A264356 Number of partitions of n*(n-2) with n parts and at least one part > n.

1, 18, 197, 1844, 16457, 143975, 1249642, 10815498, 93576157, 810347897, 7027967095, 61060521549, 531506499178, 4635288747540, 40499108355263, 354472925956809, 3107795826264979, 27290688756270363, 240010330685355235, 2113784811395623970, 18641067229072645836

Offset: 4

Author

Wolfdieter Lang, Nov 20 2015

Keywords

• nonn
• easy

Comments

This entry is motivated by A262244 by Stuart E Anderson, where the angle numbers m_j, j=1..n, for concave n-gons play the role of the parts.

The sum of the n-gon angles m_j*Pi/n, j = 1..n, is (n-2)*Pi, hence Sum_{j=1..n} m_j = n*(n-2).

For the proof of the a(n) formula sum for the part n*(n-3)+1-j the number of partitions of n-1+j with n-1 parts, for j = 0, 1, ..., n*(n-4). See an example below.

Examples

			a(4) = 1 from the partition (5,1,1,1) of 4*2 = 8 with 4 parts, at least one part > 4.
a(5) = 18 from the following partitions of 15: (11,1,1,1,1), (10,2,1,1,1), (9,3,1,1,1), (9,2,2,1,1), (8,4,1,1,1), (8,3,2,1,1),(8,2,2,1,1), (7,5,1,1,1), (7,4,2,1,1), (7,3,3,1,1), (7,3,2,2,1), (7,2,2,2,2), (6,6,1,1,1), (6,5,2,1,1), (6,4,3,1,1), (6,4,2,2,1), (6,3,3,2,1), (6,3,2,2,2).
  18 = 1 + 1 + 2 + 3 + 5 + 6.
For n=7 sum the following number of partitions: for 29 the number of 6-part partitions for 6, for 28 the 6-part partitions of 7, ..., up to the 6-part partitions of 27 for 8.
		

Crossrefs

Cf. A008284, A262244.

Formula

a(n) = Sum_{j=0..n*(n-4)} T(n-1+j, n-1), with T(n, k) = A008284(n, k) (number of partitions of n with k parts), n >= 4.