A379625 -id:A379625 - cp's OEIS frontend

A317186 One of many square spiral sequences: a(n) = n^2 + n - floor((n-1)/2).

1, 2, 6, 11, 19, 28, 40, 53, 69, 86, 106, 127, 151, 176, 204, 233, 265, 298, 334, 371, 411, 452, 496, 541, 589, 638, 690, 743, 799, 856, 916, 977, 1041, 1106, 1174, 1243, 1315, 1388, 1464, 1541, 1621, 1702, 1786, 1871, 1959, 2048, 2140, 2233, 2329, 2426

Offset: 0

N. J. A. Sloane, Jul 27 2018

Draw a square spiral on a piece of graph paper, and label the cells starting at the center with the positive (resp. nonnegative) numbers. This produces two versions of the labeled square spiral, shown in the Example section below.
The spiral may proceed clockwise or counterclockwise, and the first arm of the spiral may be along any of the four axes, so there are eight versions of each spiral. However, this has no effect on the resulting sequences, and it is enough to consider just two versions of the square spiral (starting at 1 or starting at 0).
The present sequence is obtained by reading alternate entries on the X-axis (say) of the square spiral started at 1.
The cross-references section lists many sequences that can be read directly off the two spirals. Many other sequences can be obtained from them by using them to extract subsequences from other important sequences. For example, the subsequence of primes indexed by the present sequence gives A317187.
a(n) is also the number of free polyominoes with n + 4 cells whose difference between length and width is n. In this comment the length is the longer of the two dimensions and the width is the shorter of the two dimensions (see the examples of polyominoes). Hence this is also the diagonal 4 of A379625. - Omar E. Pol, Jan 24 2025
From John Mason, Feb 19 2025: (Start)
The sequence enumerates polyominoes of width 2 having precisely 2 horizontal bars. By classifying such polyominoes according to the following templates, it is possible to define a formula that reduces to the one below:
.
 OO  O  O
 O  OO OO
 O  O  O
 O  O  OO
 OO OO  O
.
(End)

			The square spiral when started with 1 begins:
.
  100--99--98--97--96--95--94--93--92--91
                                        |
   65--64--63--62--61--60--59--58--57  90
    |                               |   |
   66  37--36--35--34--33--32--31  56  89
    |   |                       |   |   |
   67  38  17--16--15--14--13  30  55  88
    |   |   |               |   |   |   |
   68  39  18   5---4---3  12  29  54  87
    |   |   |   |       |   |   |   |   |
   69  40  19   6   1---2  11  28  53  86
    |   |   |   |           |   |   |   |
   70  41  20   7---8---9--10  27  52  85
    |   |   |                   |   |   |
   71  42  21--22--23--24--25--26  51  84
    |   |                           |   |
   72  43--44--45--46--47--48--49--50  83
    |                                   |
   73--74--75--76--77--78--79--80--81--82
.
For the square spiral when started with 0, subtract 1 from each entry. In the following diagram this spiral has been reflected and rotated, but of course that makes no difference to the sequences:
.
   99  64--65--66--67--68--69--70--71--72
    |   |                               |
   98  63  36--37--38--39--40--41--42  73
    |   |   |                       |   |
   97  62  35  16--17--18--19--20  43  74
    |   |   |   |               |   |   |
   96  61  34  15   4---5---6  21  44  75
    |   |   |   |   |       |   |   |   |
   95  60  33  14   3   0   7  22  45  76
    |   |   |   |   |   |   |   |   |   |
   94  59  32  13   2---1   8  23  46  77
    |   |   |   |           |   |   |   |
   93  58  31  12--11--10---9  24  47  78
    |   |   |                   |   |   |
   92  57  30--29--28--27--26--25  48  79
    |   |                           |   |
   91  56--55--54--53--52--51--50--49  80
    |                                   |
   90--89--88--87--86--85--84--83--82--81
.
From _Omar E. Pol_, Jan 24 2025: (Start)
For n = 0 there is only one free polyomino with 0 + 4 = 4 cells whose difference between length and width is 0 as shown below, so a(0) = 1.
   _ _
  |_|_|
  |_|_|
.
For n = 1 there are two free polyominoes with 1 + 4 = 5 cells whose difference between length and width is 1 as shown below, so a(1) = 2.
   _ _     _ _
  |_|_|   |_|_|
  |_|_|   |_|_
  |_|     |_|_|
.
(End)

Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).
Index entries for sequences related to polyominoes.

Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.
Filling in these two squares spirals with greedy algorithm: A274640, A274641.
Cf. also A317187.
Cf. A000105, A379625.

a[n_] := n^2 + n - Floor[(n - 1)/2]; Array[a, 50, 0] (* Robert G. Wilson v, Aug 01 2018 *)
LinearRecurrence[{2, 0, -2 , 1},{1, 2, 6, 11},50] (* or *)
CoefficientList[Series[(- x^3 - 2 * x^2 - 1) / ((x - 1)^3 * (x + 1)), {x, 0, 50}], x] (* Stefano Spezia, Sep 02 2018 *)

From Daniel Forgues, Aug 01 2018: (Start)
a(n) = (1/4) * (4 * n^2 + 2 * n + (-1)^n + 3), n >= 0.
a(0) = 1; a(n) = - a(n-1) + 2 * n^2 - n + 2, n >= 1.
a(0) = 1; a(1) = 2; a(2) = 6; a(3) = 11; a(n) = 2 * a(n-1) - 2 * a(n-3) + a(n-4), n >= 4.
G.f.: (- x^3 - 2 * x^2 - 1) / ((x - 1)^3 * (x + 1)). (End)
E.g.f.: ((2 + 3*x + 2*x^2)*cosh(x) + (1 + 3*x + 2*x^2)*sinh(x))/2. - Stefano Spezia, Apr 24 2024
a(n)+a(n+1)=A033816(n). - R. J. Mathar, Mar 21 2025
a(n)-a(n-1) = A042948(n), n>=1. - R. J. Mathar, Mar 21 2025

A379637 Irregular triangle read by rows: T(n,k) is the sum of the widths of the free polyominoes with n cells and width k, n >= 1, 1 <= k <= ceiling(n/2).

Original entry on oeis.org

1, 1, 1, 2, 1, 8, 1, 10, 18, 1, 24, 66, 1, 36, 213, 72, 1, 74, 579, 552, 1, 120, 1470, 2644, 365, 1, 234, 3663, 10188, 3845, 1, 400, 9033, 33668, 25945, 1530, 1, 758, 22179, 104656, 129600, 22458, 1, 1338, 54075, 312296, 544170, 192228, 6650, 1, 2500, 131541, 919524, 2041085, 1211736, 117733

Offset: 1

Omar E. Pol, Jan 16 2025

The width here is the shorter of the two dimensions.

			Triangle begins:
  1;
  1;
  1,    2;
  1,    8;
  1,   10,     18;
  1,   24,     66;
  1,   36,    213,     72;
  1,   74,    579,    552;
  1,  120,   1470,   2644,     365;
  1,  234,   3663,  10188,    3845;
  1,  400,   9033,  33668,   25945,    1530;
  1,  758,  22179, 104656,  129600,   22458;
  1, 1338,  54075, 312296,  544170,  192228,   6650;
  1, 2500, 131541, 919524, 2041085, 1211736, 117733;
  ...
Illustration for n = 5:
The free polyominoes with five cells are also called free pentominoes.
For k = 1 there is only one free pentomino of width 1 as shown below, so T(5,1) = 1.
   _
  |_|
  |_|
  |_|
  |_|
  |_|
.
For k = 2 there are five free pentominoes of width 2 as shown below, hence the sum of the widths is 2 + 2 + 2 + 2 + 2 = 5*2 = 10, so T(5,2) = 10.
   _           _         _
  |_|        _|_|      _|_|      _ _       _ _
  |_|       |_|_|     |_|_|     |_|_|     |_|_|
  |_|_      |_|         |_|     |_|_|     |_|_
  |_|_|     |_|         |_|     |_|       |_|_|
.
For k = 3 there are six free pentominoes of width 3 as shown below, hence the sum of the widths is 3 + 3 + 3 + 3 + 3 + 3 = 6*3 = 18, so T(5,3) = 18.
     _ _     _ _ _     _         _           _       _ _
   _|_|_|   |_|_|_|   |_|       |_|_       _|_|_    |_|_|
  |_|_|       |_|     |_|_ _    |_|_|_    |_|_|_|     |_|_
    |_|       |_|     |_|_|_|     |_|_|     |_|       |_|_|
.
Therefore the 5th row of the triangle is [1, 10, 18].
.

Index entries for sequences related to polyominoes.

