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%I #19 Feb 07 2014 04:08:08
%S 60,150,200,480,780,1200,1530,1600,1620,1690,1950,2040,2100,2730,2860,
%T 3570,3840,4050,4056,4200,4350,4624,5100,5400,5460,6240,7500,8120,
%U 8250,8670,8750,9600,10812,11050,11900,12180,12240,12800,12960,13260,13520,13650
%N Triangle numbers: m = a*b*c such that the integers a,b,c are the sides of a triangle with integer area.
%C A triangle number m is an integer with at least one decomposition m = a*b*c such that the area of the triangle of sides (a,b,c) is an integer. Because this property is not always unique, we introduce the notion of "triangle order" for each triangle number m, denoted by TO(m). For example, TO(60) = 1 because the decomposition 60 = 3*4*5 is unique with the triangle (3,4,5) whose area A is given by Heron's formula: A = sqrt(s*(s-a)*(s-b)*(s-c)), where s = (a+b+c)/2 => A = sqrt(6(6-3)(6-4)(6-5)) = 6, but TO(780) = 2 because 780 = 4*13*15 = 5*12*13 and the area of the triangle (4,13,15) is sqrt(16(16-4)(16-13)(16-15))=24 and the area of the triangle (5,12,13) is sqrt(15(15-5)(15-12)(15-13))=30.
%C Given an area A of A188158, there exists either a unique triangle number (for example for A = 6 => m = 60 = 3*4*5), or several triangle numbers (for example for A=60 => m1 = 4350 = 6*25*29, m2 = 2040 = 8*15*17, m3 = 1690 = 13*13*10.
%C The number of ways to write m = a*b*c with 1<=a<=b<=c<=m is given by A034836, thus: TO(m) <= A034836(m).
%C If n is in this sequence, so is nk^3 for any k > 0. Thus this sequence is infinite. - _Charles R Greathouse IV_, Oct 24 2012
%C In view of the preceding comment, one might call "primitive" the elements of the sequence for which there is no k>1 such that n/k^3 is again a term of the sequence. These elements 60, 150, 200, 780, 1530, 1690, 1950,... are listed in A218392. - _M. F. Hasler_, Oct 27 2012
%H Charles R Greathouse IV, <a href="/A218243/b218243.txt">Table of n, a(n) for n = 1..2000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Triangle.html">Triangle</a>
%e 60 is in the sequence because 60 = 3*4*5 and the corresponding area is sqrt(6(6-3)(6-4)(6-5)) = 6 = A188158(1).
%t nn = 500; lst = {}; Do[s = (a + b + c)/2; If[IntegerQ[s], area2 = s (s - a) (s - b) (s - c); If[0 < area2 <= nn^2 && IntegerQ[Sqrt[area2]], AppendTo[lst, a*b*c]]], {a, nn}, {b, a}, {c, b}]; Union[lst] (* Program from _T. D. Noe_, adapted for this sequence - see A188158 *)
%o (PARI) Heron(a,b,c)=a*=a;b*=b;c*=c;((a+b+c)^2-2*(a^2+b^2+c^2))
%o is(n)=fordiv(n,a, if(a^3<=n, next); fordiv(n/a,b, my(c=n/a/b,h); if(a>=b && b>=c && a<b+c, h=Heron(a,b,c); if(h%16==0 && issquare(h), return(1)))));0 \\ _Charles R Greathouse IV_, Oct 24 2012
%Y Subsequence of A139270.
%Y Cf. A034836, A188158, A083875.
%K nonn
%O 1,1
%A _Michel Lagneau_, Oct 24 2012
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