

A105970


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



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

1,2


COMMENTS

This count includes only primitive quadruples, i.e., with gcd=1. There may also be nonprimitive quads, e.g., with n=5, 4n2=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


LINKS

Table of n, a(n) for n=1..50.
R. L. Graham, J. C. Lagarias, C. L. Mallows, Allan Wilks, and C. H. Yan, Apollonian Circle Packings: Number Theory, arXiv:math/0009113 [math.NT], 20002003; J. Number Theory, 100 (2003), 145.
J. C. Lagarias, C. L. Mallows, and Allan Wilks, Beyond the Descartes Circle Theorem, arXiv:math/0101066 [math.MG], 2001.
J. C. Lagarias, C. L. Mallows, and Allan Wilks, Beyond the Descartes Circle Theorem, Amer. Math Monthly, 109 (2002), 338361.


EXAMPLE

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


MATHEMATICA

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];
a[n_] := Count[{a, b, c, d} /. {ToRules[r[n]]}, sol_ /; GCD @@ sol == 1];
Reap[Do[an = a[n]; Print[n, " ", an]; Sow[an], {n, 1, 50}]][[2, 1]] (* JeanFrançois Alcover, Dec 14 2018 *)


CROSSREFS

Cf. A045864.
Sequence in context: A275122 A264569 A265601 * A133950 A241512 A094265
Adjacent sequences: A105967 A105968 A105969 * A105971 A105972 A105973


KEYWORD

nonn


AUTHOR

Colin Mallows, Apr 28 2005


EXTENSIONS

More terms from Colin Mallows, May 11 2005


STATUS

approved