Row lengths give A110654.
Row sums give A379626.
Cf. A000105, A379623, A379625, A379627, A379638.

T(n,k) = k*A379623(n,k).

A380283 Irregular triangle read by rows: T(n,k) is the number of regions between the free polyominoes, with n cells and width k, and their bounding boxes, n >= 1, 1 <= k <= ceiling(n/2).

Original entry on oeis.org

0, 0, 0, 1, 0, 5, 0, 7, 14, 0, 19, 52, 0, 34, 173, 48, 0, 74, 503, 384, 0, 134, 1368, 1918, 210, 0, 282, 3642, 7742, 2307, 0, 524, 9552, 26843, 16267, 752, 0, 1064, 24889, 87343, 84789, 11556, 0, 2017, 64200, 272599, 370799, 103336, 2833, 0, 4009, 164826, 838160, 1445347, 678863, 52437

Offset: 1

Omar E. Pol, Jan 18 2025

The regions include any holes in the polyominoes.

			Triangle begins:
  0;
  0;
  0,     1;
  0,     5;
  0,     7,      14;
  0,    19,      52;
  0,    34,     173,      48;
  0,    74,     503,     384;
  0,   134,    1368,    1918,      210;
  0,   282,    3642,    7742,     2307;
  0,   524,    9552,   26843,    16267,      752;
  0,  1064,   24889,   87343,    84789,    11556;
  0,  2017,   64200,  272599,   370799,   103336,    2833;
  0,  4009,  164826,  838160,  1445347,   678863,   52437;
  0,  7663,  420373, 2539843,  5240853,  3659815,  560348,  10396;
  0, 15031, 1068181, 7631249, 18171771, 17199831, 4373770, 226716;
  ...
Illustration for n = 5:
The free polyominoes with five cells are also called free pentominoes.
For k = 1 there is only one free pentomino of width 1 as shown below, and there are no regions between the pentomino and its bounding box, so T(5,1) = 0.
   _
  |_|
  |_|
  |_|
  |_|
  |_|
.
For k = 2 there are five free pentominoes of width 2 as shown below, and from left to right there are respectively 1, 2, 2, 1, 1 regions between the pentominoes and their bounding boxes, hence the total number of regions is 1 + 2 + 2 + 1 + 1 = 7, so T(5,2) = 7.
   _           _         _
  |_|        _|_|      _|_|      _ _       _ _
  |_|       |_|_|     |_|_|     |_|_|     |_|_|
  |_|_      |_|         |_|     |_|_|     |_|_
  |_|_|     |_|         |_|     |_|       |_|_|
.
For k = 3 there are six free pentominoes of width 3 as shown below, and from left to right there are respectively 3, 2, 1, 2, 4, 2 regions between the pentominoes and their bounding boxes, hence the total number of regions is 3 + 2 + 1 + 2 + 4 + 2 = 14, so T(5,3) = 14.
     _ _     _ _ _     _         _           _       _ _
   _|_|_|   |_|_|_|   |_|       |_|_       _|_|_    |_|_|
  |_|_|       |_|     |_|_ _    |_|_|_    |_|_|_|     |_|_
    |_|       |_|     |_|_|_|     |_|_|     |_|       |_|_|
.
Therefore the 5th row of the triangle is [0, 7, 14].
.

John Mason, Table of n, a(n) for n = 1..90

Column 1 gives A000004.
Row lengths give A110654.
Row sums give A380285.
Cf. A000105, A379623, A379625, A379626, A379627, A379628, A379637, A380284.

More terms from John Mason, Feb 14 2025

A380284 Triangle read by rows: T(n,k) is the number of regions between the free polyominoes, with n cells and length k, and their bounding boxes, n >= 1, k >= 1.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 0, 5, 0, 0, 0, 16, 5, 0, 0, 0, 14, 48, 9, 0, 0, 0, 12, 145, 89, 9, 0, 0, 0, 3, 354, 453, 138, 13, 0, 0, 0, 0, 608, 1930, 876, 203, 13, 0, 0, 0, 0, 804, 6348, 4930, 1598, 276, 17, 0, 0, 0, 0, 721, 17509, 22575, 10197, 2554, 365, 17, 0, 0, 0, 0, 454, 40067, 91007, 54691, 18984, 3955, 462, 21, 0

Offset: 1

Omar E. Pol, Jan 18 2025

The regions include any holes in the polyominoes.

