Problems & Puzzles: Puzzles

Puzzle 1138 A068192 Davide Rotondo made me notice the following OEIS sequence: A068192 defined as: "Let a(1)=2, f(n) = 4*a(1)*a(2)*...*a(n-1) for n >= 1 and a(n) = f(n)-prevprime(f(n)-1) for n >= 2, where prevprime(x) is the largest prime < x." Mauro Fiorentini has computed the first 1000 temrs and observes: "The terms are easily seen to be distinct. It is conjectured that every element is prime. Do all primes occur in the sequence?" Q) Can you prove the three observations from Mauro Fiorentini about A068192: a= all terms are distinct, b) Every term is prime c) All the primes occur in this sequence?

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During the week 22-28 JUly, 2023, contributions came from Giorgos Kalogeropoulos and Alejandro Casini

 

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Giorgos wrote:

I don't have a proof but I devised a conjecture that might help in finding a proof:

Consider the following sequence S:

S(1)=2 and S(n) is the least prime not already in the sequence such that 

4*S(1)*S(2)*...*S(n-1) - S(n) is prime. In this sequence S we know that all terms are

distinct primes. 
 

My conjecture is that the sequences S and A068192 are the same.

I tested the first 1000 terms of both sequences and they are identical.

If the conjecture is true, then all terms of A068192 are distinct primes.

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Alejandro wrote:

Interesting sequence. Here’s the 200 terms following those calculated by Mauro. As expected, they’re all prime. Let me underline that the recurrence relation for a(i) can be restated in a more convenient form as “smallest a>1 such that F(i)-a is prime”.

1. Yes, all the sequence terms are distinct. In fact let a(i)=p, then every following F(j) is divisibile by p and thus F(j)-p is clearly composite, unless p is prime and F(j)=2p. The latter is impossible, given that F(j) is divisible by 8 for j>1. Consequentially, F(j)-a(i) is always composite, and so a(j) can’t be equal to a(i) for j>i.
   Actually, more can be said. Indeed, every term must be coprime with all the previous ones.
    
2. I can't see for what reasons a(n) is forced to be prime. I can only point out that if a(n) is prime, then [a(n), F(n)-a(n)] represents the minimal Goldbach partition of F(n), conjectured to be O(log(N)^2 * loglog(N)).
3. At now, the smallest missing prime is 1171, and instead the largest prime observed is 41177. Now a little probabilistic analysis for question b). At now the smallest possible composite value for a(n) is 1171^2>10^6, therefore for the moment, even removing the terms already obtained, there are plenty of primes (~10^5) before reaching this value. In other words, at this moment it is quite likely that at least one of these prime integers has F(n)-p prime. However, that doesn't mean it lasts forever and it’s only a heuristic point of view.

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