%I
%S 17,47,13,11,41,7,5,3,13,11,29,7,5,3,13,11,17,7,5,3,29,7,5,3,17,19,13,
%T 11,13,7,5,3,5,3,29,7,5,3,37,19,17,19,13,11,13,7,5,3,5,3,13,7,5,3,5,3,
%U 17,23,13,11,17,7,5,3,41,7,5,3,17,31,13,11,29,7,5,3,17,19,13,11,13,7,5,3,5
%N a(n) is the least prime k such that 2*n1+k = 2*p*q for odd primes p,q (not necessarily distinct).
%C Dickson's conjecture implies that for every prime p that does not divide 2*n1, there exist infinitely many q such that q and 2*p*q(2*n1) are prime. Thus a(n) always exists.
%H Robert Israel, <a href="/A329951/b329951.txt">Table of n, a(n) for n = 1..10000</a>
%H Mathematics StackExchange, <a href="https://math.stackexchange.com/questions/3450575">Can every odd number be represented as 2pqr where p, q, and r are distinct odd primes?</a>
%e a(3)=13 because 2*31+13 = 18 = 2*3*3 with 13, 3, 3 all primes, and 13 is the least prime for which this works.
%p f:= proc(m) local r,x;
%p r:= 2:
%p do r:= nextprime(r);
%p x:= (m+r)/2;
%p if x::odd and numtheory:bigomega(x)=2 then return r
%p fi od
%p end proc:
%p map(f, [seq(i,i=1..1000,2)]);
%K nonn
%O 1,1
%A _Robert Israel_, Nov 25 2019
