4
$\begingroup$

Send me back to mathematics stack exchange if this question is not research level.

At Terence Tao's blog post there is the expression:

$$\sum\limits_{n \leq X} \Lambda(n)\Lambda(n+h) \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$

It has been proven that the von Mangoldt function is:

$$\Lambda(n)=\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}$$

which for each $n$ has an associated Dirichlet series.

Therefore I made a program that looks at the resulting matrix of Dirichlet series of the convolutions:

$$\Lambda(n)\Lambda(n+h) = \left(\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}\right)\left( \lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n+h} \frac{\mu(d)}{d^{(s-1)}}\right)$$

Example: For $h=2$ this is a matrix starting:

$$M=\left( \begin{array}{cccccc} 1 & 1 & 1 & 1 & 1 & 1 & \dots \\ 2 & -2 & 2 & -2 & 2 & -2 & \\ -1 & 2 & -1 & -1 & 2 & -1 & \\ 3 & -1 & 3 & -1 & 3 & -1 & \\ 2 & 2 & -3 & 2 & -3 & 2 & \\ -2 & -4 & -2 & 2 & 4 & 2 & \\ 2 & 2 & 2 & 2 & -5 & 2 & \\ 4 & 0 & 4 & 0 & 4 & 0 & \\ \vdots & & & & & & \ddots \end{array} \right)$$

It appears that the rows in matrix $M$ are periodic and it therefore suffices to look at the row sums of the lower triangular matrix:

$$R=\left( \begin{array}{cccccc} 1 & & & & & & \dots \\ 2 & -2 & & & & & \\ -1 & 2 & -1 & & & & \\ 3 & -1 & 3 & -1 & & & \\ 2 & 2 & -3 & 2 & -3 & & \\ -2 & -4 & -2 & 2 & 4 & 2 & \\ \vdots & & & & & & \ddots \end{array} \right)$$

and compare them for different $h$.

In the program below I have compared $h=2$ to $h=2,3,4,5,6,7,8,...32$ and I notice that the row sums of $R$ are the same for $h$ a power of $2$ although the entries in matrices $M$ and $R$ are different for different $h$.

(*start*)
Clear[ConvolvedMangoldt];
Monitor[Table[h = 2;
  nn = 140;
  A = Table[Table[If[Mod[n, k] == 0, 1, 0], {k, 1, nn}], {n, 1, nn}];
  B = Table[
    Table[If[Mod[k, n] == 0, MoebiusMu[n]*n, 0], {k, 1, nn}], {n, 1, 
     nn}];
  MatrixForm[T = (A.B)];
  TableForm[ConvolvedMangoldt = Table[a = T[[All, kk]];
     F1 = 
      Table[Table[If[Mod[n, k] == 0, a[[n/k]], 0], {k, 1, nn}], {n, 1,
         nn}];
     b = T[[All, kk + h]];
     F2 = 
      Table[Table[If[Mod[n, k] == 0, b[[n/k]], 0], {k, 1, nn}], {n, 1,
         nn}];
     (F1.F2)[[All, 1]], {kk, 1, nn - h}]];
  Flatten[
   Table[Table[ConvolvedMangoldt[[n - k + 1, k]], {k, 1, n}], {n, 1, 
     nn - h}]];
  TableForm[
   Table[Table[ConvolvedMangoldt[[n - k + 1, k]], {k, 1, n}], {n, 1, 
     nn - h}]];
  aa = Table[
    Sum[ConvolvedMangoldt[[k, n]], {k, 1, n}], {n, 1, nn - h}];
  h = hh;
  nn = 140;
  A = Table[Table[If[Mod[n, k] == 0, 1, 0], {k, 1, nn}], {n, 1, nn}];
  B = Table[
    Table[If[Mod[k, n] == 0, MoebiusMu[n]*n, 0], {k, 1, nn}], {n, 1, 
     nn}];
  MatrixForm[T = (A.B)];
  TableForm[ConvolvedMangoldt = Table[a = T[[All, kk]];
     F1 = 
      Table[Table[If[Mod[n, k] == 0, a[[n/k]], 0], {k, 1, nn}], {n, 1,
         nn}];
     b = T[[All, kk + h]];
     F2 = 
      Table[Table[If[Mod[n, k] == 0, b[[n/k]], 0], {k, 1, nn}], {n, 1,
         nn}];
     (F1.F2)[[All, 1]], {kk, 1, nn - h}]];
  Flatten[
   Table[Table[ConvolvedMangoldt[[n - k + 1, k]], {k, 1, n}], {n, 1, 
     nn - h}]];
  TableForm[
   Table[Table[ConvolvedMangoldt[[n - k + 1, k]], {k, 1, n}], {n, 1, 
     nn - h}]];
  bb = Table[
    Sum[ConvolvedMangoldt[[k, n]], {k, 1, n}], {n, 1, nn - h}];
  {" hh = ", hh, 
   aa[[Range[100]]] - bb[[Range[100]]]}, {hh, 2, 32}], hh]
(*end*)

