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 1-4 of 4 results.

A236679 Number T(n,k) of equivalence classes of ways of placing k 2 X 2 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=2, 0<=k<=floor(n/2)^2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 4, 2, 1, 1, 3, 13, 20, 14, 1, 6, 37, 138, 277, 273, 143, 39, 7, 1, 1, 6, 75, 505, 2154, 5335, 7855, 6472, 2756, 459, 1, 10, 147, 1547, 10855, 50021, 153311, 311552, 416825, 361426, 200996, 71654, 16419, 2363, 211, 11, 1, 1, 10, 246, 3759, 39926, 291171
Offset: 2

Views

Author

Christopher Hunt Gribble, Jan 29 2014

Keywords

Comments

Changing the offset from 2 to 1, this is also the sequence: "Triangle read by rows: T(n,k) is the number of nonequivalent ways to place k non-attacking kings on an n X n board." (For if each king is represented by a 2 X 2 tile with the king in the upper left corner, the kings do not attack each other.) For example, with offset 1, T(4,3) = 20 because there are 20 nonequivalent ways to place 3 kings on a 4 X 4 chessboard so that no king threatens any other. - Heinrich Ludwig and N. J. A. Sloane, Dec 21 2016
It appears that rows 2n and 2n-1 both contain n^2 + 1 entries. Rotations and reflections of placements are not counted. If they are to be counted, see A193580. - Heinrich Ludwig, Dec 11 2016

Examples

			T(4,2) = 4 because the number of equivalence classes of ways of placing 2 2 X 2 square tiles in a 4 X 4 square under all symmetry operations of the square is 4. The portrayal of an example from each equivalence class is:
._______        _______        _______        _______
| . | . |      | . |___|      | . |   |      |_______|
|___|___|      |___| . |      |___|___|      | . | . |
|       |      |   |___|      |   | . |      |___|___|
|_______|      |_______|      |___|___|      |_______|
The first 6 rows of T(n,k) are:
.\ k  0    1    2    3    4    5    6    7    8    9
n
2     1    1
3     1    1
4     1    3    4    2    1
5     1    3   13   20   14
6     1    6   37  138  277  273  143   39    7    1
7     1    6   75  505 2154 5335 7855 6472 2756  459

Links

Crossrefs

Cf. A054252, A236560, A236757, A236800, A236829, A236865, A236915, A236936, A236939.
Row sums give A275869.
Columns 2..7: A279111, A279112, A279113, A279114, A279115, A279116.
Diagonal T(n,n) is A279117.
Cf. A193580.

Formula

It appears that:
T(n,0) = 1, n>= 2
T(n,1) = (floor((n-2)/2)+1)*(floor((n-2)/2+2))/2, n >= 2
T(c+2*2,2) = A131474(c+1)*(2-1) + A000217(c+1)*floor(2^2/4) + A014409(c+2), 0 <= c < 2, c even
T(c+2*2,2) = A131474(c+1)*(2-1) + A000217(c+1)*floor((2-1)(2-3)/4) + A014409(c+2), 0 <= c < 2, c odd
T(c+2*2,3) = (c+1)(c+2)/2(2*A002623(c-1)*floor((2-c-1)/2) + A131941(c+1)*floor((2-c)/2)) + S(c+1,3c+2,3), 0 <= c < 2 where
S(c+1,3c+2,3) =
A054252(2,3), c = 0
A236679(5,3), c = 1

Extensions

More terms from Heinrich Ludwig, Dec 11 2016 (The former entry A279118 from Heinrich Ludwig was merged into this entry by N. J. A. Sloane, Dec 21 2016)

A279872 Decimal representation of the x-axis, from the left edge to the origin, (and also from the origin to the right edge) of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 209", based on the 5-celled von Neumann neighborhood.

Original entry on oeis.org

1, 0, 7, 0, 31, 0, 127, 0, 511, 0, 2047, 0, 8191, 0, 32767, 0, 131071, 0, 524287, 0, 2097151, 0, 8388607, 0, 33554431, 0, 134217727, 0, 536870911, 0, 2147483647, 0, 8589934591, 0, 34359738367, 0, 137438953471, 0, 549755813887, 0, 2199023255551, 0
Offset: 0

Views

Author

Robert Price, Dec 21 2016

Keywords

Comments

Initialized with a single black (ON) cell at stage zero.
The nonzero bisection appears to be A083420. - Tom Copeland, Dec 27 2016

References

  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 170.

Links

Crossrefs

Cf. A279118, A083420.

Programs

  • Mathematica
    CAStep[rule_, a_] := Map[rule[[10 - #]] &, ListConvolve[{{0, 2, 0},{2, 1, 2}, {0, 2, 0}}, a, 2],{2}];
code = 209; stages = 128;
rule = IntegerDigits[code, 2, 10];
g = 2 * stages + 1; (* Maximum size of grid *)
a = PadLeft[{{1}}, {g, g}, 0,Floor[{g, g}/2]]; (* Initial ON cell on grid *)
ca = a;
ca = Table[ca = CAStep[rule, ca], {n, 1, stages + 1}];
PrependTo[ca, a];
(* Trim full grid to reflect growth by one cell at each stage *)
k = (Length[ca[[1]]] + 1)/2;
ca = Table[Table[Part[ca[[n]] [[j]],Range[k + 1 - n, k - 1 + n]], {j, k + 1 - n, k - 1 + n}], {n, 1, k}];
Table[FromDigits[Part[ca[[i]] [[i]], Range[1, i]], 2], {i, 1, stages - 1}]

Formula

Conjectures from Chai Wah Wu, Aug 02 2021: (Start)
a(n) = 5*a(n-2) - 4*a(n-4) for n > 3.
G.f.: (2*x^2 + 1)/(4*x^4 - 5*x^2 + 1). (End)

A282411 Binary representation of the x-axis, from the left edge to the origin, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 467", based on the 5-celled von Neumann neighborhood.

