-2
$\begingroup$

For $n$ a large enough positive composite integer, say $r$ is a primality radius of $n$ if both $n-r$ and $n+r$ are prime. Say $n$ is a karmic number if the following holds: $r$ is a primality radius of $n$ implies $r=1$ or $\Lambda(r)\neq 0$ where $\Lambda$ is the von Mangoldt function.

Are there infinitely many karmic numbers?

Improve this question
$\endgroup$
13
  • 7
    $\begingroup$ You are asking if there are infinitely many pairs of primes $(p_1,p_2)$ such that either $p_2=p_1+2$ or $p_2=p_1+2q^k$ for some prime $q$ and integer $k$. There is no need to introduce either a "primality radius" nor a "karmic number". $\endgroup$
    – Thomas Sauvaget
    Commented Nov 24, 2019 at 18:13
  • 6
    $\begingroup$ I disagree with your assessment. Adding unecessary terminology or hypotheses is very un-mathematical. The formulation isn't dry, it just requires to understand the object at hand by looking at examples and related theorems to patiently build a mental picture. $\endgroup$
    – Thomas Sauvaget
    Commented Nov 24, 2019 at 18:37
  • 4
    $\begingroup$ @ThomasSauvaget: I do not think your formulation is equivalent to OP's. For your pair, the number (p_1+p_2)/2$ may have another radii which fail to be of the required form. $\endgroup$
    – Ilya Bogdanov
    Commented Nov 24, 2019 at 19:16
  • 2
    $\begingroup$ I didn't understand the reason for the condition that $n$ is composite, or that it is "sufficiently large." Also, what have you done so far? $\endgroup$
    – Daniel McLaury
    Commented Nov 24, 2019 at 20:07
  • 4
    $\begingroup$ If I haven't made a mistake, the only karmic numbers $< 10^5$ are $1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 24, 30, 34, 36, 42, 60$. The sequence does not appear to be in the OEIS. I would guess that these are the only karmic numbers, but a proof would be difficult. $\endgroup$
    – Robert Israel
    Commented Nov 25, 2019 at 0:48

1 Answer 1

Reset to default
-1
$\begingroup$

Here comes an attempt of proof of the statement I made in my last comment above.

So, the considered article is "PRIME FACTORS OF ARITHMETIC PROGRESSIONS AND BINOMIAL COEFFICIENTS T. N. SHOREY AND R. TIJDEMAN". In section 2.2, it is written that $\omega(\Delta)\geq\pi(2k)-1$, where $\Delta:=\Delta(x,d,k):=x(x+d)...(x+(k-1)d)$ with the hypotheses $x\geq 1$, $k>2$, $d>1$ and $gcd(x,d)=1$.

So let $n$ be a karmic number divisible by $6$. The primality radii of $n$ are all of the form $6n+1$ or $6n+5$. An arithmetic progression of such primality radii $r_1, r_2, r_3,\cdots r_k$ of length $k$ fulfills the aforementioned hypotheses and thus $\omega(r_{1}r_{2}r_{3}\cdots r_{k})\geq\pi(2\times k)-1\gg\frac{k}{\log k}$. But the total number of primality radii of $n$ is $\ll\left(\prod_{p\mid n, p>2}\frac{p-1}{p-2}\right)\frac{n}{\log^2 n}=o(\frac{n}{\log n})$ so there exists $C.n< k_{0}<n$ with $0<C<1$ such that $k>k_{0}$ implies at least one of the $r_{i}$ with $1\leq i\leq k$ fulfills $\omega(r_{i})>1$. Hence $r_{i}$ is neither $1$ nor a prime power, in contradiction with the fact that $n$ is a karmic number beyond some threshold.

Thus there are only finitely many karmic numbers divisible by $6$.

Improve this answer
$\endgroup$
8
  • 2
    $\begingroup$ This only proves that in the sequence of primality radii of $n$ the lengths of arithmetic progressions are bounded, no? How do you conclude there are finitely many such numbers? $\endgroup$
    – Wojowu
    Commented Nov 25, 2019 at 22:01
  • $\begingroup$ $n$ being a karmic number, it implies the number of primality radii of $n$ with $\omega(n)\leq 1$ is the total number of primality radii of $n$ which is an $o(\pi(n))$. But doesn't Green-Tao's theorem imply that the length of some arithmetic progression of primality radii of an integer $n$ as $n$ tends to infinity is unbounded? $\endgroup$
    – Sylvain JULIEN
    Commented Nov 25, 2019 at 22:11
  • $\begingroup$ Indeed, Green-Tao implies that infinitely many $n$ will have long arithmetic progressions in their primality radii. But those $n$ will be sporadic, and most likely won't cover all (sufficiently large) multiples of $6$. So you can't deduce from this the claim that large multiples of $6$ won't be sporadic. $\endgroup$
    – Wojowu
    Commented Nov 25, 2019 at 22:14
  • $\begingroup$ Doesn't it follow from Dirichlet's theorem that those $n$ are equally distributed among the residual classes modulo $6$ which are not equal to the classes of $1$ or $5$? $\endgroup$
    – Sylvain JULIEN
    Commented Nov 25, 2019 at 22:21
  • 1
    $\begingroup$ I don't see how Dirichlet's theorem guarantees this, nor how this is relevant for the problem at hand. $\endgroup$
    – Wojowu
    Commented Nov 25, 2019 at 22:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .