

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).


6



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
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OFFSET

1,1


LINKS



FORMULA

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


EXAMPLE

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.


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 *)


CROSSREFS



KEYWORD

easy,nonn


AUTHOR



EXTENSIONS



STATUS

approved



