%I
%S 0,0,1,2,5,3,1,0,9,0,1,19,1,0,21,0,1,10,1,0,4,0,1,0,0,3,0,0,1,68,1,0,
%T 0,5,0,0,1,0,4,0,1,25,1,0,3,0,1,0,0,3,0,0,8,0,0,5,0,0,1,12,1,0,0,5,0,
%U 0,1,0,4,0,1,2,1,0,0,5,0,0,1,0,14,0,1,0,0,5,0,0,1,0
%N Count the squarefree numbers appearing in each interval [p,q] where (p,q) is a Goldbach partition of 2n such that all primes from p to q (inclusive) appear as a part in some Goldbach partition of p+q = 2n, and then add the results.
%C a(n) >= A279315(n).  _Wesley Ivan Hurt_, Dec 17 2016
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/GoldbachPartition.html">Goldbach Partition</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Goldbach%27s_conjecture">Goldbach's conjecture</a>
%H <a href="/index/Go#Goldbach">Index entries for sequences related to Goldbach conjecture</a>
%H <a href="/index/Par#part">Index entries for sequences related to partitions</a>
%F a(n) = Sum_{i=3..n} (A010051(i) * A010051(2ni) * (Sum_{j=i..2ni} mu(j)^2) * (Product_{k=i..n} (1abs(A010051(k)A010051(2nk))))), where mu is the MÃ¶bius function (A008683).
%p with(numtheory): A279536:=n>add( (pi(i)pi(i1)) * (pi(2*ni)pi(2*ni1)) * add(mobius(j)^2, j=i..2*ni) * (product(1abs((pi(k)pi(k1))(pi(2*nk)pi(2*nk1))), k=i..n)), i=3..n): seq(A279536(n), n=1..100);
%Y Cf. A008683, A010051, A279315.
%K nonn,easy
%O 1,4
%A _Wesley Ivan Hurt_, Dec 14 2016
