A054569 -id:A054569 - cp's OEIS frontend

A008527 Coordination sequence for body-centered tetragonal lattice.

1, 10, 34, 74, 130, 202, 290, 394, 514, 650, 802, 970, 1154, 1354, 1570, 1802, 2050, 2314, 2594, 2890, 3202, 3530, 3874, 4234, 4610, 5002, 5410, 5834, 6274, 6730, 7202, 7690, 8194, 8714, 9250, 9802, 10370, 10954, 11554, 12170, 12802, 13450, 14114, 14794, 15490, 16202, 16930, 17674

Offset: 0

N. J. A. Sloane

Also sequence found by reading the segment (1, 10) together with the line from 10, in the direction 10, 34, ..., in the square spiral whose vertices are the generalized hexagonal numbers A000217. - Omar E. Pol, Nov 02 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
M. O'Keeffe, Coordination sequences for lattices, Zeit. f. Krist., 210 (1995), 905-908.
M. O'Keeffe, Coordination sequences for lattices, Zeit. f. Krist., 210 (1995), 905-908. [Annotated scanned copy]
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Apart from leading term, same as A108100.
Cf. A206399.
Cf. A016754 (SE), A054554 (NE), A054569 (SW), A053755 (NW), A033951 (S), A054552 (E), A054556 (N), A054567 (W) (Ulam spiral spokes).
A143839 (SSE) + A143855 (ESE) = A143838 (SSW) + A143856 (ENE) = A143854 (WSW) + A143861 (NNE) = A143859 (WNW) + A143860 (NNW) = even bisection = a(2n) = A010021(n).

Concatenation([1], List([1..40], n-> 2*(1+4*n^2) )); # G. C. Greubel, Nov 09 2019

[1] cat [2*(1 + 4*n^2): n in [1..50]]; // G. C. Greubel, Nov 09 2019

Maple
```
1, seq(8*k^2+2, k=1..50);
```

a[0]:= 1; a[n_]:= 8n^2 +2; Table[a[n], {n,0,50}] (* Alonso del Arte, Sep 06 2011 *)
LinearRecurrence[{3,-3,1},{1,10,34,74},50] (* Harvey P. Dale, Feb 13 2022 *)

vector(51, n, if(n==1,1, 2*(1+(2*n-2)^2)) ) \\ G. C. Greubel, Nov 09 2019

[1]+[2*(1+4*n^2) for n in (1..40)] # G. C. Greubel, Nov 09 2019

a(0) = 1; a(n) = 8*n^2+2 for n>0.
From Gary W. Adamson, Dec 27 2007: (Start)
a(n) = (2n+1)^2 + (2n-1)^2 for n>0.
Binomial transform of [1, 9, 15, 1, -1, 1, -1, 1, ...]. (End)
From Colin Barker, Apr 13 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>3.
G.f.: (1+x)*(1+6*x+x^2)/(1-x)^3. (End)
From Bruce J. Nicholson, Jul 31 2019: (Start) Assume n>0.
a(n) = A016754(n) + A016754(n-1).
a(n) = 2 * A053755(n).
a(n) = A054554(n+1) + A054569(n+1).
a(n) = A033951(n) + A054552(n).
a(n) = A054556(n+1) + A054567(n+1). (End)
E.g.f.: -1 + 2*exp(x)*(1 + 2*x)^2. - Stefano Spezia, Aug 02 2019
Sum_{n>=0} 1/a(n) = 3/4+1/8*Pi*coth(Pi/2) = 1.178172.... - R. J. Mathar, May 07 2024

A080335 Diagonal in square spiral or maze arrangement of natural numbers.

Original entry on oeis.org

1, 5, 9, 17, 25, 37, 49, 65, 81, 101, 121, 145, 169, 197, 225, 257, 289, 325, 361, 401, 441, 485, 529, 577, 625, 677, 729, 785, 841, 901, 961, 1025, 1089, 1157, 1225, 1297, 1369, 1445, 1521, 1601, 1681, 1765, 1849, 1937, 2025, 2117, 2209, 2305, 2401, 2501

Offset: 0

Paul Barry, Mar 19 2003

Interleaves the odd squares A016754 with (1+4n^2), A053755.
Squares of positive integers (plus 1 if n is odd). - Wesley Ivan Hurt, Oct 10 2013
a(n) is the maximum total number of queens that can coexist without attacking each other on an [n+3] X [n+3] chessboard, when the lone queen is in the most vulnerable position on the board. Specifically, the lone queen will placed in any center position, facing an opponent's "army" of size a(n)-1 == A137932(n+2). - Bob Selcoe, Feb 12 2015
a(n) is also the edge chromatic number of the complement of the (n+2) X (n+2) rook graph. - Eric W. Weisstein, Jan 31 2024

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Eric Weisstein's World of Mathematics, Edge Chromatic Number
Eric Weisstein's World of Mathematics, Rook Complement Graph
Eric Weisstein's World of Mathematics, Rook Graph
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A081347, A081348.
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.

[(3+4*n+2*n^2-(-1)^n)/2: n in [0..50]]; // Vincenzo Librandi, Sep 06 2011

A080335:=n->(n mod 2) + (n+1)^2; seq(A080335(k),k=0..49); # Wesley Ivan Hurt, Oct 10 2013

With[{nn = 60}, Riffle[Range[1, nn, 2]^2, 4 Range[nn]^2 + 1]] (* Harvey P. Dale, Jan 29 2012 *)
LinearRecurrence[{2, 0, -2, 1}, {1, 5, 9, 17}, 60] (* Harvey P. Dale, Jan 29 2012 *)
Table[(3 + 4 n + 2 n^2 - (-1)^n)/2, {n, 0, 50}] (* Wesley Ivan Hurt, Oct 10 2013 *)
Table[Mod[n, 2] + (n + 1)^2, {n, 0, 20}] (* Eric W. Weisstein, Jan 31 2024 *)

a(n) = (3 + 4*n + 2*n^2 - (-1)^n)/2.
a(2*n) = A016754(n), a(2*n+1) = A053755(n+1).
E.g.f.: exp(x)*(2 + 3*x + x^2) - cosh(x). The sequence 1,1,5,9,... is given by n^2+(1+(-1)^n)/2 with e.g.f. exp(1+x+x^2)*exp(x)-sinh(x). - Paul Barry, Sep 02 2003 and Sep 19 2003
a(0)=1, a(1)=5, a(2)=9, a(3)=17, a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4). - Harvey P. Dale, Jan 29 2012
a(n)+(-1)^n = A137928(n+1). - Philippe Deléham, Feb 17 2012
G.f.: (1 + 3*x - x^2 + x^3)/((1-x)^3*(1+x)). - Colin Barker, Mar 18 2012
a(n) = A000035(n) + A000290(n+1). - Wesley Ivan Hurt, Oct 10 2013
From Bob Selcoe, Feb 12 2015: (Start)
a(n) = A137932(n+2) + 1.
a(n) = (n+1)^2 when n is even; a(n) = (n+1)^2 + 1 when n is odd.
a(n) = A002378(n+2) - A047238(n+3) + 1.
(End)
Sum_{n>=0} 1/a(n) = Pi*coth(Pi/2)/4 + Pi^2/8 - 1/2. - Amiram Eldar, Jul 07 2022

A137932 Terms in an n X n spiral that do not lie on its principal diagonals.

