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Search: id:A070201
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| A070201 |
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Number of integer triangles with perimeter n having integral inradius. |
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+0
9
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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
(list; graph; listen)
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OFFSET
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1,36
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COMMENT
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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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Eric Weisstein's World of Mathematics, Incircle.
Eric Weisstein's World of Mathematics, Heron's Formula.
R. 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 = SquareRoot((s-9)*(s-10)*(s-17)/s) = SquareRoot(9*8*1/18) = SquareRoot(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 = SquareRoot((s-9)*(s-12)*(s-15)/s) = SquareRoot(9*6*3/18) = SquareRoot(9) = 3; therefore A070200(370)=3.
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CROSSREFS
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Cf. A070209, A070202, A070208, A005044, A070140.
Sequence in context: A003475 A135767 A070203 this_sequence A070138 A024153 A079127
Adjacent sequences: A070198 A070199 A070200 this_sequence A070202 A070203 A070204
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KEYWORD
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nonn
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AUTHOR
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Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), May 05 2002
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