Original entry on oeis.org

1, 1, 111, 0, 11111, 0, 1111111, 0, 111111111, 0, 11111111111, 0, 1111111111111, 0, 111111111111111, 0, 11111111111111111, 0, 1111111111111111111, 0, 111111111111111111111, 0, 11111111111111111111111, 0, 1111111111111111111111111, 0
Offset: 0

Views

Author

Robert Price, Feb 14 2017

Keywords

Comments

Initialized with a single black (ON) cell at stage zero.

References

  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 170.

Links

Crossrefs

Cf. A282412, A282413, A282414.

Programs

  • Mathematica
    CAStep[rule_, a_] := Map[rule[[10 - #]] &, ListConvolve[{{0, 2, 0},{2, 1, 2}, {0, 2, 0}}, a, 2],{2}];
code = 467; stages = 128;
rule = IntegerDigits[code, 2, 10];
g = 2 * stages + 1; (* Maximum size of grid *)
a = PadLeft[{{1}}, {g, g}, 0,Floor[{g, g}/2]]; (* Initial ON cell on grid *)
ca = a;
ca = Table[ca = CAStep[rule, ca], {n, 1, stages + 1}];
PrependTo[ca, a];
(* Trim full grid to reflect growth by one cell at each stage *)
k = (Length[ca[[1]]] + 1)/2;
ca = Table[Table[Part[ca[[n]] [[j]],Range[k + 1 - n, k - 1 + n]], {j, k + 1 - n, k - 1 + n}], {n, 1, k}];
Table[FromDigits[Part[ca[[i]] [[i]], Range[1, i]], 10], {i, 1, stages - 1}]

Formula

Conjectures from Colin Barker, Feb 15 2017: (Start)
a(n) = (1 +(-1)^n) * (10^(1+n)-1)/18 for n>1.
a(n) = 101*a(n-2) - 100*a(n-4) for n>3.
G.f.: (1 + x + 10*x^2 - 101*x^3 + 100*x^5) / ((1 - x)*(1 + x)*(1 - 10*x)*(1 + 10*x)).
(End)
Equivalent conjecture: a(n) = A279118(n) for n>=2. - R. J. Mathar, Jun 21 2025

A282412 Binary representation of the x-axis, from the origin to the right edge, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 467", based on the 5-celled von Neumann neighborhood.

Original entry on oeis.org

1, 10, 111, 0, 11111, 0, 1111111, 0, 111111111, 0, 11111111111, 0, 1111111111111, 0, 111111111111111, 0, 11111111111111111, 0, 1111111111111111111, 0, 111111111111111111111, 0, 11111111111111111111111, 0, 1111111111111111111111111, 0
Offset: 0

Views

Author

Robert Price, Feb 14 2017

Keywords

Comments

Initialized with a single black (ON) cell at stage zero.
For n != 1, is a(n) = A279118(n) = A282411(n)? - Bruno Berselli, Feb 15 2017

References

  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 170.

Links

Crossrefs

Cf. A282411, A282413, A282414.

Programs

  • Mathematica
    CAStep[rule_, a_] := Map[rule[[10 - #]] &, ListConvolve[{{0, 2, 0},{2, 1, 2}, {0, 2, 0}}, a, 2],{2}];
code = 467; stages = 128;
rule = IntegerDigits[code, 2, 10];
g = 2 * stages + 1; (* Maximum size of grid *)
a = PadLeft[{{1}}, {g, g}, 0,Floor[{g, g}/2]]; (* Initial ON cell on grid *)
ca = a;
ca = Table[ca = CAStep[rule, ca], {n, 1, stages + 1}];
PrependTo[ca, a];
(* Trim full grid to reflect growth by one cell at each stage *)
k = (Length[ca[[1]]] + 1)/2;
ca = Table[Table[Part[ca[[n]] [[j]],Range[k + 1 - n, k - 1 + n]], {j, k + 1 - n, k - 1 + n}], {n, 1, k}];
Table[FromDigits[Part[ca[[i]] [[i]], Range[i, 2 * i - 1]], 10], {i, 1, stages - 1}]

Formula

Conjectures from Colin Barker, Feb 15 2017: (Start)
a(n) = (1 + (-1)^n)*(10^(1+n)-1) / 18 for n>1.
a(n) = 101*a(n-2) - 100*a(n-4) for n>3.
G.f.: (1+10*x+10*x^2-1010*x^3+1000*x^5) / ((1 - x)*(1 + x)*(1 - 10*x)*(1 + 10*x)).
(End)
Showing 1-4 of 4 results.

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