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 A256629 Integer areas A of integer-sided triangles such that the length of the circumradius is a prime number. 0
 24, 120, 240, 720, 840, 1320, 2520, 3360, 3960, 5280, 6240, 6840, 9360, 10920, 14280, 15600, 16320, 17160, 18480, 22440, 24360, 26520, 31920, 35880, 38760, 42840, 43680, 46200, 50160, 55200, 57960, 59280, 70200, 73920, 93840, 100800, 107640, 118320, 122400 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Subsequence of A208984. 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. 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, ... 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. Observation: - all sides of the triangles are even; - the inradius values are also even; - 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). 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: (10,24,26) => r = 4 and R = 13; (16,25,39) => r = 3 prime and R = 65/2. 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. +--------+------+-------+-------+------+-------+ |    A   |   a  |   b   |   c   |   r  |   R   | +--------+------+-------+-------+------+-------+ |    24  |   6  |    8  |   10  |   2  |    5  | |   120  |  10  |   24  |   26  |   4  |   13  | |   240  |  16  |   30  |   34  |   6  |   17  | |   720  |  18  |   80  |   82  |   8  |   41  | |   840  |  40  |   42  |   58  |  12  |   29  | |   840  |  24  |   70  |   74  |  10  |   37  | |  1320  |  22  |  120  |  122  |  10  |   61  | |  2520  |  56  |   90  |  106  |  20  |   53  | |  3360  |  30  |  224  |  226  |  14  |  113  | |  3960  |  40  |  198  |  202  |  18  |  101  | |  5280  |  96  |  110  |  146  |  30  |   73  | |  6240  |  78  |  160  |  178  |  30  |   89  | +--------+------+-------+-------+------+-------+ LINKS EXAMPLE a(1) = 24 because, for (a,b,c) = (6, 8, 10) => s= (6+8+10)/2 =12, and A = sqrt(12(12-6)(12-8)(12-10)) = sqrt(576) = 24; R = abc/4A = 480/4*24 = 5 is prime. MATHEMATICA nn=2000; Do[s=(a+b+c)/2; If[IntegerQ[s], area2=s (s-a)(s-b)(s-c); If[0

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Last modified January 27 22:22 EST 2022. Contains 350654 sequences. (Running on oeis4.)