I have put the output here at Pastebin: https://pastebin.com/WiU9GSm0

The program explains better my comparison, but I don't have the intelligence to write out in latex what the program compares.

But the question is the same anyways. Is it known that powers of $h$ a prime gap, have the same density?

The program is written in Mathematica.

Edit 18.2.2018:

An improved version of the program where one can vary the prime gap parameter "p".

(*start*)
p = 6
Do[
 nn = 140;
 h = p; (* h = 2 gives twin primes *)
 hh = hhh; (* hh = 4 gives cousin primes *)
 TableForm[
  A = Table[Table[If[Mod[n, k] == 0, 1, 0], {k, 1, nn}], {n, 1, nn}]];
 TableForm[
  B = Table[
    Table[If[Mod[k, n] == 0, MoebiusMu[n]*n, 0], {k, 1, nn}], {n, 1, 
     nn}]];
 TableForm[T = (A.B)];
 (*Print["h = ",h]*)
 TableForm[ConvolvedMangoldt = Table[
    a = T[[All, nnn]];
    b = T[[All, nnn + h]];
    Table[
     Sum[If[Mod[n, k] == 0, a[[n/k]]*b[[k]], 0], {k, 1, nn - h}], {n, 
      1, nn - h}], {nnn, 1, nn - h}]];
 TableForm[CMT = Transpose[ConvolvedMangoldt]];
 TableForm[
  AA = Table[
    Sum[CMT[[n, k]], {k, 1, n}], {n, 1, Length[CMT] - (hh - h)}]];
 (*TableForm[AAA=Table[Sort[Table[CMT[[n,k]],{k,1,n}]],{n,1,Length[\
CMT]-(hh-h)}]];*)
 (*Print["h = ",hh]*)
 h = hh;
 TableForm[ConvolvedMangoldt = Table[
    a = T[[All, nnn]];
    b = T[[All, nnn + h]];
    Table[
     Sum[If[Mod[n, k] == 0, a[[n/k]]*b[[k]], 0], {k, 1, nn - h}], {n, 
      1, nn - h}], {nnn, 1, nn - h}]];
 TableForm[CMT = Transpose[ConvolvedMangoldt]];
 TableForm[
  BB = Table[Sum[CMT[[n, k]], {k, 1, n}], {n, 1, Length[CMT]}]];
 (*TableForm[BBB=Table[Sort[Table[CMT[[n,k]],{k,1,n}]],{n,1,Length[\
CMT]}]];*)
 Print["h = ", hh, " ", Total[BB - AA]]
 (*Print["The difference between the sorted rows are zero"]*)
 (*BBB-AAA;*)
 , {hhh, p, 100}]
(*end*)

Edit 19.2.2018:

Anyways,

$$\Lambda(n)\Lambda(n+h) = \left(\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}\right)\left( \lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n+h} \frac{\mu(d)}{d^{(s-1)}}\right)$$

where $n$ is unconventionally the column index.

