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Number of ordered integral solutions of Descartes' equation 2(a^2 + b^2 + c^2 + d^2) = (a + b + c + d)^2 with a + b + c + d = 4n - 2.
2

%I #22 Dec 14 2018 07:52:49

%S 1,2,1,2,4,4,3,4,5,6,8,6,5,10,7,8,14,8,9,12,11,12,12,12,14,18,13,12,

%T 22,16,15,24,12,18,24,18,19,20,24,20,28,22,16,28,23,24,32,20,25,38

%N Number of ordered integral solutions of Descartes' equation 2(a^2 + b^2 + c^2 + d^2) = (a + b + c + d)^2 with a + b + c + d = 4n - 2.

%C This count includes only primitive quadruples, i.e., with gcd=1. There may also be nonprimitive quads, e.g., with n=5, 4n-2=18, we have 4 primitive quads (-1,2,6,11), (-2,3,7,10), (1,1,4,12), (-3,5,8,8) and also the nonprimitive (0,3,3,12) and (3,6,6,9). - _Colin Mallows_, May 11 2005

%H R. L. Graham, J. C. Lagarias, C. L. Mallows, Allan Wilks, and C. H. Yan, <a href="http://arxiv.org/abs/math/0009113">Apollonian Circle Packings: Number Theory</a>, arXiv:math/0009113 [math.NT], 2000-2003; J. Number Theory, 100 (2003), 1-45.

%H J. C. Lagarias, C. L. Mallows, and Allan Wilks, <a href="http://arxiv.org/abs/math/0101066">Beyond the Descartes Circle Theorem</a>, arXiv:math/0101066 [math.MG], 2001.

%H J. C. Lagarias, C. L. Mallows, and Allan Wilks, <a href="http://www.jstor.org/stable/2695498">Beyond the Descartes Circle Theorem</a>, Amer. Math Monthly, 109 (2002), 338-361.

%e a(5) = 4 because we have the quadruples (1,1,4,12), (-1,2,6,11), (-2,3,7,10), (3,5,8,8).

%t r[n_] := Reduce[a <= b <= c <= d && 2 (a^2 + b^2 + c^2 + d^2) == (a + b + c + d)^2 && a + b + c + d == 4 n - 2, {a, b, c, d}, Integers];

%t a[n_] := Count[{a, b, c, d} /. {ToRules[r[n]]}, sol_ /; GCD @@ sol == 1];

%t Reap[Do[an = a[n]; Print[n, " ", an]; Sow[an], {n, 1, 50}]][[2, 1]] (* _Jean-François Alcover_, Dec 14 2018 *)

%Y Cf. A045864.

%K nonn

%O 1,2

%A _Colin Mallows_, Apr 28 2005

%E More terms from _Colin Mallows_, May 11 2005