A160455 - OEIS

%I #15 Sep 05 2021 19:20:21

%S 0,2,7,12,16,27,48,70,91,127,184,243,300,385,507,631,752,919,1141,

%T 1365,1587,1875,2241,2611,2977,3434,3997,4563,5125,5808,6627,7450,

%U 8269,9241,10384,11532,12675,14008,15552,17101,18644,20419,22447

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

%C 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.

%C 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.

%H H. v. Eitzen, <a href="/A160455/b160455.txt">Table of n, a(n) for n=3..10000</a>

%H "AI", <a href="http://groups.google.com/group/sci.math/browse_frm/thread/70c1521d9143d634">(Sci.math thread that inspired investigating the sequence)</a>

%H H. v. Eitzen, <a href="http://www.von-eitzen.de/math/trianglesticks.pdf">How to Build Triangles from Integers</a>

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

%F 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.

%F Conjectures from _Colin Barker_, May 12 2019: (Start)

%F 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)).

%F 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.

%F (End)

%e 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.

%e 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.

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

%K easy,nonn

%O 3,2

%A _Hagen von Eitzen_, May 14 2009