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Number of primitive Euler bricks with side length a < b < c < 10^n, i.e., in a boxed parameter space with dimension 10^n.
1

%I #20 Jan 01 2024 19:58:42

%S 0,0,5,19,65,242,704,1884,4631

%N Number of primitive Euler bricks with side length a < b < c < 10^n, i.e., in a boxed parameter space with dimension 10^n.

%C An Euler brick is a cuboid of integer side dimensions a, b, c such that the face diagonals are integers. It is called primitive if gcd(a,b,c)=1.

%C Because the sides of a cuboid are permutable without changing its shape, the total number of primitive Euler bricks in the parameter space a, b, c < 10^n is b(n) = 6*a(n) = 0, 0, 30, 114, 390, ...

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/EulerBrick.html">Euler Brick</a>

%H <a href="/index/Br#bricks">Index entries for sequences related to bricks</a>

%e a(3) = 5, since there are the five primitive Euler bricks [44, 117, 240], [85, 132, 720], [140, 480, 693], [160, 231, 792], [240, 252, 275] with longest side length < 1000.

%o (Sage)

%o def a(n):

%o ans = 0

%o for x in range(1,10^n):

%o divs = Integer(x^2).divisors()

%o for d in divs:

%o if (d <= x^2/d): continue

%o if (d-x^2/d >= 2*x): break

%o if (d-x^2/d)%2==0:

%o y = (d-x^2/d)/2

%o for e in divs:

%o if (e <= x^2/e): continue

%o if (e-x^2/e >= 2*y): break

%o if (e-x^2/e)%2==0:

%o z = (e-x^2/e)/2

%o if (gcd([x,y,z])==1) and (y^2+z^2).is_square():

%o ans += 1

%o return ans # _Robin Visser_, Jan 01 2024

%Y Cf. A031173, A031174, A031175, A239620.

%K nonn,more

%O 1,3

%A _Martin Renner_, Mar 22 2014

%E a(6)-a(8) from _Giovanni Resta_, Mar 22 2014

%E a(9) from _Robin Visser_, Jan 01 2024