A257952 -id:A257952 - cp's OEIS frontend

A064941 Quartering a 2n X 2n chessboard (reference A257952) considering only the 90-deg rotationally symmetric results (omitting results with only 180-deg symmetry).

1, 3, 26, 596, 38171, 7083827, 3852835452, 6200587517574, 29752897658253125, 427721252609771505989, 18479976131829456895423324, 2405174963192312814001570260392, 944597040906414962273553855513194341, 1120924326970482645724785944664901286951323

Offset: 1

Walter Gilbert (Walter(AT)Gilbert.net), Oct 28 2001

nonn

Walter Gilbert, Chessboard quartering; includes generating program.

Cf. A257952, A113900.

No formula known. However, the subset of solutions consisting of "tiles" with minimum edge lengths from a corner of the board to the center is A001700.

This sequence can be computed by counting paths in a graph. To compute the n-th term a graph with n X (n-1) vertices is required. Each graph vertex corresponds to 4 intersections between grid lines on the chessboard and graph edges correspond to ways of cutting the board along the grid lines. Frontier (matrix-transfer) graph path counting methods can then be applied to the graph to get the actual count. - Andrew Howroyd, Apr 18 2016

a(7)-a(8) from Juris Cernenoks, Feb 27 2013

a(9)-a(14) from Andrew Howroyd, Apr 18 2016

A113900 Number of partitions of 2n X 2n checkerboard into two congruent edgewise-connected sets, counting partitions equal under rotation or reflection only once.

Original entry on oeis.org

1, 6, 255, 92263, 281035054, 7490694495750, 1789509008288411290

Offset: 1

Joseph Sardinha (jsardi3(AT)juno.com), Jan 29 2006

			All partitions are radially symmetric, hence can be identified by half the cut. The solution for 4 X 4 follows, with coordinates of starting point and direction of each subsequent incremental cut (North is positive Y).
(1,0)NNNES (1,0)NNE (1,0)NEN (1,0)NEENW (2,0)NN (2,0)NENW total = 6

Howard Eves, A Survey of Geometry, 1963, p265.

Giovanni Resta, Illustrations for a(2) = 6 and a(3) = 255.

Cf. A064941, A068392, A257952.

a(5) corrected by Giovanni Resta, May 14 2015

New value of a(5) confirmed by and additional values a(6) and a(7) from Andrew Howroyd, Apr 13 2016

A006067 Number of ways to quarter an n X n chessboard, with the central square removed for odd n.

Original entry on oeis.org

1, 1, 1, 5, 7, 37, 104, 766, 3970, 43318, 431932, 7695805, 137066448, 4015896016, 128095791922, 6371333036059, 355704307903818, 30153126159555641, 2952926822418475378, 431453249608567040694, 73569487283165427567144, 18558756256964594960321428

Offset: 1

N. J. A. Sloane

To "quarter" means to dissect in 4 parts, identical up to rotation, whose interior must be connected. (I.e., the parts must be polyominoes, any 1 X 1 square of which must share a side with some other 1 X 1 square of the part, unless there's only one.) Solutions that differ only by rotation or reflection are not counted separately.
See A257952 for much more information.
See A272070 for information on odd terms.

			For n = 1, we have the 1 X 1 board of which we remove the central square, so nothing is left, and the empty tiling is the only possible tiling.
For n = 2, we have the 2 X 2 board which can only be quartered using four 1 X 1 squares, so a(2) = 1 as well.
For n = 3, the 3 X 3 board without the central square can only be quartered using four 2 X 1 rectangles, so a(3) = 1 as well. (The two different solutions where the top rectangle is aligned to the left or to the right are counted as one, since they only differ by reflection.)
For n = 4 there is the trivial solution using squares, one using straight 4 X 1 tiles, one using T-shaped tiles, and two non-isomorphic ones using L-shaped tiles, one with a central symmetry and one with an axial symmetry:
          A A B B       A B C D       A B B B        A A B B        A A B B
  square: A A B B   I:  A B C D   T:  A A B C   Lc:  A C B D   La:  A C D B
          C C D D       A B C D       A D C C        A C B D        A C D B
          C C D D       A B C D       D D D C        C C D D        C C D D

M. Gardner, The Unexpected Hanging and Other Mathematical Diversions. Simon and Schuster, NY, 1969, p. 189.
T. R. Parkin, personal communication.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Andrew Howroyd, Table of n, a(n) for n = 1..28
Audrey Gruenberger (Editor), Checkerboard - Problem 15, Popular Computing, Vol. 1, No. 7 (1973), front cover and page 2. (Local copy of a scan of the cover illustration showing the a(6)-1 = 36 nontrivial solutions for n = 6, omitting the trivial solution using four squares.)

