login
The OEIS is supported by the many generous donors to the OEIS Foundation.

 

Logo
Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A219965 Number of non-intersecting unit cubes regularly packed into the tetrahedron of edge length n. 1
0, 0, 0, 1, 4, 10, 19, 30, 45, 66, 94, 130, 172, 221, 278, 344, 422, 511, 611, 723, 848, 987, 1140, 1308, 1491, 1691, 1909, 2146, 2401, 2673, 2965, 3278, 3614, 3974, 4355, 4759, 5186, 5638, 6117, 6623, 7156, 7716, 8305, 8923, 9571, 10249, 10958, 11700, 12475, 13285, 14127, 15003, 15914, 16862, 17849, 18874, 19937, 21037, 22177, 23358, 24581, 25846, 27153, 28504 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,5
COMMENTS
The tetrahedron may be aligned with the Cartesian axes by putting its triangular basis on the horizontal plane, with four vertices at (x, y, z) = (0, 0, 0), (n, 0, 0), (n/2, sqrt(3)*n/2, 0) and (n/2, n/(2*sqrt(3)), n*sqrt(2/3)) see A194082, A020769, A157697.
The volume of tetrahedron is a third times the area of the base triangle times height, (1/3) * (sqrt(3)*n^2/4) * n*sqrt(2/3) = n^3/(3*2^(2/3) = A020829*n^3. This defines an obvious upper limit of floor(n^3/sqrt(72)) = A171973(n) of placing unit cubes into this tetrahedron.
Regular packing: We place the first layer of unit cubes so they touch the floor of the tetrahedron. Their number is limited by the area of the triangular horizontal section of the plane z=1 inside the tetrahedron, which touches all of them; this isosceles horizontal triangle has edge length E(n,z) = n-z*sqrt(3/2). This edge length is a linear interpolation for triangular horizontal cuts between z=0 at the bottom and the summit of the tetrahedron at z=n*sqrt(2/3).
This first layer confined by a triangle characterized by E(n,z) may host RegSquInTri(E) := sum_{y=1..floor(E*sqrt(3)/2)} floor(E-y*2/sqrt(3)) cubes, following recursively the same regular placement and counting strategy as for squares in isosceles triangles, see A194082.
The number of unit cubes in the next layer, between z=1 and z=2, is limited by the area of the horizontal section of the triangle z=2 inside the tetrahedron, where the triangle has edge length n-z*sqrt(3/2).
So in layer z=1, 2, ... we insert ReqSquInTri(E(n,z)) cubes. a(n) is the sum over all these layers with z limited by the z-value of the vertex at the summit.
There is a generalization to placing unit cubes of higher dimensions into higher dimensional tetrahedra.
The growth is expected to be roughly equal to the growth of A000292.
LINKS
FORMULA
a(n) <= A171973(n).
MAPLE
# Number or squares in isosceles triangle of edge length n.
RegSquInTri := proc(n)
add(floor(n-2*y/sqrt(3)), y=1..floor(n*sqrt(3)/2)) ;
end proc:
A219965 := proc(n)
local a, z, triedg ;
a := 0 ;
for z from 1 to floor(n*sqrt(2/3)) do
triedg := n-z*sqrt(3/2) ;
a := a+ RegSquInTri(triedg) ;
end do:
return a;
end proc:
CROSSREFS
Sequence in context: A022785 A241239 A057312 * A008038 A301248 A160425
KEYWORD
nonn
AUTHOR
R. J. Mathar, Dec 02 2012
STATUS
approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified April 23 23:26 EDT 2024. Contains 371917 sequences. (Running on oeis4.)