3
$\begingroup$

The starting point of this post is an earlier question, where I conjectured (and GH from MO confirmed) that the von Mangoldt function is the limit at $s=1$ of a certain Dirichlet series, $$\Lambda(m)=\lim_{s\to 1+}\zeta(s)\sum_{d\mid m}\frac{\mu(d)}{d^{s-1}}=\lim_{s\to 1+}\;\sum_{n=1}^\infty\frac{1}{n^s}\left(\sum_{d\mid\gcd(n,m)}d\mu(d)\right),\qquad m>1.\tag{1}$$ The prime $k$-tuple conjecture for $k=2$ states the asymptotic formula $$\sum_{m \leq X} \Lambda(m+h) \Lambda(m) \sim {\mathfrak S}(h) X.\tag{2}$$ In the light of $(1)$, we have $$\Lambda(m+h) \Lambda(m) = \lim_{s\to 1+}\;\sum _{n=1}^{\infty}\frac{T_h(n,m)}{n^s},\qquad m>1\tag{3},$$ where $$T_{h}(n,m):=\sum_{k\mid n}\left(\sum_{c \mid \gcd (k,m+h)} c \mu (c)\right) \left(\sum_{d \mid \gcd \left(\frac{n}{k},m\right)} d \mu (d)\right) .\tag{4}$$

Conjecture 1. For any positive integers $h$, $m$, $n$, we have $$T_{h}(n,m) = T_{h}(n,m+n).\tag{5}$$

Denote the sum of the mod $n$ repeating entries by $$S_h(n) := \sum\limits_{m=1}^n T_{h}(n,m).$$

Conjecture 2. We have the asymptotic formula $$\sum\limits_{m \leq X} T_{h}(n,m)\sim\frac{S_h(n)}{n}X. \tag{6}$$

Observe that by $(3)$ we have

$$\sum_{m \leq X} \Lambda(m+h) \Lambda(m) = \lim_{s\to 1+}\;\sum\limits_{n=1}^{\infty} \frac{1}{n^s} \sum\limits_{m \leq X} T_{h}(n,m). \tag{7}$$

Main question. Is it true for all $n$ that $$S_2(n) = S_4(n)\ ? \tag{8}$$

To me this would mean that the asymptotic density of twin primes $h=2$ is equal to the asymptotic density of cousin primes $h=4$.

Associated Mathematica program

In general this should be a classification of asymptotic densities of prime gaps and not only twin primes and cousin primes. I should maybe add that it appears that $$\sum\limits_{m=1}^{m=n} T_{h_{1}}(n,m)=\sum\limits_{m=1}^{m=n} T_{h_{2}}(n,m)$$ whenever: $$\varphi^{\ast -1}(h_{1})=\varphi^{\ast -1}(h_{2})$$ where $\varphi^{\ast -1}(h)$ is the Dirichlet inverse of the Euler totient.

$\endgroup$
  • 3
    $\begingroup$ Please try to make your question more concise. That is, define the matrix $M_4$ in a few lines, and then ask the question in the title. To me it is very discouraging to see screenshots, Mathematica codes etc. instead of a clear definition and a question. I stopped reading right above your screenshot. Please don't assume anyone to spend more than 30 seconds to understand what your question is. Then, if the question is interesting, people will definitely devote considerable time to (try to) answer it! $\endgroup$ – GH from MO Aug 23 '18 at 19:05
  • 1
    $\begingroup$ I have tried to improve the quality of the question now. $\endgroup$ – Mats Granvik Aug 24 '18 at 17:26
  • $\begingroup$ Are you sure about (5)? I think you wanted to write $T_h(n,m)=T_h(n,m+n)$, which is easy to prove. Indeed, this follows immediately from $\gcd(k,h+m)=\gcd(k,h+m+n)$ and $\gcd \left(\frac{n}{k},m\right)=\gcd \left(\frac{n}{k},m+n\right)$ for any $k\mid n$. This also implies (6) readily, just distribute the sum accoring to $m\bmod{n}$. $\endgroup$ – GH from MO Aug 24 '18 at 22:34
  • 1
    $\begingroup$ Nice question! I answered it much more generally, see my response below. I also edited the question and the title for clarity, brevity, and in order to draw more attention. $\endgroup$ – GH from MO Aug 25 '18 at 3:12
4
$\begingroup$

Nice question! As I explained in the comments, Conjectures 1-2 follow easily. So let me answer your main question much more generally, and also the question that you implicitly asked at OEIS A298825. I will denote by $\tau$ the divisor function.

