cp's OEIS Frontend

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

%I A264356 #13 Nov 26 2015 06:36:31
%S A264356 1,18,197,1844,16457,143975,1249642,10815498,93576157,810347897,
%T A264356 7027967095,61060521549,531506499178,4635288747540,40499108355263,
%U A264356 354472925956809,3107795826264979,27290688756270363,240010330685355235,2113784811395623970,18641067229072645836
%N A264356 Number of partitions of n*(n-2) with n parts and at least one part > n.
%C A264356 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.
%C A264356 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).
%C A264356 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.
%F A264356 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.
%e A264356 a(4) = 1 from the partition (5,1,1,1) of 4*2 = 8 with 4 parts, at least one part > 4.
%e A264356 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).
%e A264356   18 = 1 + 1 + 2 + 3 + 5 + 6.
%e A264356 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.
%Y A264356 Cf. A008284, A262244.
%K A264356 nonn,easy
%O A264356 4,2
%A A264356 _Wolfdieter Lang_, Nov 20 2015