cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A359884 Number of 3-dimensional tilings of a 2 X 2 X n box using 2 X 2 X 1 plates and 1 X 2 X 1 dominos.

Original entry on oeis.org

1, 3, 24, 133, 839, 5056, 30969, 188603, 1150952, 7018621, 42811231, 261110416, 1592592465, 9713598835, 59245780536, 361354997685, 2203996629559, 13442737199456, 81990685695721, 500082110459883, 3050128402768520, 18603511408241453, 113467563119685583
Offset: 0

Views

Author

Gerhard Kirchner, Jan 20 2023

Keywords

Comments

The first recurrence is derived in "3d-tilings of a 2 X 2 X n box" as a special case of a more general tiling problem: III, example 4.

Examples

			a(1) = 3
      _______         _______          _______
    /       /|      /   /   /|       /______ /|
   /______ / |     /__ /__ / |      /______ /||
   |       | /     |   |   | /      |       ||/
   |_______|/      |___|___|/       |_______|/

Links

Crossrefs

Cf. A006253, A001045, A335559, A359885, A359886.

Programs

  • Mathematica
    LinearRecurrence[{5, 9, -14}, {1, 3, 24}, 25] (* Paolo Xausa, Jun 24 2024 *)
  • Maxima
    /* See link "Maxima code". */

Formula

G.f.: (1 - 2*x) / (1 - 5*x - 9*x^2 + 14*x^3).
a(n) = 3*a(n-1) + c(n-1) + 7*a(n-2) where c(n) = 8*a(n-1) + 2*c(n-1) with a(n),c(n) <= 0 for n <= 0 except for a(0)=1.
a(n) = 5*a(n-1) + 9*a(n-2) - 14*a(n-3) for n >= 3.

A359885 Number of 3-dimensional tilings of a 2 X 2 X 3n box using trominos (three 1 X 1 X 1 cubes).

Original entry on oeis.org

1, 44, 2512, 145088, 8383744, 484453376, 27994083328, 1617634967552, 93474855387136, 5401434047381504, 312121261353336832, 18035892123135377408, 1042202005934895529984, 60223526164332403490816, 3480009713100277581611008, 201091971436982107249836032
Offset: 0

Views

Author

Gerhard Kirchner, Jan 20 2023

Keywords

Comments

The first recurrence is derived in A359884, "3d-tilings of a 2 X 2 X n box" as a special case of a more general tiling problem: III, example 5.
The example uses two cross section profiles with two overstanding cubes: C (with a common square) and D (with one common edge).

Examples

			a(1)=44.
t1,t2,t3 is a tromino standing on 1,2,3 cubes respectively.
1) Two t2-tiles generate a C-profile or a D-profile in 4 ways each.
   C,D-profile: 4,2 rotation images, D-profile: 2 ways for each image.
    C-profile                      D-profiles
.     ___                      ___                   ___
.   /__ /|               ___ /__ /|            ___ /__ /|
. /__ /| |___          /__ /|   | |          /__ /|   | |
.|   | |/__ /|        |   | |___| |         |   | |___| |
.|   |/__ /| |        |   |/__ /| |         |   |/__ /  |
.|       | |/         |       | |/          |   |   |  /
.|_______|/           |_______|/            |___|___|/
2) t1+t3 generates a C-profile in 4*2=8 ways.
.     ___
.   /   /|                       ______
. /__ /  |    _______          /_____ /|    _______
.|   |  /   /__     /|        |      | |  /__     /|
.|   | |   |  /__ /  |   or   |    __|/  |  /__ /  |
.|   | |   |_|   |  /         |   | |    |_|   |  /
.|___|/      |___|/           |___|/       |___|/
1,2) There are 12 ways to generate a C-profile. The connection of two C-profiles is a 2 X 2 X 3 cuboid. Starting with a C-profile, there are 4*3*3=36 ways to generate this cuboid.
3) There are 4*2=8 ways to generate the cuboid by starting with a D-profile. Therefore, a(1)=36+8=44.
.     ___
.   /   /|          ___       ___
. /__ /  |    ___ /__ /|    /   /|
.|   |   |  /__ /|   | |  /__ /  |
.|___|/| | |   | |___| | |   |  /
.  |___|/  |   |/__ /| | |   | |    or
.          |       | |/  |   | |
.          |_______|/    |___|/
.   _______
. /______ /|          ___
.|       | |    ___ /__ /|    _______
.|    ___|/   /__ /|   | |  /______ /|
.|   | |     |   | |___| | |       | |
.|___|/      |   |/__ /| | |___    | |
.            |       | |/      |   | |
.            |_______|/        |___|/

