A283446 Integer areas of triangles with side lengths A001223(m), A001223(m+1) and A001223(m+2) for some m.

24, 96, 120, 144, 168, 216, 240, 264, 336, 360, 384, 432, 456, 480, 504, 528, 576, 600, 624, 672, 720, 792, 816, 840, 864, 936, 960, 1008, 1056, 1080, 1176, 1200, 1224, 1296, 1320, 1344, 1440, 1512, 1536, 1560, 1584, 1680, 1728, 1824, 1848, 1920, 1944, 2016

Offset: 1

Author

Michel Lagneau, Mar 08 2017

Keywords

• nonn

Comments

A001223(n) = A000040(n+1) - A000040(n) = prime(n+1) - prime(n).

The Mathematica program examines all triangles with n <= 10^8.

The sequence a(n) is a subsequence of A188158, and the lengths of the sides are even.

The area A of a triangle whose sides have lengths a, b, and c is given by Heron's formula: A = sqrt(s*(s-a)*(s-b)*(s-c)), where s = (a+b+c)/2.

a(n) == 0 mod 24 => {b(n)} = {a(n)/24} = {1, 4, 5, 6, 7, 9, 10, 11, 14, 15, 16, 18, 19, 20, 21, 22, 24, 25, 26, 28, 30, 33, 34, 35, 36, 39, 40, 42, 44, 45, 49, 51, 54, 55, 56, 60, 63, 64, 65, 70, 72, ...}. It seems that the primes > 19 are not in {b(n)}.

For the same area, the number of distinct triangles is not always unique; for example, the area 336 can be obtained with triangle (30, 28, 26) starting from prime 461983 and also from triangle (34, 20, 42) starting from prime 2473663 (Giovanni Resta, Mar 08 2017).

The following table gives the first values (A, m, sides of the triangles) where A is the area of the triangles and m is the smallest value generating A.

+-----+--------+-----------+-------------+-------------+

| A | m | A001223(m)| A001223(m+1)| A001223(m+2)|

+-----+--------+-----------+-------------+-------------+

| 24 | 123 | 6 | 8 | 10 |

| 96 | 3935 | 16 | 20 | 12 |

| 120 | 8101 | 10 | 26 | 24 |

| 144 | 13097 | 34 | 18 | 20 |

| 168 | 12226 | 30 | 40 | 14 |

| 216 | 9864 | 24 | 18 | 30 |

| 240 | 102715 | 58 | 50 | 12 |

| 264 | 98259 | 22 | 26 | 40 |

| 336 | 38604 | 30 | 28 | 26 |

+-----+--------+-----------+-------------+-------------+

Examples

			24 is in the sequence because, for the smallest value m = 123, we obtain the triangle of sides (A001223(123), A001223(124), A001223(125)) = (6, 8, 10) and the area is given by Heron's formula with s = 12 and A = sqrt(12(12-6)(12-8)(12-10)) = 24.
The set of the others values m > 123 giving the same area A = 24 starts with 127, 192, 269, 304, 417, 420, ...
		

Crossrefs

Cf. A000040, A001223, A188158, A229746.

Programs

Mathematica

nn=10^5;lst={};Do[u=Prime[a+1]-Prime[a];v=Prime[a+2]-Prime[a+1]; w=Prime[a+3]-Prime[a+2];s=(u+v+w)/2;If[IntegerQ[s],area2=s (s-u)(s-v)(s-w);If[area2>0&&IntegerQ[Sqrt[area2]],AppendTo[lst,Sqrt[area2]]]],{a,nn}];Union[lst]

Extensions

Missing terms 1200 and 1584 from Giovanni Resta, Mar 08 2017