%I
%S 1,1,2,5,13,35,93,215,437,815,1436,2413,3886,6041,9125,13436,19323,
%T 27221,37670,51293,68797,91025,118982,153797,196721,249206,312935,
%U 389761,481709,591080,720485,872763,1050980,1258565,1499351,1777462
%N Number of partitions of n*(n+1)/2 with at most four parts that can be obtained from grouping (with parentheses) a permutation of the sum 1+2+...+n
%C a(n) is the number of integer quadruples (x,y,z,w) with x >= y >= z >= w >= 0 and x+y+z+w = n*(n+1)/2 such that the set {1,2,...,n} can be partitioned into four (possibly empty) subsets with respective element sums x, y, z, w.
%H H. v. Eitzen, <a href="/A160438/b160438.txt">Table of n, a(n) for n=0..5262 (i.e. a(n) less than 2^64)</a>
%H "AI", <a href="http://groups.google.com/group/sci.math/browse_frm/thread/70c1521d9143d634">(sci.math thread that inspired investigating the sequence)</a>
%H H. v. Eitzen, <a href="http://www.voneitzen.de/math/trianglesticks.pdf">How to Build Triangles from Integers</a>
%F If n >= 8 then a(n) = A001400(n*(n+1)/2)  2*A011848(n+1)  5.
%e For n = 3 the a(3) = 5 solutions are 6 = (1+2+3), 5+1 = (2+3)+(1), 4+2 = (1+3)+(2), 3+3 = (3)+(1+2), 3+2+1 = (3)+(2)+(1). Note that 3+3 = (1+2)+(3) is the same as (3)+(1+2) as both are 3+3.
%e For n = 6 the partition 10+4+4+3 is *not* among the a(6) = 93 solutions because 4 can only come from grouping either (4) or (1+3), hence both groupings would have to occur; but (1+3) conflicts with both possible groupings (3) and (1+2) which could produce 3.
%K nonn
%O 0,3
%A _Hagen von Eitzen_, May 13 2009
