor +1 \right) !}{\lfloor \sqrt n \rfloor!}}\Biggr\} \ne 0$$

where ${\{x}\}$ denotes the fractional part of $x$.

some values evaluated: $$\frac{\left(\lfloor \ln \left( \psi_{{0}}(8)+\psi_{{1}}(8)\right) \rfloor +1 \right) !}{\lfloor \sqrt 8 \rfloor!}=12$$ $$\frac{\left(\lfloor \ln \left( \psi_{{0}}(12)+\psi_{{1}}(12)\right) \rfloor +1 \right) !}{\lfloor \sqrt 12 \rfloor!}=20$$

$$\frac{\left(\lfloor \ln \left( \psi_{{0}}(20)+\psi_{{1}}(20)\right) \rfloor +1 \right) !}{\lfloor \sqrt 20 \rfloor!}=5$$

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.enter image description here

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: $$\psi_{{0}}(x)=\ln(\operatorname{lcm}(1,2,3,...,x-1,\lfloor x\rfloor))=\alpha\,\ln \left( 2 \right) +\beta\,\ln \left( 3 \right)... +\sum _{j= 1}^{ \lfloor x \rfloor }\ln \left( p_{{j}} \right) $$ for some $$\alpha, \beta,... \in \mathbb N$$

And similarly:

$$\psi_{{1}}(x)=\sum _{j=1}^{ \lfloor x \rfloor } \sum _{i=0}^{ \Bigl\lfloor {\frac {\ln \left( x \right) }{\ln \left( p_{{j}} \right) }} \Bigr\rfloor +1}\ln \left( {p_{{j}}}^{i} \right) =\frac{1}{2}\sum _{j=1}^{ \lfloor x \rfloor }\ln \left( p_{{j}} \right) \left( \Bigl\lfloor {\frac {\ln \left( x \right) }{ \ln \left( p_{{j}} \right) }} \Bigr\rfloor +2 \right) \left( \Bigl\lfloor {\frac {\ln \left( x \right) }{\ln \left( p_{{j}} \right) }} \Bigr\rfloor +1 \right)$$ $$=\alpha'\,\ln \left( 2 \right) +\beta'\,\ln \left( 3 \right)... +\sum _{j= 1}^{ \lfloor x \rfloor }\ln \left( p_{{j}} \right) $$ 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:

$$\sum _{j=1}^{ \lfloor x \rfloor } \left( \Bigl\lfloor {\frac {\ln \left( x \right) }{\ln \left( p_{{j}} \right) }} \Bigr\rfloor +1 \right) \ln \left( p_{{j}} \right)+\psi_{{0}}(x) \sim x $$ enter image description here

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  • $\begingroup$ What is the source of this problem? Also, if I'm not mistaken, you're just asking for when $\lfloor \ln(\psi_0(n)+\psi_1(n))\rfloor < \lfloor \sqrt{n}\rfloor$, right? $\endgroup$ – Carl Schildkraut Jul 15 '18 at 22:02
  • $\begingroup$ 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. $\endgroup$ – Adam Jul 15 '18 at 22:09
  • $\begingroup$ 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 $\endgroup$ – Adam Jul 15 '18 at 22:11
  • $\begingroup$ 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 $\endgroup$ – Adam Jul 15 '18 at 22:18
  • $\begingroup$ 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). $\endgroup$ – Carl Schildkraut Jul 15 '18 at 22:21
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As