For $h=2$, the convolution above is associated with the matrix of Dirichlet series:

$$M_{h=2}\left(\begin{array}{lllllllll} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & \cdots\\ 2 & -2 & 2 & -2 & 2 & -2 & 2 & -2 \\ -1 & 2 & -1 & -1 & 2 & -1 & -1 & 2 \\ 3 & -1 & 3 & -1 & 3 & -1 & 3 & -1 \\ 2 & 2 & -3 & 2 & -3 & 2 & 2 & -3 \\ -2 & -4 & -2 & 2 & 4 & 2 & -2 & -4 \\ 2 & 2 & 2 & 2 & -5 & 2 & -5 & 2 \\ 4 & 0 & 4 & 0 & 4 & 0 & 4 & 0 \\ \vdots & & & & & & & &\ddots \\ \end{array}\right)$$

$R_{h=2}$ is equal to $M_{h=2}$ for $\text{ row index } \geq \text{ column index}$:

$$R_{h=2}=\left(\begin{array}{lllllllll} 1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} &\cdots \\ 2 & -2 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ -1 & 2 & -1 & \text{} & \text{} & \text{} & \text{} & \text{} \\ 3 & -1 & 3 & -1 & \text{} & \text{} & \text{} & \text{} \\ 2 & 2 & -3 & 2 & -3 & \text{} & \text{} & \text{} \\ -2 & -4 & -2 & 2 & 4 & 2 & \text{} & \text{} \\ 2 & 2 & 2 & 2 & -5 & 2 & -5 & \text{} \\ 4 & 0 & 4 & 0 & 4 & 0 & 4 & 0 \\ \vdots & & & & & & & & \ddots \end{array}\right)$$

Sorting the entries in the rows of $R_{h=2}$ we get:

$$R_{h=2,\text{ sorted rows}}=\left(\begin{array}{lllllllll} 1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \cdots\\ -2 & 2 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ -1 & -1 & 2 & \text{} & \text{} & \text{} & \text{} & \text{} \\ -1 & -1 & 3 & 3 & \text{} & \text{} & \text{} & \text{} \\ -3 & -3 & 2 & 2 & 2 & \text{} & \text{} & \text{} \\ -4 & -2 & -2 & 2 & 2 & 4 & \text{} & \text{} \\ -5 & -5 & 2 & 2 & 2 & 2 & 2 & \text{} \\ 0 & 0 & 0 & 0 & 4 & 4 & 4 & 4 \\ \vdots & & & & & & & & \ddots \\ \end{array}\right)$$

Again:

$$\Lambda(n)\Lambda(n+h) = \left(\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}\right)\left( \lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n+h} \frac{\mu(d)}{d^{(s-1)}}\right)$$

where $n$ is unconventionally the column index.

For $h=4$, the convolution above is associated with the matrix of Dirichlet series:

$$M_{h=4}\left(\begin{array}{lllllllll} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & \cdots\\ 2 & -2 & 2 & -2 & 2 & -2 & 2 & -2 \\ 2 & -1 & -1 & 2 & -1 & -1 & 2 & -1 \\ 3 & -1 & 3 & -1 & 3 & -1 & 3 & -1 \\ -3 & 2 & 2 & 2 & -3 & -3 & 2 & 2 \\ 4 & 2 & -2 & -4 & -2 & 2 & 4 & 2 \\ 2 & 2 & -5 & 2 & 2 & 2 & -5 & 2 \\ 4 & 0 & 4 & 0 & 4 & 0 & 4 & 0 \\ \vdots & & & & & & & & \ddots \end{array}\right)$$