Cf. A257952, A064941, A272070.

a(2n) = A257952(n), a(2n+1) = A272070(n). - Andrew Howroyd, Apr 19 2016

a(8) corrected, a(9)-a(22) from Andrew Howroyd, Apr 18 2016

Name edited to clarify definition for odd n, and other edits by M. F. Hasler, Jun 13 2025

A268606 Number of ways to trisect a hexagon with side length n exactly into three identical parts in a triangular lattice.

Original entry on oeis.org

1, 5, 116, 15785, 11599297, 47212453928, 1100377983366327, 148568527921382084692

Offset: 1

Luca Petrone, Feb 08 2016

G. P. Jelliss, Dissected Hexagons, The Games and Puzzles Journal, Issue 22, January-April 2002.
Luca Petrone, Illustration showing a(3) = 116

Cf. A257952, A271741, A271857.

Added a(6) from Jelliss's website by Vaclav Kotesovec, Mar 03 2016

a(7)-a(8) from Andrew Howroyd, Apr 11 2016

A003213 Number of ways to quarter a 2n X 2n chessboard.

Original entry on oeis.org

1, 1, 5, 37, 782, 44240

Offset: 0

N. J. A. Sloane

Warning: it now seems very likely that this is an incorrect version of A257952. - N. J. A. Sloane, Apr 17 2016
Number of ways to dissect a 2n X 2n chessboard into 4 congruent pieces.
One can ask the same question for a 2n+1 X 2n+1 board if one omits the center square: this gives A006067.
a(0)=1, since there is one way to do nothing.
Comment from Andrew Howroyd, Apr 18 2016: (Start)
This sequence is wrong because of a bug in Mr. Parkin's code, and amazingly I can pinpoint exactly what the bug is! (I can reproduce his results.)
Firstly the description of the problem and its solution in Mr. Parkin's letter is very clear -- he doesn't leave a lot of room for misinterpretation (this is hugely to his credit). He also includes a very clear description of his algorithm, so I decided I would just code it up. I obtained Giovanni Resta's results as given in A257952 -- there is nothing wrong with Mr Parkin's algorithm.
A detailed breakdown of Parkin's results is also provided in the letter. All the results match with the exception of the final line. (This would be highly improbable if there was a completely different interpretation.) In any case, one sentence stood out as a possible red flag: "Further, there are potential mirror image paths in both cases when starting on the centre lines and these are prevented by requiring a turn in one direction on the path prior to allowing a turn in the other direction" (bottom of page 6). The discrepancy in results does indeed relate to the center line and if I modify my code to lose the flag on recursion, then I get Mr. Parkin's results (so turn in one direction is only prohibited for one step). (End)

