A140761 Primes p(j) = A000040(j), j>=1, such that p(1)*p(2)*...*p(j) is an integral multiple of p(1)+p(2)+...+p(j).

2, 5, 19, 41, 83, 163, 167, 179, 191, 223, 229, 241, 263, 269, 271, 317, 337, 349, 367, 389, 433, 463, 491, 521, 701, 719, 757, 809, 823, 829, 859, 877, 883, 919, 941, 971, 991, 997, 1021, 1033, 1049, 1091, 1153, 1181, 1193, 1223, 1291, 1301, 1319, 1327, 1361

Offset: 1

Author

Enoch Haga, May 28 2008

Keywords

• easy
• nonn

Examples

			a(2) = 5 because it is the last consecutive prime in the run 2*3*5 = 30 and 2+3+5 = 10; since 30/10 = 3, it is the first integral quotient.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A002110, A007504, A051838, A116536, A140763, A159578.

Programs

Mathematica

seq = {}; sum = 0; prod = 1; p = 1; Do[p = NextPrime[p]; prod *= p; sum += p; If[Divisible[prod, sum], AppendTo[seq, p]], {200}]; seq (* Amiram Eldar, Nov 02 2020 *)

Formula

Find integral quotients of products of consecutive primes divided by their sum.

a(n) = A000040(A051838(n)). - R. J. Mathar, Jun 09 2008

Extensions

Edited by R. J. Mathar, Jun 09 2008

a(1) added by Amiram Eldar, Nov 02 2020