Theorem. Let $r=\operatorname{rad}(h)$ be the radical of $h$. Then $$S_h(n)=n\sum_{\substack{e\mid r\\e^2\mid n}}e\sum_{\substack{n=e^2fg\\\gcd(e,f)=1}}\mu(f)\tau(f)\tau(g).\tag{$\ast$}$$

Corollary 1. If $\operatorname{rad}(h)=\operatorname{rad}(j)$, then $S_h(n)=S_j(n)$.

Corollary 2. $S_h(n)$ is divisible by $n$, and it vanishes unless $n$ is powerful.

Proof of Theorem. Starting from the definitions, \begin{align*}S_h(n)&=\sum_{m=1}^n\sum_{k\mid n}\left(\sum_{c\mid\gcd(k,m+h)}c\mu(c)\right)\left(\sum_{d\mid\gcd\left(\frac{n}{k},m\right)}d\mu(d)\right)\\[6pt] &=\sum_{n=ab}\sum_{\substack{c\mid a\\d\mid b}}\mu(c)\mu(d)cd\sum_{\substack{1\leq m\leq n\\m\equiv -h\pmod{c}\\m\equiv 0\pmod{d}}} 1\\[6pt] &=n\sum_{n=ab}\sum_{\substack{c\mid a\\d\mid b}}\mu(c)\mu(d)\gcd(c,d)\sum_{\substack{1\leq m\leq\operatorname{lcm}(c,d)\\m\equiv -h\pmod{c}\\m\equiv 0\pmod{d}}} 1\\[6pt] &=n\sum_{n=ab}\sum_{\substack{c\mid a\\d\mid b\\\gcd(c,d)\mid h}}\mu(c)\mu(d)\gcd(c,d). \end{align*} Only square-free $c$'s and $d$'s contribute to this sum, hence $e:=\gcd(c,d)$ is also square-free. Therefore, $e\mid h$ is equivalent to $e\mid r$, while $e\mid c\mid a$ and $e\mid d\mid b$ imply that $e^2\mid n$. Using the notation $$a':=a/e,\qquad b':=b/e,\qquad c':=c/e,\qquad d':=d/e,$$ we get \begin{align*}S_h(n)&=n\sum_{\substack{e\mid r\\e^2\mid n}}e\sum_{n=e^2a'b'} \sum_{\substack{c'\mid a'\\d'\mid b'\\\gcd(c',d')=1}}\mu(ec')\mu(ed')\\[6pt] &=n\sum_{\substack{e\mid r\\e^2\mid n}}e\sum_{n=e^2a'b'} \sum_{\substack{c'\mid a'\\d'\mid b'\\\gcd(c'd',e)=1}}\mu(c'd'). \end{align*} Now we introduce $f:=c'd'$ and $g:=a'b'/f=(a'/c')(b'/d')$, and we obtain $(\ast)$ readily.

Proof of Corollary 1. The statement is clear, since the RHS of $(\ast)$ only depends on $r$ and $n$.

Proof of Corollary 2. It is clear from $(\ast)$ that $S_h(n)$ is divisible by $n$. For a given $e\mid r$ in $(\ast)$, let us decompose $n/e^2$ as $st$, where $\gcd(s,e)=1$ and $t\mid e^\infty$. Then, we can rewrite $(\ast)$ as $$S_h(n)=n\sum_{\substack{e\mid r\\e^2\mid n}}e\tau(t)\sum_{f\mid s}\mu(f)\tau(f)\tau\left(\frac{s}{f}\right).$$ The inner sum is the convolution of $\mu\tau$ and $\tau$ at $s$. It is multiplicative in $s$, and for $s=p^k$ a prime power, it equals $$\tau(p^k)-\tau(p)\tau(p^{k-1})=(k+1)-2k=1-k.$$ In particular, for $s=p$ a prime, it is zero, hence the inner sum is supported on powerful numbers. It follows that $S_h(n)$ vanishes unless $n$ has a square divisor $e^2$ such that the coprime-to-$e$ part of $n/e^2$ is powerful. This condition on $n$ implies that $n$ itself is powerful, hence we are done.

Added. Of course the upshot of the question, namely that the density of prime pairs for $h=2$ should be the same as for $h=4$, is not new. Indeed, by the familar product formula for $\mathfrak{S}(h)$ (see e.g. here), it is clear that $\mathfrak{S}(h)$ only depends on the radical of $h$, hence in particular $\mathfrak{S}(2)=\mathfrak{S}(4)$. The original post and my answer above should be regarded as a variant (or refinement) of this observation. Indeed, it seems likely that (cf. $(2)$, $(6)$, $(7)$ in the original post) $$\mathfrak{S}(h)=\sum_{n=1}^\infty\frac{S_h(n)}{n^2},$$ but I have not attempted to prove this (the proof seems tedious but straightforward).

$\endgroup$

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.