Prime Number Theorem Related functions Divisibility Relation Counter Sought

Question

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.

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:

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

Answer 1

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)])

Answer 2

$n=36$ is the answer to my original question

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