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 1-10 of 11 results. Next

A006076 Sequence A006075 gives minimal number of knights needed to cover an n X n board. This sequence gives number of inequivalent solutions using A006075(n) knights.

Original entry on oeis.org

1, 1, 2, 3, 8, 23, 3, 1, 1, 2, 100, 1, 20, 1, 63, 1, 29, 2551
Offset: 1

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Author

Keywords

References

  • David C. Fisher, On the N X N Knight Cover Problem, Ars Combinatoria 69 (2003), 255-274.
  • M. Gardner, Mathematical Magic Show. Random House, NY, 1978, p. 194.
  • Bernard Lemaire, Knights Covers on N X N Chessboards, J. Recreational Mathematics, Vol. 31-2, 2003, 87-99.
  • Frank Rubin, Improved knight coverings, Ars Combinatoria 69 (2003), 185-196.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A006075 (number of solutions), A098604 (rectangular board). A103315 gives the total number of solutions.

Extensions

a(11) was found in 1973 by Bernard Lemaire. (Philippe Deléham, Jan 06 2004)
a(13)-a(17) from the Morgenstern web site, Nov 08 2004
a(18) from the Morgenstern web site, Mar 20 2005

A075458 Domination number for queens' graph Q(n).

Original entry on oeis.org

1, 1, 1, 2, 3, 3, 4, 5, 5, 5, 5, 6, 7, 8, 9, 9, 9, 9, 10, 11, 11, 12, 12, 13, 13
Offset: 1

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Author

N. J. A. Sloane, Oct 16 2002

Keywords

Comments

From Dmitry Kamenetsky, Sep 03 2019: (Start)
Minimum number of queens needed to occupy or attack all squares of an n X n chessboard.
a(n) >= ceiling(n/2) for all n, except n = 3, 11. See paper by Finozhenok and Weakley below.
a(n) = p or p+1, where p = ceiling(n/2), proved for all n <= 132, except n = 3, 11. See paper by Ostergard and Weakley below. Note that this implies that a(n+4) > a(n). (End)

References

  • W. W. R. Ball and H. S. M. Coxeter, "Math'l Rec. and Essays," 13th Ed. Dover, p. 173.
  • John Watkins, Across the Board: The Mathematics of Chessboard Problems (2004), pp. 113-137

Crossrefs

A002563 gives number of solutions.
Cf. A075324 (independent domination number).

Extensions

a(19) from Peter Karpov, Mar 01 2016
a(20)-a(24) from Bird and a(25) from Dmitry Kamenetsky's file added by Andrey Zabolotskiy, Sep 03 2021

A075561 Domination number for kings' graph K(n).

Original entry on oeis.org

1, 1, 1, 4, 4, 4, 9, 9, 9, 16, 16, 16, 25, 25, 25, 36, 36, 36, 49, 49, 49, 64, 64, 64, 81, 81, 81, 100, 100, 100, 121, 121, 121, 144, 144, 144, 169, 169, 169, 196, 196, 196, 225, 225, 225, 256, 256, 256, 289, 289, 289, 324, 324, 324, 361, 361, 361, 400, 400
Offset: 1

Views

Author

N. J. A. Sloane, Oct 16 2002

Keywords

Comments

Also the lower independence number of the n X n knight graph. - Eric W. Weisstein, Aug 01 2023

References

  • John J. Watkins, Across the Board: The Mathematics of Chessboard Problems, Princeton University Press, 2004, p. 102.

Crossrefs

Programs

  • Mathematica
    Table[Floor[(n + 2)/3]^2, {n, 50}] (* Vaclav Kotesovec, May 13 2012 *)
    LinearRecurrence[{1, 0, 2, -2, 0, -1, 1}, {1, 1, 1, 4, 4, 4, 9}, 20] (* Eric W. Weisstein, Jun 20 2017 *)
    CoefficientList[Series[(-1 - x^3)/((-1 + x)^3 (1 + x + x^2)^2), {x, 0, 20}], x] (* Eric W. Weisstein, Jun 20 2017 *)
  • PARI
    Vec(-x*(x+1)*(x^2-x+1)/((x-1)^3*(x^2+x+1)^2) + O(x^100)) \\ Colin Barker, Oct 06 2014

Formula

a(n) = floor((n+2)/3)^2. - Vaclav Kotesovec, May 13 2012
G.f.: -x*(x+1)*(x^2-x+1) / ((x-1)^3*(x^2+x+1)^2). - Colin Barker, Oct 06 2014
E.g.f.: exp(-x/2)*(exp(3*x/2)*(5 + 3*x*(3 + x)) + (6*x - 5)*cos(sqrt(3)*x/2) + sqrt(3)*(3 + 2*x)*sin(sqrt(3)*x/2))/27. - Stefano Spezia, Oct 17 2022
Sum_{n>=1} 1/a(n) = Pi^2/2 (A102753). - Amiram Eldar, Nov 03 2022

Extensions

More terms added from Vaclav Kotesovec, May 13 2012

A103315 Number of minimum dominating sets for the n X n knight graph.

