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 A253686 a(n) is the smallest integer area of the triangle having the sides in the commutative ring Z[sqrt(q)] where q = A005117(n) is a squarefree number. 0
 6, 1, 3, 1, 6, 3, 3, 4, 3, 5, 3, 2, 6, 6, 6, 6, 3, 5, 6, 6, 6, 6, 6, 3, 6, 6, 4, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 4, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 3, 6, 6, 4, 6, 6, 6, 6, 6, 5, 6, 6, 6, 6, 6, 6, 6, 6, 4, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Generalized integer areas triangles in the ring Z[sqrt(q)] = {a + b sqrt(q)| a,b in Z}. Introduction: The study of triangles having their sides with values in a ring Z[sqrt(q)] and having integer area gives remarkable properties probably still unexplored today. Property: a(1) = 6 because the ring Z[sqrt(1)] = Z => the smallest area of integer sides is A188158(1) = 6 => a(n) <=6. 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. For the same area, the number of triangles is not unique, for instance, in the ring Z[sqrt(19)] the area of each following triangle: (3, 4, 5), (8, -2+sqrt(19), 2+sqrt(19)), (6-sqrt(q), 3+sqrt(19), -1+2sqrt(19)), (-3+sqrt(19), 6+sqrt(19), 1+2sqrt(19)) is A=6. Conjecture: the set of squarefree numbers q such that the integer area A of the triangles with sides in the commutative ring Z[sqrt(q)] is finite if A < 6. It follows that a(n)= 6 for n > 384 where A005117(384) = 629 and a(384)=5. The corresponding values q such that a(n)<6 are 2, 3, 5, 7, 10, 11, 13, 14, 15, 17, 26, 29, 37, 41, 65, 85, 89, 101, 113, 221 and 629 with the corresponding index in the sequence a(n): 2, 3, 4, 6, 7, 8, 9, 10, 11, 12, 17, 18, 24, 27, 40, 53, 56, 62, 71, 137 and 384. We observe a subset of numbers q such that the sides of the triangles are of the particular form (a, b, sqrt(q)) with a, b integers. This subset is {5, 13, 17, 29, 37, 41, 65, 85, 89, 101, 113, 221, 629}. See the table below for the examples. We find also five isosceles triangles with area less than 6 and it is observed that they are of the form (2,sqrt(q),sqrt(q)) with q = 2, 5, 10, 17 and 26. The corresponding areas A are 1, 2, 3, 4 and 5 respectively with the formula A = sqrt(q-1). The following table gives the first values (A, sqrt(q), a, b, c) where A is the smallest area of the triangle (a, b, c), Z[sqrt(q)] is the commutative ring and a, b, c are the sides in Z[sqrt(q)]= Z[sqrt(A005117(n))]. --------------------------------------------------------------- |  A  |   sqrt(q) |    a        |    b         |     c        | --------------------------------------------------------------| |  6  |   sqrt(1) |    2        |    3         |     4        | |  1  |   sqrt(2) |    2        |       sqrt(2)| sqrt(2)      | |  3  |   sqrt(3) | 3 - sqrt(3) |  2 + 2sqrt(3)| 1 + 3sqrt(3) | |  1  |   sqrt(5) |    1        |    2         |   sqrt(5)    | |  6  |   sqrt(6) |   2sqrt(6)  | -2 + 2sqrt(6)|  2 + 2sqrt(6)| |  3  |   sqrt(7) |    4        | -1 + sqrt(7) |  1 + sqrt(7) | |  3  |  sqrt(10) |    2        |   sqrt(10)   |   sqrt(10)   | |  4  |  sqrt(11) |    6        | -1 + sqrt(11)|  1 + sqrt(11)| |  3  |  sqrt(13) |    2        |    3         |   sqrt(13)   | |  5  |  sqrt(14) |    6        | -2 + sqrt(14)|  2 + sqrt(14)| |  3  |  sqrt(15) |    8        |  5 - sqrt(15)|  5 + sqrt(15)| |  2  |  sqrt(17) |    1        |    4         |    sqrt(17)  | |  6  |  sqrt(19) |    8        | -2 + sqrt(19)|  2 + sqrt(19)| |  6  |  sqrt(21) |    3        |    4         |    5         | |  6  |  sqrt(22) |    3        |    4         |    5         | |  6  |  sqrt(23) |    3        |    4         |    5         | |  3  |  sqrt(26) |   10        | -4 + sqrt(26)| 4 + sqrt(26) | |  5  |  sqrt(29) |    2        |    5         |    sqrt(29)  | |  6  |  sqrt(30) |    3        |    4         |    5         | ............................................................... LINKS Eric W. Weisstein, MathWorld: Triangle Wolfram MathWorld, Ring EXAMPLE a(384)=5 because q = A005117(384) = 629 and the area A of the triangle (1, 26, sqrt(629)) is given by Heron's formula: A = sqrt(s*(s-1)*(s-26)*(s-sqrt(629))) where s = (1+26+sqrt(629))/2. We find A = 5. MATHEMATICA (* take q=Sqrt(2), Sqrt(3), ..., A005117(k), ... successively *) err=1/10^10; nn=10; q=Sqrt[2]; lst={}; lst1={}; Do[If[u+q*v>0, lst=Union[lst, {u+q*v}]], {u, -nn, nn}, {v, -nn, nn}]; n1=Length[lst]; Do[a=Part[lst, i]; b=Part[lst, j]; c=Part[lst, k]; s=(a+b+c)/2; area2=s*(s-a)*(s-b)*(s-c); If[a*b*c !=0&&N[area2]>0&&Abs[N[Sqrt[area2]]-Round[N[Sqrt[area2]]]]

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Last modified July 18 19:00 EDT 2019. Contains 325144 sequences. (Running on oeis4.)