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

%I #29 Mar 21 2017 06:33:49

%S 24,96,120,144,168,216,240,264,336,360,384,432,456,480,504,528,576,

%T 600,624,672,720,792,816,840,864,936,960,1008,1056,1080,1176,1200,

%U 1224,1296,1320,1344,1440,1512,1536,1560,1584,1680,1728,1824,1848,1920,1944,2016

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

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

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

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

%C 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.

%C 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)}.

%C 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).

%C 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.

%C +-----+--------+-----------+-------------+-------------+

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

%C +-----+--------+-----------+-------------+-------------+

%C | 24 | 123 | 6 | 8 | 10 |

%C | 96 | 3935 | 16 | 20 | 12 |

%C | 120 | 8101 | 10 | 26 | 24 |

%C | 144 | 13097 | 34 | 18 | 20 |

%C | 168 | 12226 | 30 | 40 | 14 |

%C | 216 | 9864 | 24 | 18 | 30 |

%C | 240 | 102715 | 58 | 50 | 12 |

%C | 264 | 98259 | 22 | 26 | 40 |

%C | 336 | 38604 | 30 | 28 | 26 |

%C +-----+--------+-----------+-------------+-------------+

%e 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.

%e The set of the others values m > 123 giving the same area A = 24 starts with 127, 192, 269, 304, 417, 420, ...

%t 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]

%Y Cf. A000040, A001223, A188158, A229746.

%K nonn

%O 1,1

%A _Michel Lagneau_, Mar 08 2017

%E Missing terms 1200 and 1584 from _Giovanni Resta_, Mar 08 2017