Original entry on oeis.org

0, 0, 0, 4, 8, 16, 24, 36, 48, 64, 80, 100, 120, 144, 168, 196, 224, 256, 288, 324, 360, 400, 440, 484, 528, 576, 624, 676, 728, 784, 840, 900, 960, 1024, 1088, 1156, 1224, 1296, 1368, 1444, 1520, 1600, 1680, 1764, 1848, 1936, 2024, 2116, 2208, 2304, 2400, 2500, 2600, 2704, 2808

Offset: 0

William A. Tedeschi, Feb 29 2008

The count of terms not on the principal diagonals is always even.
The last digit is the repeating pattern 0,0,0,4,8,6,4,6,8,4, which is palindromic if the leading 0's are removed, 4864684.
The sum of the last digits is 40, which is the count of the pattern times 4.
A 4 X 4 spiral is the only spiral, aside from a 0 X 0, whose count of terms that do not lie on its principal diagonals equal the count of terms that do [A137932(4) = A042948(4)] making the 4 X 4 the "perfect spiral".
Yet another property is mod(a(n), A042948(n)) = 0 iff n is even. This is a large family that includes the 4 X 4 spiral.
a(n) is the maximum number of queens of one color that can coexist without attacking one queen of the opponent's color on an [n+1] X [n+1] chessboard, when the lone queen is in the most vulnerable position on the board, i.e., on a center square. - Bob Selcoe, Feb 12 2015
Also the circumference of the (n-1) X (n-1) grid graph for n > 2. - Eric W. Weisstein, Mar 25 2018
Also the crossing number of the complete bipartite graph K_{5,n}. - Eric W. Weisstein, Sep 11 2018

			a(0) = 0^2 - (2(0) - mod(0,2)) = 0.
a(3) = 3^2 - (2(3) - mod(3,2)) = 4.

Enrique Pérez Herrero, Table of n, a(n) for n = 0..5000
Kival Ngaokrajang, Illustration of initial terms.
Eric Weisstein's World of Mathematics, Complete Bipartite Graph.
Eric Weisstein's World of Mathematics, Graph Circumference.
Eric Weisstein's World of Mathematics, Graph Crossing Number.
Eric Weisstein's World of Mathematics, Grid Graph.
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A042948.
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.

A137932:=n->2*floor((n-1)^2/2); seq(A137932(n), n=0..50); # Wesley Ivan Hurt, Apr 19 2014

Table[2 Floor[(n - 1)^2/2], {n, 0, 20}] (* Enrique Pérez Herrero, Jul 04 2012 *)
2 Floor[(Range[20] - 1)^2/2] (* Eric W. Weisstein, Sep 11 2018 *)
Table[n^2 - 2 n + (1 - (-1)^n)/2, {n, 20}] (* Eric W. Weisstein, Sep 11 2018 *)
LinearRecurrence[{2, 0, -2, 1}, {0, 0, 4, 8}, 20] (* Eric W. Weisstein, Sep 11 2018 *)
CoefficientList[Series[-((4 x^2)/((-1 + x)^3 (1 + x))), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 11 2018 *)

A137932(n)={ return(n^2 - (2*n-n%2))} ;
print(vector(30,n,A137932(n-1))); /* R. J. Mathar, May 12 2014 */

Python
```
a = lambda n: n**2 - (2*n - (n%2))
```

a(n) = n^2 - (2*n - mod(n,2)) = n^2 - A042948(n).
a(n) = 2*A007590(n-1). - Enrique Pérez Herrero, Jul 04 2012
G.f.: -4*x^3 / ( (1+x)*(x-1)^3 ). a(n) = 4*A002620(n-1). - R. J. Mathar, Jul 06 2012
From Bob Selcoe, Feb 12 2015: (Start)
a(n) = (n-1)^2 when n is odd; a(n) = (n-1)^2 - 1 when n is even.
a(n) = A002378(n) - A047238(n+1). (End)
From Amiram Eldar, Mar 20 2022: (Start)
Sum_{n>=3} 1/a(n) = Pi^2/24 + 1/4.
Sum_{n>=3} (-1)^(n+1)/a(n) = Pi^2/24 - 1/4. (End)
E.g.f.: x*(x - 1)*cosh(x) + (x^2 - x + 1)*sinh(x). - Stefano Spezia, Oct 17 2022

A156859 The main column of a version of the square spiral.

