

A224894


a(1) = 1, a(n+1) = smallest prime divisor of 1 + product of all the primes p <= a(n).


0




OFFSET

1,2


COMMENTS

Following Euclid's proof that there are infinitely many primes.
For example, 211 is 2*3*5*7 + 1 and 1051 is the smallest prime divisor of 2*3*5*...*211 + 1. This differs from the EuclidMullin sequence (A000945) because all the primes between a(n1) and a(n) are used in calculating a(n+1).


LINKS



EXAMPLE

a(5) = 2*3*5*7 + 1 = 211.
a(6) = 1051 because 1051 is the smallest prime divisor of 2*3*5*...*211 + 1.


MATHEMATICA

a[1] = 1; a[n_] := a[n] = Block[{pr = 1 + Product[Prime[k], {k, PrimePi@a[n  1]}], p = NextPrime@a[n  1]}, While[Mod[pr, p] > 0, p = NextPrime@p]; p]; Array[a, 7] (* Giovanni Resta, Jul 24 2013 *)


CROSSREFS



KEYWORD

nonn,hard


AUTHOR



EXTENSIONS



STATUS

approved



