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A160455
Number of triangles that can be built from rods with lengths 1,2,...,n by using and concatenating all rods.
2
0, 2, 7, 12, 16, 27, 48, 70, 91, 127, 184, 243, 300, 385, 507, 631, 752, 919, 1141, 1365, 1587, 1875, 2241, 2611, 2977, 3434, 3997, 4563, 5125, 5808, 6627, 7450, 8269, 9241, 10384, 11532, 12675, 14008, 15552, 17101, 18644, 20419, 22447
OFFSET
3,2
COMMENTS
a(n) is the number of triples (a,b,c) with b+c > a >= b >=c > 0 such that three disjoint subsets A,B,C of {1,2,...,n} with respective element sums a,b,c exist.
a(n) is also the number of partitions as counted in A160438 with additional constraint that there are only three parts and these satisfy the triangle inequality.
FORMULA
If n<=2, a(n)=0 trivially because three edges need at least three rods.
If n>=4, then a(n) = A005044(n*(n+1)/2), i.e. for n big enough all triangles of suitable perimeter can be obtained.
Conjectures from Colin Barker, May 12 2019: (Start)
G.f.: x^4*(2 - x + 2*x^2 - 3*x^3 + 6*x^4 - 5*x^5 + 8*x^6 - 9*x^7 + 11*x^8 - 11*x^9 + 11*x^10 - 10*x^11 + 10*x^12 - 8*x^13 + 5*x^14 - 3*x^15 + x^16) / ((1 - x)^5*(1 + x^2)^3*(1 + x + x^2)*(1 + x^4)).
a(n) = 4*a(n-1) - 9*a(n-2) + 17*a(n-3) - 27*a(n-4) + 37*a(n-5) - 47*a(n-6) + 55*a(n-7) - 59*a(n-8) + 59*a(n-9) - 55*a(n-10) + 47*a(n-11) - 37*a(n-12) + 27*a(n-13) - 17*a(n-14) + 9*a(n-15) - 4*a(n-16) + a(n-17) for n>20.
(End)
EXAMPLE
For n=3, only one integer-sided triangle with perimeter 1+2+3=6 exists, namely (2,2,2). This cannot be built from rods of length 1,2 and 3. Therefore a(3)=0.
For n=4, two triangles with perimeter 1+2+3+4=10 exist: (4,4,2) and (4,3,3); both can be built from the available rods: (4,1+3,2) and (4,3,1+2). Therefore a(4)=2.
CROSSREFS
A002623 is a similar problem where one rod per edge is to be used.
Sequence in context: A190453 A320900 A215247 * A045929 A277598 A105501
KEYWORD
easy,nonn
AUTHOR
Hagen von Eitzen, May 14 2009
STATUS
approved