Problems & Puzzles: Puzzles

Puzzle 118. Primorial product numbers

Very recently I received a copy of the book "Florentin Smarandache: Definitions, solved & unsolved conjectures and theorems in number theory and geometry", Xiquan Publishing House, 2000, kindly sent by Dr. M.L.Perez.

From the many many interesting issues compiled there, I selected one of them to construct the puzzle of this week.

The Definition 37 (p. 36) introduces the "factorial product" Fn numbers defined as: Fn = f1.f2...fn +1, where fk is the k-th factorial number.

Mutatis mutandis I would like to define the P#n numbers as P#n = p#1.p#2 ...p#n +1, where p#k is the k-th primorial number.

The first P#n numbers are: 3, 13, 361, ...

361 = (2).(2.3).(2.3.5)+1

I have found that P#n is prime for n = 1, 2, 5, 12, 15 & 35

Question 1: Find 3 more P#n prime numbers
Question 2: Redo the exercise for Q#n, where Q#n = P#n - 2

Solution

Jeff Heleen notices that P#3 is 19^2 and ask if there are more perfect powers in this sequence of numbers.

***

Dr. M.L.Perez remind to me the fact that the original Definition 37 for the numbers Fn was followed by the Problem 14 that asked "how many of them are primes". Maybe you would like to try also this question.

***

Patrick De Geest tackled the question 2 and found that Q#n  is prime for n= 2, 3, & 33. (See the Neil Sloane's sequences submitted by him: A066268 & A066269). The third prime (n=33) has 759 digits.

Who will get the fourth member of this sequence?

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