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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.

%I A253686 #22 Feb 16 2025 08:33:24
%S A253686 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,
%T A253686 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,
%U A253686 6,6,4,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6
%N 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.
%C A253686 Generalized integer areas triangles in the ring Z[sqrt(q)] = {a + b sqrt(q)| a,b in Z}.
%C A253686 Introduction:
%C A253686 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.
%C A253686 Property:
%C A253686 a(1) = 6 because the ring Z[sqrt(1)] = Z => the smallest area of integer sides is A188158(1) = 6 => a(n) <=6.
%C A253686 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:
%C A253686 (3, 4, 5),
%C A253686 (8, -2+sqrt(19), 2+sqrt(19)),
%C A253686 (6-sqrt(q), 3+sqrt(19), -1+2*sqrt(19)),
%C A253686 (-3+sqrt(19), 6+sqrt(19), 1+2*sqrt(19)) is A=6.
%C A253686 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.
%C A253686 It follows that a(n)= 6 for n > 384 where A005117(384) = 629 and a(384)=5.
%C A253686 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.
%C A253686 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.
%C A253686 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).
%C A253686 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))].
%C A253686 +---+----------+-------------+--------------+---------------+
%C A253686 | A |  sqrt(q) |      a      |       b      |       c       |
%C A253686 +---+----------+-------------+--------------+---------------+
%C A253686 | 6 |  sqrt(1) |      2      |       3      |       4       |
%C A253686 | 1 |  sqrt(2) |      2      |    sqrt(2)   |    sqrt(2)    |
%C A253686 | 3 |  sqrt(3) | 3 - sqrt(3) | 2 + 2*sqrt(3)| 1 + 3*sqrt(3) |
%C A253686 | 1 |  sqrt(5) |      1      |       2      |    sqrt(5)    |
%C A253686 | 6 |  sqrt(6) |  2*sqrt(6)  |-2 + 2*sqrt(6)| 2 + 2*sqrt(6) |
%C A253686 | 3 |  sqrt(7) |      4      | -1 + sqrt(7) |  1 + sqrt(7)  |
%C A253686 | 3 | sqrt(10) |      2      |   sqrt(10)   |    sqrt(10)   |
%C A253686 | 4 | sqrt(11) |      6      |-1 + sqrt(11) |  1 + sqrt(11) |
%C A253686 | 3 | sqrt(13) |      2      |       3      |    sqrt(13)   |
%C A253686 | 5 | sqrt(14) |      6      |-2 + sqrt(14) |  2 + sqrt(14) |
%C A253686 | 3 | sqrt(15) |      8      | 5 - sqrt(15) |  5 + sqrt(15) |
%C A253686 | 2 | sqrt(17) |      1      |       4      |    sqrt(17)   |
%C A253686 | 6 | sqrt(19) |      8      |-2 + sqrt(19) |  2 + sqrt(19) |
%C A253686 | 6 | sqrt(21) |      3      |       4      |       5       |
%C A253686 | 6 | sqrt(22) |      3      |       4      |       5       |
%C A253686 | 6 | sqrt(23) |      3      |       4      |       5       |
%C A253686 | 3 | sqrt(26) |     10      |-4 + sqrt(26) | 4 + sqrt(26)  |
%C A253686 | 5 | sqrt(29) |      2      |       5      |    sqrt(29)   |
%C A253686 | 6 | sqrt(30) |      3      |       4      |       5       |
%C A253686 .............................................................
%H A253686 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Triangle.html">Triangle</a>.
%H A253686 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Ring.html">Ring</a>.
%e A253686 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.
%t A253686 (* take q=sqrt(2), sqrt(3), ..., A005117(k), ... successively *)
%t A253686 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]]]]<err,AppendTo[lst1,Round[Sqrt[N[area2]]]];Print[Round[Sqrt[N[area2]]]," ",a," ",b," ",c]],{i,1,n1},{j,i,n1},{k,j,n1}];Union[lst1]
%Y A253686 Cf. A005117, A188158, A238369.
%K A253686 nonn
%O A253686 1,1
%A A253686 _Michel Lagneau_, Jan 09 2015