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Let p(1)=2, p(2)=3, p(3)=5, ... denote the primes and let E(n) = 1 + p(1) * p(2) * ... * p(n). Sequence gives primes p such that p(n+2) | E(n).
1

%I #12 Feb 06 2021 21:42:43

%S 7,271,307,673

%N Let p(1)=2, p(2)=3, p(3)=5, ... denote the primes and let E(n) = 1 + p(1) * p(2) * ... * p(n). Sequence gives primes p such that p(n+2) | E(n).

%C No other terms for p < 80000000.

%e 7 is a term of the sequence, because it is the 4th prime and divides E(2)=2*3+1=7 trivially. - _Martin Ehrenstein_, Feb 05 2021

%Y See A066735 for further information.

%Y Cf. A006862 Euclid numbers.

%K nonn,more

%O 1,1

%A Lévai Gábor (gablevai(AT)vipmail.hu), Nov 23 2004

%E a(2) corrected by _Martin Ehrenstein_, Feb 05 2021