Original entry on oeis.org

1, 1, 8, 9, 47, 127, 10, 2, 2, 4, 800, 2, 152, 4, 504, 2, 212, 19562
Offset: 1

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Author

N. J. A. Sloane, Mar 20 2005, following a suggestion from Lee Morgenstern

Keywords

Comments

In other words, as made explicit in the old name: Sequence A006075 gives minimum number of knights needed to cover an n X n board (i.e., the domination number of the n X n knight graph). This sequence (A103315) gives total number of solutions using A006075(n) knights (compare A006076).

Crossrefs

Cf. A006075 (domination number of the n X n knight graph).
Cf. A006076 (inequivalent number of minimum dominating sets).
Cf. A098604.

Extensions

New name from Eric W. Weisstein, Sep 06 2021

A098604 Triangle T(n,k) read by rows, for 1 <= k <= n: minimal number of knights needed to cover a k X n board.

Original entry on oeis.org

1, 2, 4, 3, 4, 4, 4, 4, 4, 4, 5, 4, 4, 4, 5, 6, 4, 4, 4, 6, 8, 7, 6, 6, 6, 7, 8, 10, 8, 8, 8, 8, 8, 8, 11, 12, 9, 8, 8, 8, 8, 10, 12, 13, 14, 10, 8, 8, 8, 9, 12, 14, 14, 15, 16, 11, 8, 8, 8, 10, 12, 15, 16, 17, 19, 21, 12, 8, 8, 8, 10, 12, 16, 16, 18, 20, 22, 24, 13, 10, 10, 10, 12, 14
Offset: 1

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Author

Keywords

Comments

How many knights are needed to occupy or attack every square of a k X n board?
I do not know how many of these numbers have been proved to be optimal. - N. J. A. Sloane, Nov 08 2004

Examples

			Triangle (with rows n >= 1 and columns k >= 1) begins as follows:
  1
  2 4
  3 4 4
  4 4 4 4
  5 4 4 4 5
  6 4 4 4 6 8
  7 6 6 6 7 8 10
  ...
		

Crossrefs

See A006075 for the n X n case (the main diagonal). A006076 gives number of ways to cover an n X n board using the minimal number of knights.

Extensions

Morgenstern's table extends a long way beyond what is shown here.

A110214 Minimal number of knights to cover a cubic board.

Original entry on oeis.org

1, 8, 6, 8, 13
Offset: 1

Views

Author

Nikolaus Meyberg (Nikolaus.Meyberg(AT)t-online.de), Jul 17 2005

Keywords

Examples

			Illustration for n = 3, 4, 5 ( O = empty field, K = knight ):
n = 3: OOO KKK OOO n = 4: OOOO OKOO OOOO OOOO
...... OKO OKO OKO ...... OOOO OKKK OOOO OOOO
...... OOO OOO OOO ...... OOOO KKKO OOOO OOOO
......................... OOOO OOKO OOOO OOOO
n = 5: 1, 2, 4 and 5 planes empty, 3 plane: OKOKO OKOKO KKKKK KOKOK OOKOO.
		

Crossrefs

This is a 3-dimensional version of A006075. a(n) = A110217(n, n, n). A110215 gives number of inequivalent ways to cover the board using a(n) knights, A110216 gives total number.

Formula

Generalize a knight for a spatial board: a move consists of two steps in the first, one step in the second and no step in the third dimension. How many of such knights are needed to occupy or attack every field of an n X n X n board? Knights may attack each other.

A110217 Cone C(n,m,k) read by planes and rows, for 1 <= k <= m <= n: minimal number of knights needed to cover a k X m X n board.

Original entry on oeis.org

1, 2, 4, 8, 3, 4, 8, 4, 6, 6, 4, 4, 8, 4, 6, 6, 4, 6, 7, 8, 5, 4, 8, 4, 6, 6, 4, 6, 7, 8, 5, 6, 8, 10, 13, 6, 4, 6, 4, 7, 6, 4, 8, 8, 12, 6, 8, 10, 12
Offset: 1

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Author

Nikolaus Meyberg (Nikolaus.Meyberg(AT)t-online.de), Jul 17 2005

Keywords

Examples

			Cone starts:
1..2....3......4........5............6.................
...4.8..4.8....4.8......4.8..........4..6
........4.6.6..4.6.6....4.6.6........4..7..6
...............4.6.7.8..4.6.7..8.....4..8..8.12
........................5.6.8.10.13..6..8.10.12.?
.....................................8.11.12..?....
		