Links

Crossrefs

Cf. A006253, A001045, A335559, A359884, A359886.

Programs

  • Mathematica
    LinearRecurrence[{60, -128}, {1, 44}, 20] (* Paolo Xausa, Jun 24 2024 *)
  • Maxima
    /* See A359884. */

Formula

G.f.: (1 - 16*x) / (1 - 60*x + 128*x^2).
a(n) = 44*a(n-1) + 6*e(n-1) where e(n) = 96*a(n-1) + 16*e(n-1) with a(n),e(n) <= 0 for n < =0 except for a(0)=1.
a(n) = 60*a(n-1) - 128*a(n-2) for n >= 2.
E.g.f.: exp(30*x)*cosh(2*sqrt(193)*x) + 7*exp(30*x)*sinh(2*sqrt(193)*x)/sqrt(193). - Stefano Spezia, Jan 21 2023

A360064 Number of 3-dimensional tilings of a 2 X 2 X n box using 1 X 1 X 1 cubes and trominos (L-shaped connection of 3 cubes).

Original entry on oeis.org

1, 5, 89, 1177, 16873, 237977, 3366793, 47599097, 673035625, 9516252633, 134553882441, 1902506043833, 26900227288361, 380352114739609, 5377937177440009, 76040613721296249, 1075165950495479017, 15202163218500810073, 214948926180739194569
Offset: 0

Views

Author

Gerhard Kirchner, Jan 30 2023

Keywords

Comments

Recurrence 1 is derived in A359884, "3d-tilings of a 2 X 2 X n box" as a special case of a more general tiling problem: III, example 8.

Examples

			4 rotations:
   ___ ___     ___ ___
  |   |   |   |   |   | (cross sections)
  |   |___|   |___|___|
  |       |   |   |   |
  |_______|   |___|___| a(1) = 4 + 1 = 5.

Links

Crossrefs

Cf. A006253, A001045, A033516, A335559, A359885, A359886, A360065, A360066.

Programs

  • Mathematica
    LinearRecurrence[{13, 20, -64, 112, 224, -128}, {1, 5, 89, 1177, 16873, 237977}, 25] (* Paolo Xausa, Oct 02 2024 *)

Formula

G.f.: (1 - 8*x + 4*x^2 - 16*x^3) / (1 - 13*x - 20*x^2 + 64*x^3 - 112*x^4 - 224*x^5 + 128*x^6).
Recurrence 1:
a(n) = 5*a(n-1) + 2*b(n-1) + c(n-1) + d(n-1) + e(n-1) + 8*a(n-2) + 4*b(n-2) + c(n-2) + 2*d(n-2),
b(n) = 8*a(n-1) + 4*b(n-1) + 2*c(n-1),
c(n) = 20*a(n-1) + 6*b(n-1) + 4*c(n-1) + 4*d(n-1) + 2*e(n-1),
d(n) = 4*a(n-1), e(n) = 16*a(n-1) + 4*b(n-1),
with a(n), b(n), c(n), d(n), e(n) = 0 for n <= 0 except for a(0)=1.
Recurrence 2:
a(n) = 13*a(n-1) + 20*a(n-2) - 64*a(n-3) + 112*a(n-4) + 224*a(n-5) - 128*a(n-6) for n >= 6. For n < 6, recurrence 1 can be used.
Showing 1-3 of 3 results.

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