$R_{h=4}$ is equal to $M_{h=4}$ for $\text{ row index } \geq \text{ column index}$:

$$R_{h=4}=\left(\begin{array}{lllllllll} 1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \cdots \\ 2 & -2 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 2 & -1 & -1 & \text{} & \text{} & \text{} & \text{} & \text{} \\ 3 & -1 & 3 & -1 & \text{} & \text{} & \text{} & \text{} \\ -3 & 2 & 2 & 2 & -3 & \text{} & \text{} & \text{} \\ 4 & 2 & -2 & -4 & -2 & 2 & \text{} & \text{} \\ 2 & 2 & -5 & 2 & 2 & 2 & -5 & \text{} \\ 4 & 0 & 4 & 0 & 4 & 0 & 4 & 0 \\ \vdots & & & & & & & & \ddots \end{array}\right)$$

Sorting the entries in the rows of $R_{h=4}$ we get:

$$R_{h=4,\text{ sorted rows}}=\left(\begin{array}{lllllllll} 1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \cdots \\ -2 & 2 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ -1 & -1 & 2 & \text{} & \text{} & \text{} & \text{} & \text{} \\ -1 & -1 & 3 & 3 & \text{} & \text{} & \text{} & \text{} \\ -3 & -3 & 2 & 2 & 2 & \text{} & \text{} & \text{} \\ -4 & -2 & -2 & 2 & 2 & 4 & \text{} & \text{} \\ -5 & -5 & 2 & 2 & 2 & 2 & 2 & \text{} \\ 0 & 0 & 0 & 0 & 4 & 4 & 4 & 4 \\ \vdots & & & & & & & & \ddots \end{array}\right)$$

which looks very much as:

$$R_{h=2,\text{ sorted rows}}=\left(\begin{array}{lllllllll} 1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \cdots\\ -2 & 2 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ -1 & -1 & 2 & \text{} & \text{} & \text{} & \text{} & \text{} \\ -1 & -1 & 3 & 3 & \text{} & \text{} & \text{} & \text{} \\ -3 & -3 & 2 & 2 & 2 & \text{} & \text{} & \text{} \\ -4 & -2 & -2 & 2 & 2 & 4 & \text{} & \text{} \\ -5 & -5 & 2 & 2 & 2 & 2 & 2 & \text{} \\ 0 & 0 & 0 & 0 & 4 & 4 & 4 & 4 \\ \vdots & & & & & & & & \ddots \\ \end{array}\right)$$

So proving that the asymptotic density of the twin primes $(h=2)$ is equal to asymptotic density of the cousin primes $(h=4)$, appears to be equivalent to proving:

$$R_{h=2,\text{ sorted rows}}=R_{h=4,\text{ sorted rows}}$$

Edit 28.2.2018:

In an attempt to verify $R_{h=2}=R_{h=4}$ I wrote this program that compares rows in $R_{h=2}$ and $R_{h=4}$. The shifts are found in the output from the program. Squares of primes is a problem in the comparison, but inspection shows that they sofar match entrywise between $R_{h=2}$ and $R_{h=4}$.

(*start*)
nn = 230;
h = 2; (* h = 2 gives twin primes *)
hh = 4; (* hh = 4 gives cousin primes *)
(* TableForm[A=Table[Table[If[Mod[n,k]==0,1,0],{k,1,nn}],{n,1,nn}]];
TableForm[B=Table[Table[If[Mod[k,n]==0,MoebiusMu[n]*n,0],{k,1,nn}],{n,\
1,nn}]];
TableForm[T=(A.B)]; *)
Clear[a, n, d];
a[n_] := If[n < 1, 0, 
  Sum[d MoebiusMu@d, {d, Divisors[n]}]] (*Michael Somos,Jul 18 2011*)


