A070201 Number of integer triangles with perimeter n having integral inradius.

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 2, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 3, 0, 0, 0, 2, 0, 1, 0, 1, 0, 2, 0, 2, 0, 0, 0, 1, 0, 1, 0, 2, 0, 0, 0, 8, 0, 0, 0, 1, 0, 3

Offset: 1

Author

Reinhard Zumkeller, May 05 2002

Keywords

• nonn

Comments

a(n) = #{k | A070083(k) = n and A070200(k) = exact inradius};

a(n) = A070203(n) + A070204(n);

a(n) = A070205(n) + A070206(n) + A024155(n);

a(odd) = 0.

Examples

			a(36)=2, as there are two integer triangles with integer inradius having perimeter=32:
First: [A070080(368), A070081(368), A070082(368)] = [9,10,17], for s = A070083(368)/2 = (9+10+17)/2 = 18: inradius = sqrt((s-9)*(s-10)*(s-17)/s) = sqrt(9*8*1/18) = sqrt(4) = 2; therefore A070200(368) = 2.
2nd: [A070080(370), A070081(370), A070082(370)] = [9,12,15], for s = A070083(370)/2 = (9+12+15)/2 = 18: inradius = sqrt((s-9)*(s-12)*(s-15)/s) = sqrt(9*6*3/18) = sqrt(9) = 3; therefore A070200(370) = 3.
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..5000
• Mohammad K. Azarian, Solution to Problem S125: Circumradius and Inradius, Math Horizons, Vol. 16, Issue 2, November 2008, p. 32.
• Eric Weisstein's World of Mathematics, Incircle.
• Eric Weisstein's World of Mathematics, Heron's Formula.
• Reinhard Zumkeller, Integer-sided triangles

Crossrefs

Cf. A070209, A070202, A070208, A005044, A070140.

Cf. A120062, A120572, A331040.

Programs

Ruby

def A(n)
  cnt = 0
  (1..n / 3).each{|a|
    (a..(n - a) / 2).each{|b|
      c = n - a - b
      if a + b > c
        s = n / 2r
        t = (s - a) * (s - b) * (s - c) / s
        if t.denominator == 1
          t = t.to_i
          cnt += 1 if Math.sqrt(t).to_i ** 2 == t
        end
      end
    }
  }
  cnt
end
def A070201(n)
  (1..n).map{|i| A(i)}
end
p A070201(100) # Seiichi Manyama, Oct 06 2017