1,1

No more primes. Starting with a(14) = (1! + ... + 43!)/3 the sum always has a factor of 47.

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.

Cf. A000040, A000142.

Adjacent sequences: A114334 A114335 A114336 this_sequence A114338 A114339 A114340

Sequence in context: A093163 A141060 A136648 this_sequence A009720 A076361 A130408

bref,easy,fini,nonn

Jonathan Vos Post (jvospost2(AT)yahoo.com), Feb 07 2006