The first 28 terms were calculated by hand.

			Triangle begins:
  0;
  0,  0;
  0,  1,  0;
  0,  0,  5,   0;
  0,  0, 16,   5,  0;
  0,  0, 14,  48,  9,  0;
  0,  0, 12, 145, 89,  9,  0;
  ...
Illustration for n = 5:
The free polyominoes with five cells are also called free pentominoes.
For k = 1 there are no free pentominoes of length 1, hence there are no regions, so T(5,1) = 0.
For k = 2 there are no free pentominoes of length 2, hence there are no regions, so T(5,2) = 0.
For k = 3 there are eight free pentominoes of length 3 as shown below, and the number of regions between the pentominoes and their bounding boxes are from left to right respectively 1, 1, 3, 2, 1, 2, 4, 2, hence the total number of regions is 1 + 1 + 3 + 2 + 1 + 2 + 4 + 2 = 16,  so T(5,3) = 16.
   _ _     _ _       _ _     _ _ _     _         _           _       _ _
  |_|_|   |_|_|    _|_|_|   |_|_|_|   |_|       |_|_       _|_|_    |_|_|
  |_|_|   |_|_    |_|_|       |_|     |_|_ _    |_|_|_    |_|_|_|     |_|_
  |_|     |_|_|     |_|       |_|     |_|_|_|     |_|_|     |_|       |_|_|
.
For k = 4 there are three free pentominoes of length 4 as shown below, and the number of regions between the pentominoes and their bounding boxes are from left to right respectively 1, 2, 2, hence the total number of regions is 1 + 2 + 2 = 5,  so T(5,4) = 5.
   _         _       _
  |_|      _|_|    _|_|
  |_|     |_|_|   |_|_|
  |_|_    |_|       |_|
  |_|_|   |_|       |_|
.
For k = 5 there is only one free pentomino of length 5 as shown below, and there are no regions between the pentomino and its bounding box, so T(5,5) = 0.
   _
  |_|
  |_|
  |_|
  |_|
  |_|
.
Therefore the 5th row of the triangle is [0, 0, 16, 5, 0].
.

John Mason, Table of n, a(n) for n = 1..171 (first 18 rows)
Index entries for sequences related to polyominoes.

Column 1 and leading diagonal give A000004.
Column 2 gives A063524.
Row sums give A380285.
Cf. A000105, A379624, A379625, A379627, A379628, A379629, A379638, A380282, A380283.

More terms from John Mason, Feb 14 2025

A379626 Sum of the widths of the free polyominoes with n cells.

Original entry on oeis.org

1, 1, 3, 9, 29, 91, 322, 1206, 4600, 17931, 70577, 279652, 1110758, 4424120, 17647314, 70484576, 281750598, 1127181327

Offset: 1

Omar E. Pol, Jan 16 2025

The width here is the shorter of the two dimensions.

			For n = 4 the free polyominoes with four cells are also called free tetrominoes.
The five free tetrominoes are as shown below:
    _
   |_|     _       _       _
   |_|    |_|     |_|_    |_|_     _ _
   |_|    |_|_    |_|_|   |_|_|   |_|_|
   |_|    |_|_|     |_|   |_|     |_|_|
.
There is only one free tetromino of width 1 and there are four free tetrominoes of width 2, hence the sum of the widths is 1 + 2 + 2 + 2 + 2 = 9, so a(4) = 9.

Index entries for sequences related to polyominoes.

Row sums of A379637.

Cf. A000105, A379623, A379625, A379627, A379629.

a(15)-a(16) from Pontus von Brömssen, Jan 17 2025

a(17)-a(18) from John Mason, Feb 16 2025

A379629 Sum of the lengths of the free polyominoes with n cells.

Original entry on oeis.org

1, 2, 5, 15, 41, 139, 474, 1773, 6686, 26043, 101814, 402297, 1593124, 6332329, 25200575, 100462874, 400908688, 1601541747

Offset: 1

Omar E. Pol, Jan 16 2025

The length here is the longer of the two dimensions.

			For n = 4 the free polyominoes with four cells are also called free tetrominoes.
The five free tetrominoes are as shown below:
                                     _
             _       _       _      |_|
    _ _     |_|     |_|_    |_|_    |_|
   |_|_|    |_|_    |_|_|   |_|_|   |_|
   |_|_|    |_|_|     |_|   |_|     |_|
.
There are no free tetrominoes of length 1, there is only one free tetromino of length 2, there are three free tetrominoes of length 3 and there is only one free tetromino of length 4, hence the sum of the lengths is 0 + 2 + 3 + 3 + 3 + 4 = 15, so a(4) = 15.

