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A070103
Number of obtuse integer triangles with perimeter n and prime side lengths.
6
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
1,27
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
EXAMPLE
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.
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 *)
KEYWORD
nonn
AUTHOR
Reinhard Zumkeller, May 05 2002
STATUS
approved