10
$\begingroup$

This question is inspired by Fortune's conjecture .

Can you provide proofs or counterexamples for the following two claims :

First claim

If $q$ is the smallest prime greater than $\displaystyle\prod_{i=1}^n C_i+1$ , where $\displaystyle\prod_{i=1}^n C_i$ is the product of the first $n$ composite numbers , then $q-\displaystyle\prod_{i=1}^n C_i$ is prime .

The first few such differences are :

3,5,5,5,11,7,23,11,29,17,31...

Try it Yourself !

Second claim

If $q$ is the greatest prime less than $\displaystyle\prod_{i=1}^n C_i-1$ , where $\displaystyle\prod_{i=1}^n C_i$ is the product of the first $n$ composite numbers , then $\displaystyle\prod_{i=1}^n C_i-q$ is prime .

The first few such differences are :

2,5,11,5,23,17,13,11,13,23,53...

Try it Yourself !

I have tested both claims up to $n=660$ and there were no counterexamples .

$\endgroup$
1
  • $\begingroup$ Very interesting conjectures! Unless, or until, there are counter-examples within computing range this could be deep. $\endgroup$ Jul 10, 2017 at 2:11

2 Answers 2

2
$\begingroup$

here's a list of things we know ( it's a hint or help not an strictly an answer ,partly because I don't know ):

  1. q-product for n>1 will have the same modular remainder mod 6 as q.

  2. When we take the product up to 2n, we get at least all the primes less than or equal to n up to exponents of 1 or more.

  3. all primes less than ${2\over 3}n$ appear at least twice in the product in 2.

  4. Generally,all primes less than ${2\over a}n$ will appear at least a-1 times in the product in 2.

    This all means the difference can't divide by any product within the product because then q would not be prime. n# is within this product. so the difference ( again 2n case) has already been trial factored up to n.

$\endgroup$
1
$\begingroup$

For the "first claim" we have...

(perhaps the notation is not correct)

Given

$$1 \le k \le n \textrm{ and } 1 < \ell < q,$$

then

$$q > \prod_{\imath=1}^n C_\imath \Rightarrow (q{ \not\mid}C_k) \wedge (C_k{\not\mid}q) \Rightarrow \ell{\not\mid}\left(q - \prod_{\imath=1}^n C_\imath\right),$$

whence

$$q - \prod_{\imath=1}^n C_\imath$$

is a prime.

I am working on the "second claim" :)

(corrected the typo... thanks to Peđa Terzić)

$\endgroup$
2

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.