A066360 - OEIS

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A066360

Number of unordered solutions in positive integers of xy + xz + yz = n with gcd(x,y,z) = 1.

0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 2, 0, 1, 1, 2, 1, 2, 0, 2, 1, 2, 0, 3, 2, 1, 2, 2, 0, 3, 0, 3, 2, 2, 1, 4, 1, 1, 2, 4, 2, 4, 0, 2, 2, 2, 1, 5, 2, 2, 2, 4, 1, 3, 2, 4, 4, 2, 0, 6, 0, 3, 3, 4, 2, 4, 2, 2, 3, 4, 0, 7, 2, 2, 4, 4, 2, 4, 0, 5, 4, 3, 1, 6, 2, 2, 4, 6, 2, 6, 2, 4, 2, 2, 3, 8, 4, 2, 3, 4, 1

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OFFSET

1,11

COMMENTS

These correspond to Descartes quadruples (-s, s+x+y, s+x+z, s+y+z) where s = sqrt(n), which are primitive if n is a perfect square.

Many empirical regularities are known, e.g., for n = 2^(2k) or n=2^(2k-1), (2 <= k <= 10 and even k <= 20), a(n) = 2^(k-2).

It appears that a(n) > 0 for n > 462. An upper bound on the number of solutions appears to be 1.5*sqrt(n). - T. D. Noe, Jun 14 2006

LINKS

T. D. Noe, Table of n, a(n) for n = 1..10000

EXAMPLE

a(81) = 3 because we have the triples (x,y,z) = (1,1,40),(2,3,15),(3,6,7) (and not (3,3,12) because this is not primitive).

MATHEMATICA

Table[cnt=0; Do[z=(n-x*y)/(x+y); If[IntegerQ[z] && GCD[x, y, z]==1, cnt++ ], {x, Sqrt[n/3]}, {y, x, Sqrt[x^2+n]-x}]; cnt, {n, 100}] (* T. D. Noe, Jun 14 2006 *)

PROG

(Haskell)

a066360 n = length [(x, y, z) | x <- [1 .. a000196 n],

y <- [x .. div n x],

z <- [y .. n - x*y],

x*y+(x+y)*z == n, gcd (gcd x y) z == 1]

-- Reinhard Zumkeller, Mar 23 2012

CROSSREFS

Cf. A060790, A062536 (and A007875 for xy = n).

Cf. A000196, A066955.

Sequence in context: A055639 A156542 A307990 * A061358 A025866 A259920

Adjacent sequences: A066357 A066358 A066359 * A066361 A066362 A066363