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Number of obtuse integer triangles with perimeter n.
17

%I #14 May 14 2019 10:24:02

%S 0,0,0,0,0,0,1,0,1,0,2,0,2,2,3,2,3,3,5,3,7,4,8,5,9,7,10,8,11,9,14,11,

%T 16,12,18,14,19,17,21,18,23,21,27,22,30,24,32,27,34,30,37,33,40,35,44,

%U 37,47,40,50,44,53,49,56,52,60,55,64,57,68

%N Number of obtuse integer triangles with perimeter n.

%C An integer triangle [A070080(k) <= A070081(k) <= A070082(k)] is obtuse iff A070085(k) < 0.

%H Seiichi Manyama, <a href="/A070101/b070101.txt">Table of n, a(n) for n = 1..1000</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/ObtuseTriangle.html">Obtuse Triangle</a>.

%H R. Zumkeller, <a href="/A070080/a070080.txt">Integer-sided triangles</a>

%F a(n) = A005044(n) - A070093(n) - A024155(n).

%F a(n) = A024156(n) + A070106(n).

%F a(n) = Sum_{k=1..floor(n/3)} Sum_{i=k..floor((n-k)/2)}

%F (1-sign(floor((i^2 + k^2)/(n-i-k)^2))) * sign(floor((i+k)/(n-i-k+1))). - _Wesley Ivan Hurt_, May 12 2019

%e For n=14 there are A005044(14)=4 integer triangles: [2,6,6], [3,5,6], [4,4,6] and [4,5,5]; two of them are obtuse, as 3^2+5^2<36=6^2 and 4^2+4^2<36=6^2, therefore a(14)=2.

%t Table[Sum[Sum[(1 - Sign[Floor[(i^2 + k^2)/(n - i - k)^2]]) Sign[Floor[(i + k)/(n - i - k + 1)]], {i, k, Floor[(n - k)/2]}], {k, Floor[n/3]}], {n, 100}] (* _Wesley Ivan Hurt_, May 12 2019 *)

%Y Cf. A070102, A070103, A070127.

%Y Cf. A005044, A024155, A070093.

%K nonn

%O 1,11

%A _Reinhard Zumkeller_, May 05 2002