1,1
No more primes. Starting with a(14) = (1! + ... + 43!)/3 the sum always has a factor of 47.
Table of n, a(n) for n=1..4.
Defining prime(0)= 1: a(n) = (1/3)*SUM[i=0 to n]A000142(A000040(i+1)) iff in A000040. a(n) = (1/3)*SUM[i=0 to n]prime(i+1)! iff in A000040.
prime(0)! = 1! = 1; prime(1)! = 2! = 2.
a(1) = (1! + 2! + 3!)/3 = 9/3 = 3.
a(2) = (1! + 2! + 3! + 5!)/3 = 129/3 = 43.
a(3) = (1! + 2! + 3! + 5! + 7!)/3 = 5169/3 = 1723.
a(4) = (1! + 2! + 3! + 5! + 7! + 11!)/3 = 39921969/3 = 13307323.
f[n_] := (1 + Plus @@ ((Prime@ Range@ n)!))/3; Select[f /@ Range@ 43, PrimeQ@# &] [From Robert G. Wilson v, Apr 30 2009]
Cf. A000040, A000142.
Sequence in context: A141060 A201784 A136648 * A009720 A201173 A190637
Adjacent sequences: A114334 A114335 A114336 * A114338 A114339 A114340
fini,full,nonn
Jonathan Vos Post, Feb 07 2006
approved
