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%I #12 Nov 02 2020 20:38:54
%S 2,5,19,41,83,163,167,179,191,223,229,241,263,269,271,317,337,349,367,
%T 389,433,463,491,521,701,719,757,809,823,829,859,877,883,919,941,971,
%U 991,997,1021,1033,1049,1091,1153,1181,1193,1223,1291,1301,1319,1327,1361
%N 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).
%H Amiram Eldar, <a href="/A140761/b140761.txt">Table of n, a(n) for n = 1..10000</a>
%F Find integral quotients of products of consecutive primes divided by their sum.
%F a(n) = A000040(A051838(n)). - _R. J. Mathar_, Jun 09 2008
%e 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.
%t 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 *)
%Y Cf. A002110, A007504, A051838, A116536, A140763, A159578.
%K easy,nonn
%O 1,1
%A _Enoch Haga_, May 28 2008
%E Edited by _R. J. Mathar_, Jun 09 2008
%E a(1) added by _Amiram Eldar_, Nov 02 2020
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