A256629 - OEIS

%I #13 Jan 14 2025 23:15:38

%S 24,120,240,720,840,1320,2520,3360,3960,5280,6240,6840,9360,10920,

%T 14280,15600,16320,17160,18480,22440,24360,26520,31920,35880,38760,

%U 42840,43680,46200,50160,55200,57960,59280,70200,73920,91080,93840,100800,107640,117600,118320,122400

%N Integer areas A of integer-sided triangles such that the length of the circumradius is a prime number.

%C Subsequence of A208984.

%C For the same area, the number of triangles such that the length of the circumradius is a prime number is not unique; for example, from a(5)= 840 we find two triangles of sides (a,b,c)=(40,42,58) and (24,70,74) where R = 29 and 37, respectively.

%C The circumradius R values corresponding to the terms of the sequence are 5, 13, 17, 41, (29 or 37), 61, 53, 113, 101, 73, 89, 181, 97, (109 or 197), 149, ...

%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. The circumradius R is given by R = abc/4A.

%C Observation:

%C - all sides of the triangles are even;

%C - the inradius values are also even;

%C - the first triangle, of sides (6,8,10), is the unique triangle in which the lengths of the inradius and the circumradius are both prime numbers (r = A/p = 24/12 = 2 and R = abc/4A = 480/4*24 = 5).

%C For the same area, it is possible to find a prime inradius (see A230195), but the corresponding circumradius is generally rational. For example, for a(2) = 120, we find two triangles:

%C (10,24,26) => r = 4 and R = 13;

%C (16,25,39) => r = 3 prime and R = 65/2.

%C The following table gives the first values (A, a, b, c, r, R) where A is the integer area, a,b,c are the sides and r = A/p, R = a*b*c/4*A are respectively the values of the inradius and the circumradius.

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

%C |    A   |   a  |   b   |   c   |   r  |   R   |

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

%C |    24  |   6  |    8  |   10  |   2  |    5  |

%C |   120  |  10  |   24  |   26  |   4  |   13  |

%C |   240  |  16  |   30  |   34  |   6  |   17  |

%C |   720  |  18  |   80  |   82  |   8  |   41  |

%C |   840  |  40  |   42  |   58  |  12  |   29  |

%C |   840  |  24  |   70  |   74  |  10  |   37  |

%C |  1320  |  22  |  120  |  122  |  10  |   61  |

%C |  2520  |  56  |   90  |  106  |  20  |   53  |

%C |  3360  |  30  |  224  |  226  |  14  |  113  |

%C |  3960  |  40  |  198  |  202  |  18  |  101  |

%C |  5280  |  96  |  110  |  146  |  30  |   73  |

%C |  6240  |  78  |  160  |  178  |  30  |   89  |

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

%e a(1) = 24 because, for (a,b,c) = (6, 8, 10) => s= (6+8+10)/2 =12, and

%e A = sqrt(12(12-6)(12-8)(12-10)) = sqrt(576) = 24;

%e R = abc/4A = 480/4*24 = 5 is prime.

%Y Cf. A188158, A208984, A230195.

%K nonn

%O 1,1

%A _Michel Lagneau_, Apr 05 2015

%E Missing terms 91080 and 117600 added by _Zachary Sizer_, Jan 02 2025