A337509 - OEIS

%I #21 Oct 05 2020 12:19:59

%S 0,0,0,0,0,0,0,1,1,0,0,2,0,0,2,2,0,1,0,2,2,0,0,4,1,0,2,2,0,2,0,3,2,0,

%T 2,4,0,0,2,4,0,2,0,2,4,0,0,6,1,1,2,2,0,2,2,4,2,0,0,6,0,0,4,4,2,2,0,2,

%U 2,2,0,7,0,0,4,2,2,2,0,6,3,0,0,6,2,0,2,4,0,4,2,2,2,0,2,8

%N Number of partitions of n into two distinct parts (s,t), such that (t-s) | n, and where n/(t-s) <= s < t.

%C If n is prime, then a(n) = 0.

%F a(n) = Sum_{i=1..floor((n-1)/2)} Sum_{k=1..i} [k*(n-2*i) = n], where [ ] is the Iverson bracket.

%e a(8) = 1; There are 3 partitions of 8 into two distinct parts: (7,1), (6,2), (5,3), with differences 6, 4 and 2. Only the partition (6,2) satisfies (6-2) | 8 where 8/4 = 2 <= 2, so a(8) = 1.

%e a(9) = 1; There are 4 partitions of 9 into two distinct parts: (8,1), (7,2), (6,3), (5,4) with differences 7, 5, 3 and 1. Only the partition (6,3) satisfies (6-3) | 9 where 9/3 = 3 <= 3, so a(9) = 1.

%e a(10) = 0; The partition (6,4) has difference of (6-4) = 2 | 10, but 10/2 = 5 > 4. So a(10) = 0.

%e a(11) = 0; No difference divides 11 (prime), so a(11) = 0.

%e a(12) = 2; Check (9,3), (8,4) and (7,5) since 9-3 = 6, 8-4 = 4 and 7-5 = 2 all divide 12. Then we have 12/6 = 2 < 3 and 12/4 = 3 < 4, but 12/2 = 6 > 5 so a(12) = 2.

%t Table[Sum[Sum[KroneckerDelta[k (n - 2 i), n], {k, i}], {i, Floor[(n - 1)/2]}], {n, 100}]

%Y Cf. A337976 (same with s | t).

%K nonn

%O 1,12

%A _Wesley Ivan Hurt_, Oct 05 2020