Problem
of the
Week
PROBLEM 187

Define a sequence of natural numbers p0, p1, p2,... as follows: 

Put p0 = 2, and for k > 0 define pk+1 = p0p1×××pk+ 1; 

in words, the next term in the sequence is one more than the product of all the previous terms.  The sequence begins 

p0 = 2
p1 = 3 = 2 + 1
p2 = 7 = 2×3 + 1
p3 = 43 = 2×3×7 + 1
p4 = 1807 = 2×3×7×43 + 1
p5 = 3263443 = 2×3×7×43×1807 + 1
p6 = 10650056950807 = 2×3×7×43×1807×3263443 + 1

Euclid proved there are infinitely many integer prime numbers by showing that pk must have a prime divisor that does not divide any pi for any 0 £ i < k.  Let's call a prime number a Euclid prime if it divides some pk.   Clearly 2, 3, 7 are Euclid primes, but so is 13 since p4 = 1807 = 13 × 139.  

Which of 5, 11, 17 are Euclid primes?  

For the wizards: Find infinitely many Euclid primes.  

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ã2004 Alberto L. Delgado