Original entry on oeis.org

0, 3, 7, 14, 22, 33, 45, 60, 76, 95, 115, 138, 162, 189, 217, 248, 280, 315, 351, 390, 430, 473, 517, 564, 612, 663, 715, 770, 826, 885, 945, 1008, 1072, 1139, 1207, 1278, 1350, 1425, 1501, 1580, 1660, 1743, 1827, 1914, 2002, 2093, 2185, 2280, 2376, 2475, 2575

Offset: 0

Emilio Apricena (emilioapricena(AT)yahoo.it), Feb 17 2009

This spiral is sometimes called an Ulam spiral, but square spiral is a better name. - N. J. A. Sloane, Jul 27 2018
It is easy to see that the only two primes in the sequence are 3, 7. Therefore the primes of the version of Ulam spiral are divided into four parts (see also A035608): northeast (NE), northwest (NW), southwest (SW), and southeast (SE).
Number of pairs (x,y) having x and y of opposite parity with x in {0,...,n} and y in {0,...,2n}. - Clark Kimberling, Jul 02 2012
Partial Sums of A014601(n). - Wesley Ivan Hurt, Oct 11 2013

E. Apricena, A version of Ulam Spiral divided into four parts.
Minh Nguyen, 2-adic Valuations of Square Spiral Sequences, Honors Thesis, Univ. of Southern Mississippi (2021).
Marco Ripà, The n x n x n Points Problem Optimal Solution, viXra.org.
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A000290, A000384, A004526, A014601 (first differences), A115258.
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.

A156859:=n->n^2+n+floor((n+1)/2); seq(A156859(k), k=0..100); # Wesley Ivan Hurt, Oct 11 2013

Table[n^2 + n + Floor[(n+1)/2], {n, 0, 100}] (* Wesley Ivan Hurt, Oct 11 2013 *)

a(n) = n^2 + n + floor((n+1)/2) = A002378(n) + A004526(n+1) = A002620(n+1) + 3*A002620(n).
From R. J. Mathar, Feb 20 2009: (Start)
G.f.: x*(3+x)/((1+x)*(1-x)^3).
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4). (End)
a(n-1) = floor(n/(e^(1/n)-1)). - Richard R. Forberg, Jun 19 2013
a(n) = A000290(n+1) + A004526(-n-1). - Wesley Ivan Hurt, Jul 15 2013
a(n) + a(n+1) = A014105(n+1). - R. J. Mathar, Jul 15 2013
a(n) = floor(A000384(n+1)/2). - Bruno Berselli, Nov 11 2013
E.g.f.: (x*(5 + 2*x)*cosh(x) + (1 + 5*x + 2*x^2)*sinh(x))/2. - Stefano Spezia, Apr 24 2024
Sum_{n>=1} 1/a(n) = 4/9 + 2*log(2) - Pi/3. - Amiram Eldar, Apr 26 2024

More terms added by Wesley Ivan Hurt, Oct 11 2013

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

Original entry on oeis.org

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

A267682 a(n) = 2a(n-1) - 2a(n-3) + a(n-4) for n > 3, with initial terms 1, 1, 4, 8.

Original entry on oeis.org

1, 1, 4, 8, 15, 23, 34, 46, 61, 77, 96, 116, 139, 163, 190, 218, 249, 281, 316, 352, 391, 431, 474, 518, 565, 613, 664, 716, 771, 827, 886, 946, 1009, 1073, 1140, 1208, 1279, 1351, 1426, 1502, 1581, 1661, 1744, 1828, 1915, 2003, 2094, 2186, 2281, 2377, 2476

Offset: 0

Robert Price, Jan 19 2016

Also, total number of ON (black) cells after n iterations of the "Rule 201" elementary cellular automaton starting with a single ON (black) cell.

S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.

