A336331 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.

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

Offset

1,1

Comments

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.

References

Martin Gardner, Mathematical Circus, Elegant triangles, First Vintage Books Edition, 1979, p. 65.

Links

Table of n, a(n) for n=1..49.

Project Euler, Problem 143 - Investigating the Torricelli point of a triangle.

Formula

a(n) = A336328(n, 2).

Example

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.

Crossrefs

Cf. A336328 (triples), A336329 (FA + FB + FC), A336330 (smallest side), this sequence (middle side), A336332 (largest side), A336333 (perimeter).

Cf. A072053 (middle sides: primitives and multiples).

Sequence in context: A299247 A015788 A072053 * A160055 A250642 A280755

Adjacent sequences: A336328 A336329 A336330 * A336332 A336333 A336334

Keyword

nonn

Author

Bernard Schott, Jul 20 2020

Status

approved