A070103 Number of obtuse integer triangles with perimeter n and prime side lengths.

0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 2, 0, 1, 0, 0, 0, 0, 0, 2, 0, 1, 0, 3, 0, 2, 0, 2, 0, 1, 0, 3, 0, 2, 0, 1, 0, 2, 0, 0, 0, 0, 0, 3, 0, 1, 0, 4, 0, 5, 0, 4, 0, 2, 0, 1, 0, 1, 0, 2, 0, 2, 0, 3, 0, 1, 0, 6, 0, 4, 0, 6, 0, 6, 0

Offset: 1

Author

Reinhard Zumkeller, May 05 2002

Keywords

• nonn

Examples

			For n=11 there are A005044(11)=4 integer triangles: [1,5,5], [2,4,5], [3,3,5] and [3,4,4]; only one of the two obtuses ([2,4,5] and [3,3,5]) consists of primes, therefore a(11)=1.
		

Links

• R. Zumkeller, Integer-sided triangles

Crossrefs

Cf. A070080, A070081, A070082, A070093, A070098, A070101, A070088, A070105, A070108, A070129.

Programs

Mathematica

Table[Sum[Sum[(PrimePi[i] - PrimePi[i - 1]) (PrimePi[k] - PrimePi[k - 1]) (PrimePi[n - i - k] - PrimePi[n - i - k - 1]) (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 13 2019 *)

Formula

a(n) = A070093(n) - A070098(n).

a(n) = Sum_{k=1..floor(n/3)} Sum_{i=k..floor((n-k)/2)} (1 - sign(floor((i^2 + k^2)/(n-i-k)^2))) * sign(floor((i + k)/(n-i-k+1))) * A010051(i) * A010051(k) * A010051(n-i-k). - Wesley Ivan Hurt, May 13 2019