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A336331
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Middle side of primitive integer-sided triangles with A < B < C < 2*Pi/3 and such that FA + FB + FC is an integer where F is the Fermat point of the triangle.
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10
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65, 88, 147, 168, 185, 221, 285, 312, 315, 343, 392, 377, 464, 343, 408, 589, 725, 901, 931, 1085, 1120, 1240, 1147, 1128, 1208, 1400, 1105, 1099, 1005, 1464, 1541, 1544, 1635, 1403, 1463, 1387, 1744, 1891, 1617, 1505, 1769, 1885, 2072, 2345, 1713, 2165, 2667, 2784, 2855
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OFFSET
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1,1
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COMMENTS
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Inspired by Project Euler, Problem 143 (see link).
For the corresponding primitive triples and miscellaneous properties and references, see A336328.
If FA + FB + FC = d, then we have this "beautifully symmetric equation" between a, b, c and d (see Martin Gardner):
3*(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.
This sequence is not increasing. For example, a(11) = 392 for triangle with largest side = 407 while a(12) = 377 for triangle with largest side = 437.
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REFERENCES
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Martin Gardner, Mathematical Circus, Elegant triangles, First Vintage Books Edition, 1979, p. 65.
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LINKS
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FORMULA
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EXAMPLE
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a(1) = 65 because the first triple is (57, 65, 73) with corresponding d = FA + FB + FC = 264/7 + 195/7 + 325/7 = 112 and the symmetric relation satisfies: 3*(57^4 + 65^4 + 73^4 + 112^4) = (57^2 + 65^2 + 73^2 + 112^2)^2 = 642470409.
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CROSSREFS
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Cf. A072053 (middle sides: primitives and multiples).
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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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