

A228383


Area A of the triangle such that A, the sides, and the inradius are integers.


4



6, 24, 30, 36, 42, 48, 54, 60, 66, 84, 96, 108, 114, 120, 126, 132, 144, 150, 156, 168, 180, 192, 198, 210, 216, 240, 252, 264, 270, 294, 300, 324, 330, 336, 360, 378, 384, 390, 396, 408, 420, 432, 456, 462, 468, 480, 486, 504, 510, 522, 528, 540, 546, 570
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OFFSET

1,1


COMMENTS

The sequences A208984 and A185210 are subsequences of this sequence. The corresponding inradius r are 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 3, 4, 3, ...
The area A of a triangle whose sides have lengths a, b, and c is given by Heron's formula: A = sqrt(s*(sa)*(sb)*(sc)), where s = (a+b+c)/2. The inradius r is given by r = A/s.
a(n) is divisible by 6 and the squares of the form 36k^2 are in the sequence.


LINKS

Table of n, a(n) for n=1..54.
Mohammad K. Azarian, Solution of problem 125: Circumradius and Inradius, Math Horizons, Vol. 16, No. 2 (Nov. 2008), pp. 3234.
Eric W. Weisstein, MathWorld: Inradius


EXAMPLE

24 is in the sequence because for (a, b, c) = (6, 8, 10) => s =(6 + 8 + 10)/2 = 12; A = sqrt(12*(126)*(128)*(1210)) = sqrt(576)= 24; r = A/s = 2.


MATHEMATICA

nn = 1000; 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]] && IntegerQ[Sqrt[area2]/s], AppendTo[lst, Sqrt[area2]]]], {a, nn}, {b, a}, {c, b}]; Union[lst]


CROSSREFS

Cf. A188158, A120572, A210250, A208984, A185210.
Sequence in context: A216793 A294900 A064510 * A249667 A114274 A292985
Adjacent sequences: A228380 A228381 A228382 * A228384 A228385 A228386


KEYWORD

nonn


AUTHOR

Michel Lagneau, Aug 21 2013


STATUS

approved



