
COMMENTS

a(n) == 0 (mod 6).
Empirically, 3*sqrt(3) < a(n)/n^2 <= 6. The lower bound is provably tight, the upper bound seems to be achieved infinitely often, e.g., for prime n >= 5.
From Michel Lagneau, Mar 02 2012: (Start)
Subset of A188158.
The area A of a triangle whose sides have lengths a, b, and c is given by Heron's formula: A = sqrt(s(sa)(sb)(sc)), where s = (a+b+c)/2. The radius of the incircle or inscribed circle (also known as the inradius, r) is given by r = A/s.
From n = 17, it is possible to find couples of triangles with the property: r1 > r2 and A1 < A2 where A1, A2 are the consecutive areas corresponding to the inradius r1, r2. For example, a(17) = 1734 with (a,b,c) = (51, 68, 85) and a(18) = 1710 with (a,b,c) = (57, 65, 68).
Another interesting property of this sequence is that a(n) is divisible by 6 and, except n = 3, a(n)/6 = n^2 if n prime, hence the proposition:
Among the set of triangles whose area and sides are integers and whose inradius r is a prime number other than 3, the smallest area A is given by A = 6r^2.
Example: if r = 5, the areas of the triangles are {150, 210, 270, 330, 390, 510, ...} and the smallest area of them is A = 6*5^2 = 150 because 5 is prime.
Proof: Let r be a number such that the sides of a triangle are a = 3r, b = 4r, c = 5r. Then s = (a+b+c)/2 = 6r and A = sqrt(s(sa)(sb)(sc) = sqrt(36r^4) = 6r^2 is a possible area. Is 6r^2 the smallest area? The response is no in the general case for the composite numbers.
Writing a = (m+n)/2, b = (n+l)/2, c = (l+m)/2, and using rs = A and Heron's formula for A, we find lmn = 4r^2(l+m+n). Since m, n and l have to be of the same parity for a, b and c to be integral, they must therefore be even. Setting l = 2u, m = 2v, and n = 2w, we have a = v+w, b = w+u, c = u+v, and uvw = r^2(u+v+w). Then r^2 = uvw/(u+v+w).
First case: If r = p is prime, we prove that A = 6p^2 is the smallest area of all the triangles whose inradius is p. Suppose A' < A with inradius(A') = p. The area A is the corresponding value of the triangle (u,v,w) = (1*p, 2*p, 3*p) because 6p^3/6p = p^2. However, inradius(A') = p => u'v'w'/(u'+v'+w') = p^2 => (u',v',w') = (u,v,w) and A is the smallest area.
Second case: If r = q is composite, the triangle (u,v,w) = (1*q, 2*q, 3*q) gives an area A with inradius(A) = q, but it is possible to find A' < A with inradius(A') = q; for example, if q = 10, the triangle (u,v,w) = (30, 20, 10) whose area is A = 600 gives sqrt{(30*20*10)/(30+20+10)} = sqrt(6000/60) = 10 and the triangle(u',v',w') = (24,15,15) whose area is A' = 540 gives sqrt{(24*15*15)/(24+15+15)} = sqrt(5400/54) = 10.
(End)
