|
| |
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
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
|
|
|
|
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.
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.
|
|
|
KEYWORD
|
easy,nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
