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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*)

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

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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 $r(n)=|\{ (a,b) \in \mathbb{Z}^2: n=a^2+b^2 \}|$.
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  • $\begingroup$ 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

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