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A070201
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Number of integer triangles with perimeter n having integral inradius.
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11
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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
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OFFSET
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1,36
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COMMENTS
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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.
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LINKS
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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
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EXAMPLE
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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.
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PROG
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(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
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CROSSREFS
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Cf. A070209, A070202, A070208, A005044, A070140.
Cf. A120062, A120572, A331040.
Sequence in context: A135767 A208575 A070203 * A070138 A024153 A341523
Adjacent sequences: A070198 A070199 A070200 * A070202 A070203 A070204
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KEYWORD
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nonn
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AUTHOR
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Reinhard Zumkeller, May 05 2002
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STATUS
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approved
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