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A218243 Triangle numbers: m = a*b*c such that the integers a,b,c are the sides of a triangle with integer area. 2
60, 150, 200, 480, 780, 1200, 1530, 1600, 1620, 1690, 1950, 2040, 2100, 2730, 2860, 3570, 3840, 4050, 4056, 4200, 4350, 4624, 5100, 5400, 5460, 6240, 7500, 8120, 8250, 8670, 8750, 9600, 10812, 11050, 11900, 12180, 12240, 12800, 12960, 13260, 13520, 13650 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
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.
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.
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).
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
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
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..2000
Eric Weisstein's World of Mathematics, Triangle
EXAMPLE
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).
MATHEMATICA
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 *)
PROG
(PARI) Heron(a, b, c)=a*=a; b*=b; c*=c; ((a+b+c)^2-2*(a^2+b^2+c^2))
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
CROSSREFS
Subsequence of A139270.
Sequence in context: A044692 A211332 A039497 * A218392 A135199 A112065
KEYWORD
nonn
AUTHOR
Michel Lagneau, Oct 24 2012
STATUS
approved

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Last modified July 27 11:17 EDT 2024. Contains 374647 sequences. (Running on oeis4.)