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A070103
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Number of obtuse integer triangles with perimeter n and prime side lengths.
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6
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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
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OFFSET
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1,27
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LINKS
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FORMULA
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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
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EXAMPLE
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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.
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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