A337509 Number of partitions of n into two distinct parts (s,t), such that (t-s) | n, and where n/(t-s) <= s < t.

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, 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, 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

Offset: 1

Author

Wesley Ivan Hurt, Oct 05 2020

Keywords

• nonn

Comments

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

Examples

			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.
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.
a(10) = 0; The partition (6,4) has difference of (6-4) = 2 | 10, but 10/2 = 5 > 4. So a(10) = 0.
a(11) = 0; No difference divides 11 (prime), so a(11) = 0.
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.
		

Crossrefs

Cf. A337976 (same with s | t).

Programs

Mathematica

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

Formula

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