M. Gardner, The Unexpected Hanging and Other Mathematical Diversions. Simon and Schuster, NY, 1969, p. 189.
Popular Computing (Calabasas, CA), Vol. 1 (No. 7, 1973), Problem 15, front cover and page 2.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. R. Parkin, Letter to N. J. A. Sloane, Feb 01, 1974. This letter contained as an attachment the following 11-page letter to Fred Gruenberger.
T. R. Parkin, Letter to Fred Gruenberger, Jan 29, 1974, Page 1.
T. R. Parkin, Letter to Fred Gruenberger, Jan 29, 1974, Page 2.
T. R. Parkin, Letter to Fred Gruenberger, Jan 29, 1974, Page 3.
T. R. Parkin, Letter to Fred Gruenberger, Jan 29, 1974, Page 4.
T. R. Parkin, Letter to Fred Gruenberger, Jan 29, 1974, Page 5.
T. R. Parkin, Letter to Fred Gruenberger, Jan 29, 1974, Page 6.
T. R. Parkin, Letter to Fred Gruenberger, Jan 29, 1974, Page 7.
T. R. Parkin, Letter to Fred Gruenberger, Jan 29, 1974, Page 8.
T. R. Parkin, Letter to Fred Gruenberger, Jan 29, 1974, Page 9.
T. R. Parkin, Letter to Fred Gruenberger, Jan 29, 1974, Page 10.
T. R. Parkin, Letter to Fred Gruenberger, Jan 29, 1974, Page 11.
T. R. Parkin, Discussion of Problem 15, Popular Computing (Calabasas, CA), Vol. 2, Number 15 (June 1974), page PC15-4.
T. R. Parkin, Discussion of Problem 15, Popular Computing (Calabasas, CA), Vol. 2, Number 15 (June 1974), page PC15-5.
T. R. Parkin, Discussion of Problem 15, Popular Computing (Calabasas, CA), Vol. 2, Number 15 (June 1974), page PC15-6.
T. R. Parkin, Discussion of Problem 15, Popular Computing (Calabasas, CA), Vol. 2, Number 15 (June 1974), page PC15-7.
T. R. Parkin, Discussion of Problem 15, Popular Computing (Calabasas, CA), Vol. 2, Number 15 (June 1974), page PC15-8.
Popular Computing (Calabasas, CA), Illustration showing that a(3) = 37, Vol. 1 (No. 7, 1973), front cover. (One of the 37 is simply the square divided into four quadrants.)

Bisection of A006067. Cf. A064941.

See A257952 for another version.

A272070 Number of ways to quarter a 2n+1 X 2n+1 chessboard with central square removed.

Original entry on oeis.org

1, 1, 7, 104, 3970, 431932, 137066448, 128095791922, 355704307903818, 2952926822418475378, 73569487283165427567144, 5515501712040561162370942752, 1246743475892797935712690352483842, 850999652841310762943520023896881419780

Offset: 0

Andrew Howroyd, Apr 19 2016

nonn

The chessboard must be dissected into four identical pieces. All solutions have 90-degree rotational symmetry and solutions that differ only by rotation or reflection are considered equivalent.
The pieces must be polyominoes, without holes. They can't have parts that are only connected through one or more corners: if the piece is larger than one square, each square must share at least one side with some other square of the same piece. - M. F. Hasler, Jun 13 2025
The solutions can be constructed as self-avoiding paths made of horizontal and vertical unit segments, that connect one corner of the central square to some point on the border, also avoiding the 3 other paths obtained by rotation by +-90 and 180 degrees around the center. To avoid counting solutions that are equivalent modulo reflection, it is enough to take the first segment in a fixed direction (e.g., from the upper left corner, always first go up by one unit). - M. F. Hasler, Jun 14 2025

			For n = 1, we have the 3 X 3 board, with the central square removed.
  In order to split this into four congruent parts, we must use four 2 X 1 tiles.
  Up to reflections, the only solution is :
                                   _____
   A A B    or, in a different    |___| |
   D   B       representation:    | |_|_|
   D C C                          |_|___|
More generally, without loss of generality (i.e., up to reflections), any solution restricted to this 3 X 3 area squares will always be of that form; the entire set of (equivalent) solutions can be obtained by "growing" the four areas (symmetrically) to fill the entire board.
Thus, for n = 2, the 7 solutions for the 5 X 5 board are obtained by filling the outer ring with 4 groups of 2n = 4 adjacent identical letters so that they are in contact with a same letter in the "central ring":
   B B B B C   A B B B B   A A B B B   A A A B B   A A A A B   D A A A A   D D A A A
   A A A B C   A A A B C   A A A B B   A A A B B   D A A B B   D A A B B   D A A B A
   A D   B C   A D   B C   A D   B C   D D   B B   D D   B B   D D   B B   D D   B B
   A D C C C   A D C C C   D D C C C   D D C C C   D D C C B   D D C C B   C D C C B
   A D D D D   D D D D C   D D D C C   D D C C C   D C C C C   C C C C B   C C C B B
(Observe how the groups of four identical letters "travel clockwise" in the outer ring, while the inner ring remains the same. Given that we have 2(n-1) = 2 'A's in the previously outer ring (which always include one in the corner), and we have 2n = 4 'A's to (dis)place in the outer ring, there are 2n+1+2(n-1) = 4n-1 = 7 possibilities here.)
For n = 3, each of the previous 7 solutions gives rise to 4n-1 = 11 distinct solutions for the 7 X 7 board, by again filling the outer ring with 2n = 6 adjacent identical letters A, B, C and D so that at least one is in contact with one of the 2n-2 = 4 identical letters in the second ring. But this doesn't give all of the 104 solutions for the 7 X 7 board. We can get additional solutions by incrementing/decrementing (A > B > C > D > A) squares in the n=2 ring which couldn't be changed before (because it would have created "disconnected components" which are now connected through the outer ring). For example, "incrementing" the letters in the corners of the 3rd solution for n=2 gives the solution (3,78) depicted below, plus 3 other solutions by rotating the outer ring clockwise.  Similarly, we can increment or also decrement the letter in the corner and/or the next one in the 4th solution for n=2, and, e.g., the fourth letter of the 5th n=2 solution, see solution (3,104) below:
             B B B B C C C              D D D A A A A
  solution   B B A B B C C    solution  D D A A B A A
   (3,78):   B A A A B B C    (3,104):  D A A A B B A
  (from      A A D   B C C    (from     D D D   B B B
  solution   A D D C C C D    solution  C D D C C C B
   (2,3))    A A D D C D D     (2,6)    C C D C C B B
             A A A D D D D              C C C C B B B