TableForm[T = Table[Table[a[GCD[n, k]], {k, 1, nn}], {n, 1, nn}]];
nnn = 12;
TableForm[AA = Table[
    Divisors[n];
    Clear[a, b];
    TableForm[c1 = Table[a = T[[Reverse[Divisors[n]], k]];
       b = T[[Divisors[n], k + h]];
       a*b, {k, 1, n}]];
    TableForm[A = Reverse[Transpose[c1]]];
    TableForm[c1 = Table[a = T[[Reverse[Divisors[n]], k]];
       b = T[[Divisors[n], k + hh]];
       a*b, {k, 1, 2*n - 1}]];
    TableForm[B = Reverse[Transpose[c1]]];
    TableForm[Table[A[[1]] -
       B[[1]][[n + Range[Length[A[[1]]]]]], {n, 
       0, -n + Length[B[[1]]]}]];
    Table[Min[Flatten[Position[Table[A[[k]] -
          B[[k]][[n + Range[Length[A[[k]]]]]], {n, 
          0, -n + Length[B[[k]]]}],
        Range[n]*0]]], {k, 1, Length[Divisors[n]]}], {n, 1, nnn}]];
Table[AA[[n]][[1]], {n, 1, nnn}]
AA = Table[Reverse[AA[[n]]], {n, 1, nnn}]
TableForm[
  BB = Table[
    Reverse[Accumulate[
       Reverse[Table[If[Mod[n, k] == 0, 1, 0], {k, 1, nnn}]]]]*
     Table[If[Mod[n, k] == 0, 1, 0], {k, 1, nnn}], {n, 1, nnn}]];
TableForm[
  CC = Table[
    Table[If[Mod[n, k] == 0, AA[[n, BB[[n, k]]]], 0], {k, 1, 
      nnn}], {n, 1, nnn}]];
MatrixForm[ReplaceAll[CC, Infinity -> 0]]
(*end*)

$\text{Shifts between rows in } R_{h=2} and R_{h=4}$ $$=\left( \begin{array}{cccccccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 4 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 5 & 5 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 6 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 2 & 0 & 3 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 9 & 9 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 10 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 5 & 2 & 1 & 5 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \end{array} \right)$$

Improve this question
$\endgroup$
10
$\begingroup$

Yes - conjecturally.

A special case of the Hardy-Littlewood prime-tuple conjecture states that, for any even $\Delta$, $$\lim_{x \to \infty} \frac{\sum_{1 \le n\le x} \Lambda(n)\Lambda(n+\Delta)}{x} =C(\Delta),$$ where $$C(\Delta) = \prod_{p \text{ a prime}} \frac{1-\frac{| \{0 \mod p,\, \Delta \mod p\}|}{p}}{(1-\frac{1}{p})^2} = 2 \prod_{p \text{ an odd prime}} \frac{1-\frac{2}{p}}{(1-\frac{1}{p})^2} \prod_{p \text{ an odd prime dividing }\Delta} \frac{1-\frac{1}{p}}{1-\frac{2}{p}}.$$

As can be seen from the definition, $C(\Delta)$ only depends on the set of primes dividing $\Delta$ (one can rephrase that by saying that $C(\Delta)$ is a function of the radical of $\Delta$). In particular, it does not depend on the multiplicities of the primes in the factorization of $\Delta$. An immediate consequence is that $C(h^i)=C(h)$ for any $i\ge 1$ and any nonzero $h$.

Some remarks:

  1. Although the Hardy-Littlewood prime-tuple conjecture is completely open, it is known (by sieve theory) that $$\frac{\sum_{1 \le n\le x} \Lambda(n)\Lambda(n+\Delta)}{x} \le 4C(\Delta) (1+o(1)).$$
  2. There are some arithmetic functions $\alpha$ for which the mean value $C_{\alpha}(\Delta):=\lim_{x \to \infty} \frac{\sum_{1 \le n\le x} \alpha(n)\alpha(n+\Delta)}{x}$ depends in a more complicated on the prime factorization of $\Delta$ - for instance, the function $r(n)=|\{ (a,b) \in \mathbb{Z}^2: n=a^2+b^2 \}|$.
Improve this answer
$\endgroup$
1
  • $\begingroup$ This may be a very naive question, but can this $ \Delta $ be interpreted as the conductor of some L-function ? $\endgroup$ – Sylvain JULIEN Feb 18 '18 at 12:49

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.