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%I #12 Aug 01 2015 11:52:58
%S 3,43,1723,13307323
%N Primes which are 1/3 of the cumulative sum of factorials of primes, if 1 is used as the zeroth prime.
%C No more primes. Starting with a(14) = (1! + ... + 43!)/3 the sum always has a factor of 47.
%F Defining prime(0)= 1: a(n) = (1/3)*Sum_{i=0..n}A000142(A000040(i+1)) iff in A000040. a(n) = (1/3)*Sum_{i=0..n}prime(i+1)! iff in A000040.
%e prime(0)! = 1! = 1; prime(1)! = 2! = 2.
%e a(1) = (1! + 2! + 3!)/3 = 9/3 = 3.
%e a(2) = (1! + 2! + 3! + 5!)/3 = 129/3 = 43.
%e a(3) = (1! + 2! + 3! + 5! + 7!)/3 = 5169/3 = 1723.
%e a(4) = (1! + 2! + 3! + 5! + 7! + 11!)/3 = 39921969/3 = 13307323.
%t f[n_] := (1 + Plus @@ ((Prime@ Range@ n)!))/3; Select[f /@ Range@ 43, PrimeQ@# &] (* _Robert G. Wilson v_, Apr 30 2009 *)
%Y Cf. A000040, A000142.
%K fini,full,nonn
%O 1,1
%A _Jonathan Vos Post_, Feb 07 2006
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