3

How to prove the following claims ?

Let $b_n=b_{n-1}+\operatorname{lcm}(\lceil \pi \cdot n \rceil , b_{n-1})$ with $b_1=1$ and $n>1$ . Let $a_n=b_{n+1}/b_n-1$ .

  1. Every term of this sequence $a_i$ is either prime or $1$ .

  2. Every prime of the form $\lceil \pi \cdot n \rceil$ is a term of this sequence.

  3. At the first appearance of each prime of the form $\lceil \pi \cdot n \rceil$ greater than $7$, it is the next prime of the given form after the largest prime that has already appeared.

Implementation of this generator in PARI/GP can be found here.

Pedja
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  • This is tangible related. – rtybase Jul 19 '19 at 19:09
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    source of problem ? –  Jul 19 '19 at 20:36
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    @RoddyMacPhee I made it by myself. – Pedja Jul 20 '19 at 07:17
  • @PeđaTerzić How far did cou check the sequence ? In particular, which is the smallest prime that has not yet appeared ? – Peter Jul 20 '19 at 08:59
  • $1.$ is true upto $n=3\cdot 10^5$ – Peter Jul 20 '19 at 09:15
  • It seems that if $p_n:=\lceil \pi\cdot n \rceil$ is prime then $a_{n-1}=p_n$. This holds upto $n=10^4$. – Peter Jul 20 '19 at 09:26
  • @Peter I must admit that I checked the third claim only for the first hundred primes of the form $\lceil \pi \cdot n \rceil$. – Pedja Jul 20 '19 at 09:46
  • @PeđaTerzić What about arbitary primes ? Does $a_n$ contain every prime, not only the special ones ? – Peter Jul 20 '19 at 09:48
  • @Peter I don't know. – Pedja Jul 20 '19 at 09:49
  • $a_{n-1}=p_n$ from above holds upto $n=10^6$ – Peter Jul 20 '19 at 10:57
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  • is true as long as $b_{n-1}$ is coprime to the prime. After step 2 it is equivalent to: $$b_n=b_{n-2}\cdot (a_{n-2}+1)+lcm(\lceil \pi\cdot n\rceil , b_{n-2}\cdot (a_{n-2}+1))$$ a few steps later you'll be able to replace the $b_{n-2}$ values with $a_i$ value expressions. At that point, the sequence created for a values, is nearly self contained.
    • –  Jul 20 '19 at 12:03
  • How did you make this by yourself? You just played with recurrence relations and plugged $\pi$ inside and after that you realized that this produces primes (or $1$)? You can't be that lucky. – Konstantinos Gaitanas Jul 22 '19 at 09:42
  • @KonstantinosGaitanas I was inspired by Cloitre's prime number generator. – Pedja Jul 22 '19 at 10:49