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Conjecture

For any sufficiently large integer $kn$ , the sequence representing the number of primes in each block obtained by splitting $kn$ into $k$ equal blocks, is a strictly decreasing sequence, i.e:

$$\pi\left(n\right)>\pi\left(2n\right)-\pi\left(n\right)>\pi\left(3n\right)-\pi\left(2n\right)\ldots>\pi\left(kn\right)-\pi\left(\left(k-1\right)n\right)$$

My approach

For any given $k$, we need to prove that for all sufficiently large $n$, the last block contains less primes than the block before:

$$\underbrace{\pi\left(\left(k-1\right)n\right)-\pi\left(\left(k-2\right)n\right)}_{\text{before last}}>\underbrace{\pi\left(kn\right)-\pi\left(\left(k-1\right)n\right)}_{\text{last}}$$

Let $f( k,n )$ denote the difference between the prime count of before last and last block of $kn$:

$$f\left(k,n\right) = \left(\pi\left(\left(k-1\right)n\right)-\pi\left(\left(k-2\right)n\right)\right)-\left(\pi\left(kn\right)-\pi\left(\left(k-1\right)n\right)\right)$$ $$ = 2\pi\left(\left(k-1\right)n\right)-\pi\left(\left(k-2\right)n\right)-\pi\left(kn\right)$$

Therefore, an equivalent formulation of the conjecture is that $f(k,n)>0$ for any integers $k,n$ with sufficiently large $n$.

Does anybody have ideas or references on how to prove this? $$$$ Computation of $f( k,n )$

I made some computation that strongly suggests the conjecture is true, at least for small $k$:

enter image description here

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Failing case for small $n$

Let's take the case $k=2$. We see that the conjecture fails for $n=1,2,4$ & $10$:

enter image description here

I computed $f( k,n )$ for $2 ≤ k ≤ 30$ with $1 ≤ n ≤ 10^{8}$, in order to find what appears to be the last failing case of each $k$ , after which $f( k,n ) >0$ for all $n$:

enter image description here

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    $\begingroup$ Prime gaps increases unboundedly(one can see that considering all integers n!+2,....n!+n are composites) is this a consequence of that? $\endgroup$ Aug 9, 2021 at 9:54
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    $\begingroup$ I think this should follow from the Prime Number Theorem (with error terms). For example, the explicit version of the PNT mentioned here: mathoverflow.net/questions/373737/… $\endgroup$
    – Tony Huynh
    Aug 9, 2021 at 11:26
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    $\begingroup$ Note that the second Hardy Littlewood conjecture is likely false, and it is a statement of a similar flavor en.wikipedia.org/wiki/… . $\endgroup$
    – JoshuaZ
    Aug 9, 2021 at 11:51
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    $\begingroup$ To make sure I understand the order of your quantifiers? One is given an $n$ and then is allowed to choose $k$ based on $n$? $\endgroup$
    – JoshuaZ
    Aug 9, 2021 at 11:56
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    $\begingroup$ @JoshuaZ: I agree that this is similar, and yet while I would bet $1000 against $1 that the 2nd Hardy Littlewood conjecture is false, this seems reasonable. Being tied to the beginning rather than floating gives a LOT less flexibility. $\endgroup$
    – Charles
    Aug 9, 2021 at 16:44

1 Answer 1

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The conjecture is true, and it follows from the following form of the Prime Number Theorem: $$\pi(x)=\frac{x}{\log x}+(1+o(1))\frac{x}{\log^2 x},\qquad x\to\infty.$$ Indeed, this implies for any fixed $m\geq 1$ and sufficiently large $n$ that $$2\pi(mn)>\pi(mn+n)+\pi(mn-n).$$ Why? We have \begin{align*}\pi(mn)&=\frac{mn}{\log(mn)}+(1+o(1))\frac{mn}{\log^2(mn)}\\ &=m\frac{n}{\log n}\frac{1}{1+\frac{\log m}{\log n}}+(m+o(1))\frac{n}{\log^2 n}\\ &=m\frac{n}{\log n}+(m-m\log m+o(1))\frac{n}{\log^2 n}. \end{align*} Similarly (replacing $m$ by $m-1$ and $m+1$, respectively), \begin{align*} \pi(mn+n)&=(m+1)\frac{n}{\log n}+(m+1-(m+1)\log(m+1)+o(1))\frac{n}{\log^2 n},\\ \pi(mn-n)&=(m-1)\frac{n}{\log n}+(m-1-(m-1)\log(m-1)+o(1))\frac{n}{\log^2 n}. \end{align*} Here, in case of $m=1$, we understand $(m-1)\log(m-1)$ as zero. It follows that $$2\pi(mn)-\pi(mn+n)-\pi(mn-n)=(f(m)+o(1))\frac{n}{\log^2 n},$$ where $$f(m):=(m+1)\log(m+1)+(m-1)\log(m-1)-2m\log m.$$ So we only need to verify that $f(m)$ is positive, that is, $$(m+1)\log(m+1)+(m-1)\log(m-1)>2m\log m.$$ However, this one is clear, because the function $x\mapsto x\log x$ is strictly convex (its derivative is strictly increasing).

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    $\begingroup$ The way I read the question, OP does not ask that $n$ be sufficiently large, only that $kn$ be sufficiently large. If $n=1$ and $k$ is sufficiently large, then $kn$ is sufficiently large, but what OP wants to conclude is quite false. $\endgroup$ Aug 14, 2021 at 13:05
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    $\begingroup$ @GerryMyerson: The OP wrote: "For any given $k$, we need to prove that for all sufficiently large $n$, the last block contains less primes than the block before". This is what I proved. $\endgroup$
    – GH from MO
    Aug 14, 2021 at 17:39
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    $\begingroup$ Gerry Myerson nonetheless has a point, in that the main body of the question should contain all the prerequisites, rather one of them being made clear only in the My approach section. $\endgroup$ Aug 14, 2021 at 21:26
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    $\begingroup$ @YaakovBaruch: I agree. $\endgroup$
    – GH from MO
    Aug 14, 2021 at 21:27
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    $\begingroup$ @FrançoisHuppé: The statement should begin with: for fixed $k$ and sufficiently large $n$ etc. More formally: for any integer $k\geq 1$ there exists an integer $N\geq 1$ such that for any integer $n\geq N$ etc. $\endgroup$
    – GH from MO
    Aug 15, 2021 at 4:47

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