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.
LINKS
H. v. Eitzen, Table of n, a(n) for n=3..10000
H. v. Eitzen, How to Build Triangles from Integers
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
KEYWORD
easy,nonn
AUTHOR
Hagen von Eitzen, May 14 2009
STATUS
approved