Robert Price, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Elementary Cellular Automaton
S. Wolfram, A New Kind of Science
Index entries for sequences related to cellular automata
Index to Elementary Cellular Automata
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A267679.
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.

rule=201; rows=20; ca=CellularAutomaton[rule,{{1},0},rows-1,{All,All}]; (* Start with single black cell *) catri=Table[Take[ca[[k]],{rows-k+1,rows+k-1}],{k,1,rows}]; (* Truncated list of each row *) nbc=Table[Total[catri[[k]]],{k,1,rows}]; (* Number of Black cells in stage n *) Table[Total[Take[nbc,k]],{k,1,rows}] (* Number of Black cells through stage n *)
LinearRecurrence[{2, 0, -2, 1}, {1, 1, 4, 8}, 60] (* Vincenzo Librandi, Jan 19 2016 *)

Vec((1-x+2*x^2+2*x^3)/((1-x)^3*(1+x)) + O(x^100)) \\ Colin Barker, Jan 19 2016

print([n*(n-1)+n//2+1 for n in range(51)]) # Karl V. Keller, Jr., Jul 14 2021

G.f.: (1 - x + 2*x^2 + 2*x^3) / ((1-x)^3*(1+x)). - Colin Barker, Jan 19 2016
a(n) = n*(n-1) + floor(n/2) + 1. - Karl V. Keller, Jr., Jul 14 2021
E.g.f.: (exp(x)*(2 + x + 2*x^2) - sinh(x))/2. - Stefano Spezia, Jul 16 2021

Edited by N. J. A. Sloane, Jul 25 2018, replacing definition with simpler formula provided by Colin Barker, Jan 19 2016.

A082040 a(n) = 9n^2 + 3n + 1.

Original entry on oeis.org

1, 13, 43, 91, 157, 241, 343, 463, 601, 757, 931, 1123, 1333, 1561, 1807, 2071, 2353, 2653, 2971, 3307, 3661, 4033, 4423, 4831, 5257, 5701, 6163, 6643, 7141, 7657, 8191, 8743, 9313, 9901, 10507, 11131, 11773, 12433, 13111, 13807, 14521, 15253, 16003, 16771, 17557

Offset: 0

Paul Barry, Apr 02 2003

4th row of A082039, case k = 3 of family T(n,k) = k^2*n^2 + k*n + 1.
a(n)^2 = 81*n^4 + 54*n^3 + 27*n^2 + 6*n + 1 = (24*((3*((3*n^2 + n)/2)^2 + ((3*n^2 + n)/2))/2) + 1). Therefore, (a(n)^2 - 1)/24 is a second pentagonal number (A005449) of index number equal to the n-th second pentagonal number. For example, a(30) = 8191 and (8191^2 - 1)/24 = (67092481 - 1)/24 = 2795520, the 1365th second pentagonal number. 1365 is the 30th second pentagonal number. - Raphie Frank, Sep 19 2012
For n >= 1, a(n) is the number of vertices in the hex derived network HDN1(n+1) from the Manuel et al. reference (see HFN1(4) in Fig. 8). - Emeric Deutsch, May 21 2018
4*a(n) - 3 is a square. - Muniru A Asiru, May 24 2018

Muniru A Asiru, Table of n, a(n) for n = 0..5000
Nicolay Avilov, Graphic illustration of sequence members
P. Manuel, R. Bharati, I. Rajasingh, and Chris Monica M, On minimum metric dimension of honeycomb networks, Journal of Discrete Algorithms, Vol. 6 (2008), pp. 20-27.
Leo Tavares, Snowflake illustration.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A002061, A005449, A045945, A054569, A082039.
Partial sums of A019557.
Cf. A000290, A003215.