Index entries for sequences related to polyominoes.

Row sums of A379638.

Cf. A000105, A379624, A379625, A379626, A379627.

a(13)-a(16) from Pontus von Brömssen, Jan 17 2025

a(17)-a(18) (based on A379624 b-file) from Pontus von Brömssen, Feb 21 2025

A379638 Triangle read by rows: T(n,k) is the sum of the lengths of the free polyominoes with n cells and length k, n >= 1, k >= 1.

Original entry on oeis.org

1, 0, 2, 0, 2, 3, 0, 2, 9, 4, 0, 0, 24, 12, 5, 0, 0, 24, 84, 25, 6, 0, 0, 21, 236, 180, 30, 7, 0, 0, 9, 548, 835, 324, 49, 8, 0, 0, 3, 892, 3345, 1842, 539, 56, 9, 0, 0, 0, 1148, 10445, 9762, 3773, 824, 81, 10, 0, 0, 0, 1020, 27360, 42756, 22659, 6712, 1206, 90, 11, 0, 0, 0, 676, 59595, 165024, 116942, 46808, 11439, 1680, 121, 12

Offset: 1

Omar E. Pol, Jan 16 2025

The length here is the longer of the two dimensions.

			Triangle begins:
  1;
  0,  2;
  0,  2,   3;
  0,  2,   9,    4;
  0,  0,  24,   12,     5;
  0,  0,  24,   84,    25,      6;
  0,  0,  21,  236,   180,     30,      7;
  0,  0,   9,  548,   835,    324,     49,     8;
  0,  0,   3,  892,  3345,   1842,    539,    56,     9;
  0,  0,   0, 1148, 10445,   9762,   3773,   824,    81,   10;
  0,  0,   0, 1020, 27360,  42756,  22659,  6712,  1206,   90,   11;
  0,  0,   0,  676, 59595, 165024, 116942, 46808, 11439, 1680,  121,  12;
  ...
Illustration for n = 5:
The free polyominoes with five cells are also called free pentominoes.
For k = 1 there are no free pentominoes of length 1, so T(5,1) = 0.
For k = 2 there are no free pentominoes of length 2, so T(5,2) = 0.
For k = 3 there are eight free pentominoes of length 3 as shown below, hence the sum of the lengths is 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = 8*3 = 24, so (5,3) = 24.
   _ _     _ _       _ _     _ _ _     _         _           _       _ _
  |_|_|   |_|_|    _|_|_|   |_|_|_|   |_|       |_|_       _|_|_    |_|_|
  |_|_|   |_|_    |_|_|       |_|     |_|_ _    |_|_|_    |_|_|_|     |_|_
  |_|     |_|_|     |_|       |_|     |_|_|_|     |_|_|     |_|       |_|_|
.
For k = 4 there are three free pentominoes of length 4 as shown below, hence the sum of the lengths is 4 + 4 + 4 = 3*4 = 12, so T(5,4) = 12.
   _         _       _
  |_|      _|_|    _|_|
  |_|     |_|_|   |_|_|
  |_|_    |_|       |_|
  |_|_|   |_|       |_|
.
For k = 5 there is only one free pentomino of length 5 as shown below, so T(5,5) = 5.
   _
  |_|
  |_|
  |_|
  |_|
  |_|
.
Therefore the 5th row of the triangle is [0, 0, 24, 12, 5].

Index entries for sequences related to polyominoes.

Row sums give A379629.

Cf. A000105, A379624, A379625, A379627, A379637.

T(n,k) = k*A379624(n,k).

A380285 Total number of regions between the free polyominoes with n cells and their bounding boxes.

Original entry on oeis.org

0, 0, 1, 5, 21, 71, 255, 961, 3630, 13973, 53938, 209641, 815784, 3183642, 12439291, 48686549, 190787588, 748645732

Offset: 1

Omar E. Pol, Jan 18 2025

The regions include any holes in the polyominoes.