Cf. A006067, A064941, A257952.

A006067 = Import["https://oeis.org/A006067/b006067.txt", "Table"][[All, 2]];
a[n_] := A006067[[2n+1]];
a /@ Range[1, 13] (* Jean-François Alcover, Sep 14 2019 *)

# program is unoptimized, slow for n > 5, rather for illustration
def A272070(n, sol=None): # return solution (path) with index sol if given
    if not hasattr(A:=A272070,'sol'): A.sol=[[[]]]
    while n >= len(A.sol):
        L = len(A.sol); width = 2*L+1; A.sol.append(N := [])
        path = [pos := 1+1j]; todo = [pos+2]; steps = 1,1j,-1j,-1
        used = lambda pos: any(pos*s in path for s in steps)
        while todo:
            if pos := todo.pop():
                if width in (abs(pos.real),abs(pos.imag)):
                    N . append(tuple(path+[pos]))
                elif go := [p for s in steps if not used(p := pos+2*s)]:
                    path += [pos]; todo += [0]; todo += go
            else: path.pop()
    return len(A.sol[n]) if sol is None else A.sol[n][sol]
from matplotlib.pyplot import plot,show # only needed for plotting
def A272070_plot(n, sol): # to plot the solution w/ index sol
    for s in (1, 2*n+1): # outer frame and central square
        plot(*([(s:=-s)if(i-j)&1 else s for i in range(5)]for j in range(2)))
    for i in range(4):
        path = [1j*p for p in path] if i else A272070(n,sol)
        plot([p.real for p in path], [p.imag for p in path])
    show(); return path # M. F. Hasler, Jun 14 2025

a(n) = A006067(2*n+1).

a(0) = 1 prepended by M. F. Hasler, Jun 13 2025

cp's OEIS Frontend

A064941 Quartering a 2n X 2n chessboard (reference A257952) considering only the 90-deg rotationally symmetric results (omitting results with only 180-deg symmetry).

Views

Author

Keywords

Links

Crossrefs

Formula

Extensions

A113900 Number of partitions of 2n X 2n checkerboard into two congruent edgewise-connected sets, counting partitions equal under rotation or reflection only once.

Views

Author

Keywords

Examples

References

Links

Crossrefs

Extensions

A006067 Number of ways to quarter an n X n chessboard, with the central square removed for odd n.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Formula

Extensions

A268606 Number of ways to trisect a hexagon with side length n exactly into three identical parts in a triangular lattice.

Views

Author

Keywords

Links

Crossrefs

Extensions

A003213 Number of ways to quarter a 2n X 2n chessboard.

Views

Author

Keywords

Comments

References

Links

Crossrefs

A272070 Number of ways to quarter a 2n+1 X 2n+1 chessboard with central square removed.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Python

Formula

Extensions

A003155 Number of ways to halve an n X n chessboard.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Formula

Extensions

A271858 Number of ways to trisect a triangle with side length n exactly into three identical parts in a triangular lattice.

Views

Author

Keywords

Comments

Links

Crossrefs