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" F_{n} numbers defined as: F_{n} = f_{1}.f_{2}...f_{n} +1, where f_{k} is the kth 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 kth 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 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 F_{n} 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? ***






