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A229941 Sequence of triples: the 10 solutions of 1/p + 1/q + 1/r = 1/2 with 0 < p <= q <= r, lexicographically sorted. 2

%I #33 Apr 29 2019 01:34:45

%S 3,7,42,3,8,24,3,9,18,3,10,15,3,12,12,4,5,20,4,6,12,4,8,8,5,5,10,6,6,6

%N Sequence of triples: the 10 solutions of 1/p + 1/q + 1/r = 1/2 with 0 < p <= q <= r, lexicographically sorted.

%C As noted by John Baez, "each of [the 10 solutions of 1/p + 1/q + 1/r = 1/2] gives a way for three regular polygons to snugly meet at a point".

%C Among the 14 4-term Egyptian fractions with unit sum, there are 10 of the form 1/2 + 1/p + 1/q + 1/r.

%C Also integer values of length, width and height of a rectangular prism whose surface area is equal to its volume: pqr = 2(pq+pr+qr). - _John Rafael M. Antalan_, Jul 05 2015

%H John Baez, <a href="http://www.math.ucr.edu/home/baez/42.html">The answer is 42.</a>

%H J. F. T. Rabago and R. P. Tagle, <a href="http://nntdm.net/volume-19-2013/number-3/28-32/">On the Area and Volume of a certain Rectangular Solid and the Diophantine Equation 1/2=1/x+1/y+1/z</a>, Notes on Number Theory and Discrete Mathematics, 19-3 (2013), 28-32.

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Hurwitz%27s_automorphisms_theorem">Hurwitz's automorphisms theorem.</a>

%e a(1) = 3, a(2) = 7, a(3) = 42, since 1/3 + 1/7 + 1/42 = 1/2.

%e The 10 solutions are:

%e 3, 7, 42;

%e 3, 8, 24;

%e 3, 9, 18;

%e 3, 10, 15;

%e 3, 12, 12;

%e 4, 5, 20;

%e 4, 6, 12;

%e 4, 8, 8;

%e 5, 5, 10;

%e 6, 6, 6

%t {p, q, r} /. {ToRules[Reduce[0 < p <= q <= r && 1/p + 1/q + 1/r == 1/2, {p, q, r}, Integers] ]} // Flatten

%Y Cf. A230400, A260819.

%K easy,fini,nonn,full,tabf

%O 1,1

%A _Jean-François Alcover_, Oct 04 2013

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