A160438 -id:A160438 - cp's OEIS frontend

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

Hagen von Eitzen, May 14 2009

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.

			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.

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

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

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)

A160456 Number of triangles that can be built from rods with lengths 1,2,...,n by using and concatenating not necessarily all rods.

Original entry on oeis.org

0, 3, 20, 70, 172, 366, 709, 1274, 2166, 3537, 5573, 8494, 12588, 18227, 25846, 35942, 49124, 66138, 87827, 115132, 149166, 191238, 242800, 305447, 381012, 471602, 579518, 707254, 857627, 1033812, 1239238, 1477589, 1752963

Offset: 3

Hagen von Eitzen, May 14 2009

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.

			For n = 4, there are 10 triangles with perimeter at most 1+2+3+4 = 10: (1,1,1), (2,2,1), (2,2,2), (3,2,2), (3,3,2), (3,3,3), (4,3,2), (4,3,3), (4,4,1) and (4,4,2). We have a(4)=3 because only 3 of these can be built from rods among 1,2,3,4: (4,3,2), (4,3,3)=(4,3,1+2) and (4,4,2)=(4,1+3,2). For example, it is not possible to build (4,4,1) because the 1-rod must be used for one of the 4-edges.

H. v. Eitzen, Table of n, a(n) for n=3..5262 (i.e. a(n) less than 2^64)
"AI", (Sci.math thread)
H. v. Eitzen, How to Build Triangles from Integers

A002623 is a similar problem where one rod per edge is to be used.
A160455 is a similar problem where all rods must be used.
A160438 is related to this if one drops the triangle inequality condition.

If n<=2, then trivially a(n)=0 because three edges need at least three rods.

If n>=8 then a(n) = A001400(n*(n+1)/2 - 3) - 11 - A133872(n+1).

cp's OEIS Frontend

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

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