A160455 Number of triangles that can be built from rods with lengths 1,2,...,n by using and concatenating all rods.

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

Author

Hagen von Eitzen, May 14 2009

Keywords

• easy
• nonn

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.

Examples

			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.
		

Links

• H. v. Eitzen, Table of n, a(n) for n=3..10000
• "AI", (Sci.math thread that inspired investigating the sequence)
• H. v. Eitzen, How to Build Triangles from Integers

Crossrefs

A002623 is a similar problem where one rod per edge is to be used.

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)