$$\psi_0(x)=\sum_{p^i\leq x}\ln(p),$$

we have

\begin{align} \psi_0(x)+\psi_1(x) &=\sum_{p\leq x}\left[\lfloor\log_p(x)\rfloor\ln(p)+\sum_{i=0}^{\lfloor\log_p(x)\rfloor+1}i\ln(p)\right]\\ &=\frac{1}{2}\sum_{p\leq x}\ln(p)\left[\lfloor\log_p(x)\rfloor^2+5\lfloor\log_p(x)\rfloor+2\right]\\ &=\vartheta(x)+3\psi_0(x)+\frac{1}{2}\left[\sum_{p\leq \sqrt{x}} \ln(p)\left(\lfloor \log_p(x)\rfloor^2-\lfloor \log_p(x)\rfloor\right)\right]\\ &=\vartheta(x)+3\psi_0(x)+\frac{1}{2}\sum_{p\leq \sqrt{x}} \ln(p)\lfloor \log_p(x)\rfloor\left(\lfloor \log_p(x)\rfloor-1\right)\\ &\leq \vartheta(x)+3\psi_0(x)+\frac{\lfloor \log_2(x)\rfloor-1}{2}\sum_{p\leq \sqrt{x}} \ln(p)\lfloor \log_p(x)\rfloor\\ &= \vartheta(x)+3\psi_0(x)+\frac{\lfloor \log_2(x)\rfloor-1}{2}\psi_0(\sqrt{x})\\ &\leq \vartheta(x)+3\psi_0(x)+\log(x)\psi_0(\sqrt{x}). \end{align}

As $\psi_0(x)\sim x$ by the Prime Number Theorem, this is asymptotic to

$$4x+c\sqrt{x}\log(x),$$

so

$$\log(\psi_0(x)+\psi_1(x))\sim \log(4x) << \sqrt{x}.$$

As your condition is

$$\lfloor \ln(\psi_0(n)+\psi_1(n))\rfloor < \lfloor \sqrt{n}\rfloor,$$

this is true for all sufficiently large $n$.


We claim that, for all $n>100$,

$$\ln(\psi_0(n)+\psi_1(n)) + 1 < \sqrt{n}.$$

Indeed, we already know

$$\psi_0(n)+\psi_1(n) \leq \vartheta(x)+3\psi_0(x)+\log(x)\psi_0(\sqrt{x});$$

using bounds from here we have that

$$\psi_0(x)+\psi_1(x)\leq 1.000028x+3\cdot 1.03883x+1.03883\left(\log(x)\sqrt{x}\right);$$

as

$$\frac{1}{2}\log(x)=\log(\sqrt{x})\sqrt{x}\leq x,$$

we then have

$$\psi_0(x)+\psi_1(x)\leq (4.116518)x+(2\cdot 1.03883)x < 7x.$$

So, we only need to prove that

$$\ln(7x)+1<\sqrt{x} \Leftrightarrow 0<\sqrt{x}-1-\ln(7x)$$

for $x>100$. At $x=100$, this is true. Its derivative is

$$\frac{1}{2\sqrt{x}}-\frac{\ln(7)}{x}=\frac{1}{2x}\left(\sqrt{x}-2\ln(7)\right),$$

which is $>0$ if $x>4\ln(7)^2$, which is true for all $x>100$. This finishes the proof.

I have numerically verified that your condition is true for all $n\geq 25$ but no $n\leq 24$ 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)])
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  • $\begingroup$ Wow! It may take me some time to see how you were able to change things to a single sum $\endgroup$ – Adam Jul 15 '18 at 23:13
  • $\begingroup$ 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! $\endgroup$ – Adam Jul 15 '18 at 23:22
  • $\begingroup$ 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. $\endgroup$ – Carl Schildkraut Jul 15 '18 at 23:34
  • $\begingroup$ I'm sorry sir don't you mean $n$ up to 100, rather than beyond 100? $\endgroup$ – Adam Jul 15 '18 at 23:43
  • $\begingroup$ @Adam This should be a complete answer now; see my latest edits. $\endgroup$ – Carl Schildkraut Jul 15 '18 at 23:50
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$n=36$ is the answer to my original question

$$ {\Biggl\{\frac{\left(\lfloor \ln \left( \psi_{{0}}(36)+\psi_{{1}}(36)\right) \rfloor +1 \right) !}{\lfloor \sqrt 36 \rfloor!}}\Biggr\} =\frac{1}{6}$$

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