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A160438
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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
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3
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1, 1, 2, 5, 13, 35, 93, 215, 437, 815, 1436, 2413, 3886, 6041, 9125, 13436, 19323, 27221, 37670, 51293, 68797, 91025, 118982, 153797, 196721, 249206, 312935, 389761, 481709, 591080, 720485, 872763, 1050980, 1258565, 1499351, 1777462
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,3
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COMMENTS
| 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.
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LINKS
| H. v. Eitzen, Table of n, a(n) for n=0..5262 (i.e. a(n) less than 2^64)
"AI", (sci.math thread that inspired investigating the sequence)
H. v. Eitzen, How to Build Triangles from Integers
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FORMULA
| If n >= 8 then a(n) = A001400(n*(n+1)/2) - 2*A011848(n+1) - 5.
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EXAMPLE
| 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.
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.
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CROSSREFS
| Sequence in context: A137674 A048781 A097919 * A054657 A024576 A057960
Adjacent sequences: A160435 A160436 A160437 * A160439 A160440 A160441
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KEYWORD
| nonn
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AUTHOR
| Hagen von Eitzen (math(AT)von-eitzen.de), May 13 2009
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