			Illustration for n = 4:
The free polyominoes with four cells are also called free tetrominoes.
The five free tetrominoes are as shown below:
    _
   |_|     _       _       _
   |_|    |_|     |_|_    |_|_     _ _
   |_|    |_|_    |_|_|   |_|_|   |_|_|
   |_|    |_|_|     |_|   |_|     |_|_|
.
The bounding boxes are respectively as shown below:
    _
   | |     _ _     _ _     _ _
   | |    |   |   |   |   |   |    _ _
   | |    |   |   |   |   |   |   |   |
   |_|    |_ _|   |_ _|   |_ _|   |_ _|
.
From left to right the number of regions between the free tetrominoes and their bounding boxes are respectively 0, 1, 2, 2, 0. Hence the total number of regions is 0 + 1 + 2 + 2 + 0 = 5, so a(4) = 5.
.

Index entries for sequences related to polyominoes.

Row sums of A380283 and of A380284.
Cf. A379628 (total area of the regions).
Cf. A000105, A057766, A379623, A379624, A379625, A379626, A379627, A379629, A379637, A379638, A380282.

a(n) = Sum_{k>0} k*A380282(n,k). - Pontus von Brömssen, Jan 24 2025

a(8)-a(16) from Pontus von Brömssen, Jan 24 2025

a(17)-a(18) from John Mason, Feb 14 2025

A381703 Irregular triangle read by rows in which every row of length A071764(n) lists A(n,w,h) = the number of free polyominoes of size n, width w and height h (for w <= h, and all possible w,h pairs).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 3, 1, 2, 3, 6, 1, 1, 6, 5, 7, 15, 1, 2, 11, 5, 7, 39, 25, 18, 1, 1, 10, 19, 7, 3, 59, 96, 35, 77, 61, 1, 3, 22, 28, 7, 1, 42, 210, 188, 49, 181, 383, 97, 73, 1, 1, 15, 52, 40, 9, 21, 255, 550, 332, 63, 266, 1304, 822, 155, 529, 240, 1, 3, 45, 90, 53, 9, 4, 212, 954, 1231, 529, 81, 251, 2847, 3548, 1551, 220, 2413, 2366, 410, 255

Offset: 1

John Mason, Mar 04 2025

			Triangle begins:
   n
   1:  1
   2:  1
   3:  1  1
   4:  1  1   3
   5:  1  2   3   6
   6:  1  1   6   5   7  15
   7:  1  2  11   5   7  39  25   18
   8:  1  1  10  19   7   3  59   96   35   77   61
   9:  1  3  22  28   7   1  42  210  188   49  181  383    97   73
  10:  1  1  15  52  40   9  21  255  550  332   63  266  1304  822  155  529  240
  ...
Any row contains an irregular array that shows the number of polyominoes having width w and height h. E.g., row 6 contains the array:
  h/w 1  2  3
  1
  2
  3      1  7
  4      6 15
  5      5
  6   1
.
There are 5 polyominoes of size 6 with width 2 and height 5, so A(6,2,5)=5:
.
  OO O  O  O  O
  O  OO O  O  O
  O  O  OO O  OO
  O  O  O  OO  O
  O  O  O   O  O

John Mason, Table of n, a(n) for n = 1..386

Row sums give A000105.
Row lengths give A071764.
Cf. A379623, A379624, A379625, A317186.

More terms from John Mason, Mar 07 2025

cp's OEIS Frontend

A317186 One of many square spiral sequences: a(n) = n^2 + n - floor((n-1)/2).

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A379637 Irregular triangle read by rows: T(n,k) is the sum of the widths of the free polyominoes with n cells and width k, n >= 1, 1 <= k <= ceiling(n/2).

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A380283 Irregular triangle read by rows: T(n,k) is the number of regions between the free polyominoes, with n cells and width k, and their bounding boxes, n >= 1, 1 <= k <= ceiling(n/2).

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A380284 Triangle read by rows: T(n,k) is the number of regions between the free polyominoes, with n cells and length k, and their bounding boxes, n >= 1, k >= 1.

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A379626 Sum of the widths of the free polyominoes with n cells.

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A379629 Sum of the lengths of the free polyominoes with n cells.

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A379638 Triangle read by rows: T(n,k) is the sum of the lengths of the free polyominoes with n cells and length k, n >= 1, k >= 1.

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A380285 Total number of regions between the free polyominoes with n cells and their bounding boxes.

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A381703 Irregular triangle read by rows in which every row of length A071764(n) lists A(n,w,h) = the number of free polyominoes of size n, width w and height h (for w <= h, and all possible w,h pairs).

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