Crossrefs

C(n, n, 1) = A006075(n), C(n, k, 1) = A098604(n, k), C(n, n, n) = A110214(n). A110218 gives number of inequivalent ways to cover the board using C(n, m, k) knights, A110219 gives total number.

Formula

How many knights with move vector (2, 1, 0) are needed to occupy or attack every field of a k X m X n board? Knights may attack each other.

A279407 Domination number for knight graph on an n X n toroidal board.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 9, 8, 12, 15, 18, 21, 25, 28, 33, 32
Offset: 1

Views

Author

Andrey Zabolotskiy, Dec 12 2016

Keywords

Comments

That is, the minimal number of knights needed to cover an n X n toroidal chessboard so that every square either has a knight on it, or is under attack by a knight, or both.

Examples

			For an 8 X 8 board, the solution is:
  N . . . . . . N
  . . . . . . . .
  . . N . . N . .
  . . . . . . . .
  . . . N N . . .
  . . . . . . . .
  . N . . . . N .
  . . . . . . . .
		

References

  • John J. Watkins, Across the Board: The Mathematics of Chessboard Problem, Princeton University Press, 2004, pages 140-144.

Crossrefs

Extensions

a(9)-a(16) from Andy Huchala, Mar 03 2024

A261752 Minimum number of knights on an n X n chessboard such that every square is attacked.

Original entry on oeis.org

6, 7, 8, 10, 14, 18, 22, 25, 28, 32, 36, 43, 48, 54, 58, 66, 70, 78, 84, 91, 98, 107, 112, 123, 128, 139, 146, 156, 164
Offset: 4

Views

Author

Matthew Conroy, Aug 31 2015

Keywords

Comments

Total domination number of n X n knight graph.
Distinct from A006075 since here all squares must be attacked, whereas, in A006075, all squares are either attacked or occupied.
a(34) = 182, a(36) = 202, a(38) = 224. - Andy Huchala, Jun 04 2021

Examples

			An example for the 4 X 4 case:
  ....
  .NNN
  .N..
  NN..
and for the 5 x 5 case:
  .....
  ..N..
  .NN..
  NNN..
  N....
		

Crossrefs

Cf. A006075.

Extensions

a(15)-a(18) from Giovanni Resta, Aug 31 2015
a(19)-a(26) from Andy Huchala, Oct 16 2017
a(27)-a(30) from Andy Huchala, Oct 18 2017
a(31)-a(32) from Andy Huchala, Jun 04 2021

A342576 Independent domination number for knight graph on an n X n board.

Original entry on oeis.org

1, 4, 4, 4, 5, 8, 13, 14, 14, 16, 22, 24, 29, 33, 36, 40, 47, 52, 58, 63, 68
Offset: 1

Views

Author

Andrey Zabolotskiy, Mar 15 2021

Keywords

References

  • Sandra M. Hedetniemi, Stephen T. Hedetniemi, Robert Reynolds, Combinatorial Problems on Chessboards: II, in: Domination in Graphs - Advanced Topics, Marcel Dekker, 1998. See p. 141.

Crossrefs

Programs

  • Maple
    f:= proc(N)
      local verts,Rverts,edg,cons,i,j,e;
      verts:= [seq(seq([i,j],i=1..N),j=1..N)]:
      for i from 1 to N^2 do Rverts[op(verts[i])]:= i od:
      edg:= {seq(seq({Rverts[i,j],Rverts[i+1,j+2]},i=1..N-1),j=1..N-2),
           seq(seq({Rverts[i,j],Rverts[i+2,j+1]},i=1..N-2),j=1..N-1),
           seq(seq({Rverts[i,j],Rverts[i+1,j-2]},i=1..N-1),j=3..N),
           seq(seq({Rverts[i,j],Rverts[i+2,j-1]},i=1..N-2),j=2..N)}:
      cons:= {seq(x[e[1]]+x[e[2]]<=1, e=edg),
        seq(x[i]+add(`if`(member({i,j},edg),x[j],0),j=1..N^2)>=1, i=1..N^2)}:
      Optimization:-Minimize(add(x[i],i=1..N^2),cons,assume=binary)[1]
    end proc:
    map(f, [$1..13]); # Robert Israel, Mar 17 2021

Extensions

a(11) to a(14) from Robert Israel, Mar 17 2021
a(15)-a(18) from Eric W. Weisstein, Aug 01 2023
a(19) from Eric W. Weisstein, Jan 14 2024
a(20)-a(21) from Andy Huchala, Mar 10 2024
Showing 1-10 of 11 results. Next