7
$\begingroup$

From Terry Tao's post here there is the statement:

"Conversely, if one can somehow establish a bound of the form

$$\displaystyle \sum_{n \leq x} \Lambda(n) = x + O( x^{1/2+\epsilon} ) \tag{1}$$

for any fixed ${\epsilon}$, then the explicit formula can be used to..."

I don't know about the word "fixed", but the irregular behaviour of the blue curve below gives plenty of room for an ${\epsilon}$, if it is true that the asymptotic is $(8x)^{1/2}$, and if it is also true that it bounds the partial sums of the Möbius transform of the Harmonic numbers minus $x$. But we don't know and can't conclude any such bounds from this question. I am only asking about the asymptotics of a certain sum that is connected to / a truncated absolute value version of the numerators of the expansion of the primes.

Let:

$$\varphi^{-1}(n) = \sum_{d \mid n} \mu(d)d \tag{2}$$

Then for $n>1$:

$$\Lambda(n) = \sum\limits_{k=1}^{\infty}\frac{\varphi^{-1}(\gcd(n,k))}{k} \tag{3}$$

Form the table: $$A(n,k)=\sum_{\substack{i=k\\\ n \geq k}}^n \varphi^{-1}(\gcd (i,k)) \tag{4}$$

From numerical evidence it appears that:

$$\sum _{k=1}^{x} \text{sgn}\left(\left(\text{sgn}\left(x+\sum _{j=2}^k -|A(x,j)|\right)+1\right)\right)+1 \sim (8x)^{1/2} \tag{5}$$

Is it true or is the asymptotic something else?

Question:

The complicated sign formula in $(5)$ comes from what we are really doing which is to ask: What is the asymptotic of the least $k$ for which the function $F(x)$:

$$F(x)=x+\sum _{j=2}^k -|A(x,j)| \tag{6}$$

is negative? For $k=1..x$.

Plot of the numerical evidence where the irregular blue curve is that least $k$ for which the function $F(x)$ is negative and thereby also the LHS of (5) while the smooth red curve is the conjectured asymptotic $(8x)^{1/2}$:

what asymptotics sqrt 8x

Efficient Mathematica program to generate the plot. Setting nn=10000 gives the plot above:

(*start*)
(*Mathematica*)
Clear[a, f, p];
nn = 1000;
p = 0;
f[n_] := n*Log[n]^p;
(*Clear[f];*)
(*f[n_] := n*Log[n]^4/(Pi*8)^2/8;*)
a[n_] := DivisorSum[n, MoebiusMu[#] # &];
Monitor[TableForm[
   A = Accumulate[
     Table[Table[If[n >= k, a[GCD[n, k]], 0], {k, 1, nn}], {n, 1, 
       nn}]]];, n]
TableForm[B = -Abs[A]];
Clear[A];
B[[All, 1]] = N[Table[f[n], {n, 1, nn}]];
TableForm[B];
TableForm[B1 = Sign[Transpose[Accumulate[Transpose[B]]]]];
Clear[B];
Quiet[Show[
  ListLinePlot[
   v = ReplaceAll[
     Flatten[Table[First[Position[B1[[n]], -1]], {n, 1, nn}]], 
     First[{}] -> 1], PlotStyle -> Blue], 
  Plot[Sqrt[8*f[n]], {n, 1, nn}, PlotStyle -> {Red, Thick}], 
  ImageSize -> Large]]
ListLinePlot[v/Table[Sqrt[8*f[n]], {n, 1, nn}]]
(*end*)

Variant of the Mathematica program above: https://pastebin.com/GJ81MQez

Inefficient Mathematica program to generate the LHS in (5):

Clear[varphi];
nn = 20;
constant = 2*Sqrt[2];
varphi[n_] := Total[Divisors[n]*MoebiusMu[Divisors[n]]];
Monitor[TableForm[
   A = Table[
     Table[Sum[If[n >= k, varphi[GCD[i, k]], 0], {i, k, n}], {k, 1, 
       nn}], {n, 1, nn}]];, n]
Table[1 + 
  Sum[Sign[(1 + Sign[x + Sum[-Abs[A[[x, j]]], {j, 2, k}]])], {k, 1, 
    x}], {x, 1, nn}]

which starts: {2, 3, 4, 5, 6, 5, 7, 7, 10, 7, 11, 10, 11, 10, 11, 11, 14, 13, 14, 13}

For my own memory to remember where to start editing tomorrow I write this Mathematica program:

Clear[varphi];
nn = 40;
varphi[n_] := Total[Divisors[n]*MoebiusMu[Divisors[n]]];
Table[1 + 
  Sum[Sign[(1 + 
      Sign[x + 
        Sum[-Abs[
           Sum[If[x >= j, varphi[GCD[i, j]], 0], {i, j, x}]], {j, 2, 
          k}]])], {k, 1, x}], {x, 1, nn}]

There are previous efforts related to this question. Here is one of them.

A construction: $$\sqrt{x} \log ^2(x)=\sqrt{x} \left(x-\left(\sqrt{x}-\log (x)\right) \left(\sqrt{x}+\log (x)\right)\right)$$

$\endgroup$
  • $\begingroup$ Proof of von Mangoldt function formula: math.stackexchange.com/a/51708/8530, mathoverflow.net/a/162214/25104 $\endgroup$ – Mats Granvik May 1 at 11:05
  • $\begingroup$ Sqrt(8) in the OEIS: oeis.org/A010466 $\endgroup$ – Mats Granvik May 1 at 11:08
  • 2
    $\begingroup$ Could you say clearly if the "blue curve" represents $\sum_{n \leq x} \Lambda(n) -x$ or something else? $\endgroup$ – YCor May 28 at 20:11
  • $\begingroup$ The irregular blue curve is not $\sum_{n \leq x} \Lambda(n) -x$. I will try to edit and clarify. It uses negated absolute values of the terms in the same expansion as $\sum_{n \leq x} \Lambda(n)$ though. $\endgroup$ – Mats Granvik May 28 at 20:33
  • 3
    $\begingroup$ You don't have to give a complicated formula with signs, you can just say "Let $F(n)$ be the least $k$ such that .... is negative" $\endgroup$ – Will Sawin May 28 at 21:02
9
$\begingroup$

Let us denote the left hand side of $(1)$ by $\psi(x)$. It is known that $|\psi(x)-x|$ is not bounded by a constant times $x^{1/2}$. In fact Littlewood (1914) proved that $$\psi(x)-x=\Omega_{\pm}(x^{1/2}\log\log\log x).$$ This is Theorem 15.11 in Montgomery-Vaughan: Multiplicative number theory I.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Setting:

$$F(x)=x\log(x)+\sum _{j=2}^k -|A(x,j)| \tag{1}$$

appears to give the asymptotic $\sqrt{8x\log(x)}$ for the least $k$ such that $F(x)$ is negative.

In general it appears that the least $k$ such that:

$$F(x)=f(x)+\sum _{j=2}^k -|A(x,j)| \tag{2}$$

is negative, has the asymptotic: $\sqrt{8f(x)}$.

See the Mathematica program in the question, by setting p=1.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.