A340575 - OEIS

%I #44 Jan 20 2021 17:56:47

%S 2,2,3,5,5,3,2,11,7,7,5,2,3,0,3,7,11,7,31,2,2,2,5,13,3,3,2,71,2,3,31,

%T 109,71,7,3,0,2,59,2,17,3,5,61,0,29,17,41,29,5,3,5,3,11,79,0,41,3,11,

%U 5,5,5,37,37,59,13,3,3,3,31,31,61,2,2,3,2,2,2,23,2,5,11,5,7,2,5,5,67,5,0

%N a(n) is the least prime p such that the product of 2*n-1 consecutive primes starting with p is divisible by the sum of those same primes, or 0 if there is no such prime.

%H Robert Israel, <a href="/A340575/b340575.txt">Table of n, a(n) for n = 1..10000</a>

%e For n=4, the product of 2*4-1 = 7 primes starting with a(4) = 5 is 5*7*11*13*17*19*23 which is divisible by 5+7+11+13+17+19+23 = 95 = 5*19.

%p plow:= n -> n*(ln(n)+ln(ln(n))-1):

%p phi:= n -> n*(ln(n)+ln(ln(n))):

%p f:= proc(n) local i,k,m,q;

%p   m:= 2*n-1;

%p   for k from 0 do

%p     if mul(ithprime(k+i),i=1..2*n-1) mod add(ithprime(k+i),i=1..2*n-1) = 0 then return ithprime(k+1) fi;

%p     if k >= 5 and is(m*phi(k+m-1) < plow(k)^2) then return 0 fi;

%p   od;

%p end proc:

%p map(f, [$1..100]);

%K nonn

%O 1,1

%A _J. M. Bergot_ and _Robert Israel_, Jan 19 2021