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..n}A000142(A000040(i+1)) iff in A000040. a(n) = (1/3)*Sum_{i=0..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@# &] (* Robert G. Wilson v, Apr 30 2009 *)

Cf. A000040, A000142.

Sequence in context: A274387 A300988 A136648 * A317343 A307248 A009720

Adjacent sequences: A114334 A114335 A114336 * A114338 A114339 A114340

fini,full,nonn

Jonathan Vos Post, Feb 07 2006

approved