A334171 - OEIS

%I #7 Apr 27 2020 11:51:50

%S 0,1,1,2,6,4,5,4,6,5,5,8,9,3,6,5,6,10,8,3,8,7,6,11,11,4,9,7,5,11,9,7,

%T 13,6,5,12,12,7,14,12,5,14,13,5,15,11,6,11,6,6,15,12,8,14,13,9,20,16,

%U 7,20,14,5,16,8,7,17,13,6,11,15,10,21,17,5,19,11,10

%N Maximum number of rectangles with prime side lengths that can be made with up to 2 perpendicular cuts from all p X q rectangles (p, q prime) such that p + q = 2n.

%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=1..n} c(i) * c(2*n-i) * ((1 - c(i-2))*(1 - c(2*n-i-2)) + 2*(1 - [c(i-2) = c(2*n-i-2)]) + 4*c(i-2)*c(2*n-i-2), where [] is the Iverson bracket and c is the prime characteristic (A010051).

%e a(5) = 6; 2*5 = 10 has 2 Goldbach partitions, (7,3) and (5,5), so we start with two rectangles of size 7 X 3 and 5 X 5. Note that the 7 X 3 rectangle can be cut into two rectangles of size 5 X 3 and 3 X 2 (both prime), but since the cuts must be perpendicular to each other (and not parallel), no smaller rectangles with prime side lengths can be made. Furthermore, the 5 X 5 rectangle can be cut into 4 rectangles with prime side lengths using two perpendicular cuts: 3 X 3, 3 X 2, 2 X 3, and 2 X 2. This makes 6 total rectangles.

%t Table[Sum[((1 - PrimePi[i - 2] + PrimePi[i - 3]) (1 - PrimePi[2 n - i - 2] + PrimePi[2 n - i - 3]) + 2 (1 - KroneckerDelta[PrimePi[i - 2] - PrimePi[i - 3], PrimePi[2 n - i - 2] - PrimePi[2 n - i - 3]]) + 4 (PrimePi[i - 2] - PrimePi[i - 3]) (PrimePi[2 n - i - 2] - PrimePi[2 n - i - 3])) (PrimePi[i] - PrimePi[i - 1]) (PrimePi[2 n - i] - PrimePi[2 n - i - 1]), {i, n}], {n, 100}]

%Y Cf. A010051, A045917.

%K nonn,easy

%O 1,4

%A _Wesley Ivan Hurt_, Apr 17 2020