Problems & Puzzles: Conjectures On August 19, 2026 Alain Rochelli sent the following Conjecture: Let A(n) be the largest prime of the form prime(n)#/d + d where prime(n)# denotes the product of the first n primes and d divides prime(n)#. A(n) is growing exponentially (i.e. 3, 7, 31, 211, 2311, 15017, 102107, 1616621, 22309297, 3234846617, 200560490131). Obviously the corresponding value of d is the smallest squarefree number such that A(n) = d + prime(n)#/d is prime. Using Michael Branicky's table up to 2000 (cf. A295741), we can formulate the conjecture d < prime(n). For example, for values n > 1000 in A395095, we obtain with d increasing the following results: n / prime(n) / d 1046 / 8353 / 5065 1167 / 9431 / 6098 1450 / 12109 / 6637 1460 / 12211 / 6982 1580 / 13309 / 7417 1745 / 14891 / 12419 This conjecture makes it possible to obtain primes of the order of magnitude approximating exp(prime(n)) quite easily. Q1. Can you find a counterexample with d > prime(n)? Q2. Can you find an extension to A395095 (32nd term)? Q3. Can you find a heuristic proof of the conjecture d < prime(n) for all n > 0? Q4. Could you specify the reliability level of the primality test?
From Set 5 to 11, 2026, contributions came from Alain Rochelli, *** Alain wrote:I only have an answer for Q3:Assuming d is small enough compared to prime(n), we can write: prime(n-1)# < d + prime(n)#/d < prime(n)#.We can therefore deduce that the probability that A(n) is prime is on the order of 1/log(prime(n)#), which is also equivalent to 1/prime(n) using the prime number theorem (cf. Tschebycheff’s function). Moreover, A(n) has no prime factors less than prime(n). Consequently, the previous probability must be multiplied by exp(gamma)*log(prime(n)) with gamma as Euler's constant (cf. Mertens’s formula). Then comes the result: Proba (A(n) prime) ~ exp(gamma) * log(prime(n)) / prime(n)Furthermore, the probability that a random natural number is squarefree is equal to 6 / Pi^2 (cf. A005117). Therefore, there are on average 6*prime(n)/Pi^2 squarefree numbers between 0 and prime(n).It follows that the average number of values of d (less than prime(n), cf. A395412) such that A(n) is prime is equal to (6*exp(gamma)/Pi^2) * log(prime(n)). The previous equivalence relating to the number of d’s (i.e. ~ 1.083 * log(prime(n))) seems to agree with a calculation made with the first 500 terms of the sequence A395412. The growth of log(prime(n)) and the verification up to n=2000 by Michael Branicky show the statistical truth of the conjecture d < prime(n). ***
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