List([0..50], n->9*n^2+3*n+1); # Muniru A Asiru, May 21 2018

seq(9*n^2+3*n+1,n=0..50); # Muniru A Asiru, May 21 2018

a(n)=9*n^2+3*n+1 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 18*n + a(n-1) - 6 with n > 0, a(0)=1. - Vincenzo Librandi, Aug 08 2010
a(n) = A045945(n) + 1: subsequence of A002061. - Muniru A Asiru, May 26 2018
a(n) = A003215(n) + 6*A000290(n). - Leo Tavares, Jul 14 2023
From Elmo R. Oliveira, Oct 23 2024: (Start)
G.f.: (1 + 10*x + 7*x^2)/(1 - x)^3.
E.g.f.: (1 + 12*x + 9*x^2)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A081352 Main diagonal of square maze arrangement of natural numbers A081349.

Original entry on oeis.org

1, 7, 11, 21, 29, 43, 55, 73, 89, 111, 131, 157, 181, 211, 239, 273, 305, 343, 379, 421, 461, 507, 551, 601, 649, 703, 755, 813, 869, 931, 991, 1057, 1121, 1191, 1259, 1333, 1405, 1483, 1559, 1641, 1721, 1807, 1891, 1981, 2069, 2163, 2255, 2353, 2449, 2551

Offset: 0

Paul Barry, Mar 19 2003

Conjecture: let a and b be integers such that 0 < a < b so that 0 < a/b is a proper fraction. Define the map f(a,b,D) = a/b + gcd(a,b)/D. Of course, all such a/b can be partially ordered by value, i.e., 1/2 = 0.5 < 2/3 = 4/6 = 6/9 = 0.6666... < 3/4 = 6/8 = 0.75 < 4/5 = 0.8 etc. The map f appears to specify a total strict order on the co-domain for all a/b that is consistent with the given partial order of the domain, i.e., the partial order remains intact, while equivalent fractions are given a total strict order themselves. Moreover, equivalent fractions are strictly ordered by numerator (or denominator), e.g., 1/2 < 2/4 < 3/6 etc. The conditions are that for n >= 4 all of the fractions with denominator b <= n are listed and the minimum integer value of D to achieve the total strict order of the co-domain is 2*C(n-1,2) - (-1)^(n-1). So, a(n-3) = D for n >= 4. Example: given n = 4, we have D = 2*(4-1,2) - (-1)^(4-1) = 2*3 + 1 = 7 = a(4-3) = a(1). Partial order of domain. 1/4 < 1/3 < 1/2 = 2/4 < 2/3 < 3/4. Total order of co-domain. f(1,4,7) = 1/4 + 1/7 = 33/84 < f(1,3,7) = 1/3 + 1/7 = 40/84 < f(1,2,7) = 1/2 + 1/7 = 54/84 < f(2,4,7) = 2/4 + 2/7 = 66/84 < f(2,3,7) = 2/3 + 2/7 = 68/84 < f(3,4,7) = 3/4 + 1/7 = 75/84. Observe that if D = 6, then f(2,4,6) = 2/4 + 2/6 = 10/12 = f(2,3,6) = 2/3 + 1/6. Computation shows the same failure to achieve total strict order of the co-domain for D = 2..5. (As a >= 1, then b >=2, from the above). Computation also shows that the conjecture holds for n = 4..17. - Ross La Haye, Oct 02 2016

B. Berselli, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A081350, A081351, A054569.

I:=[1,7,11,21]; [n le 4 select I[n] else 2*Self(n-1)-2*Self(n-3)+Self(n-4): n in [1..50]]; // Vincenzo Librandi, Aug 08 2013

A081352:=n->(n + 1)*(n + 2) - (-1)^n; seq(A081352(n), n=0..50); # Wesley Ivan Hurt, Feb 26 2014

CoefficientList[Series[(1 + 5 x - 3 x^2 + x^3) / ((1 + x) (1 - x)^3), {x, 0, 60}], x] (* Vincenzo Librandi, Aug 08 2013 *)

x='x+O('x^99); Vec((1+5*x-3*x^2+x^3)/((1+x)*(1-x)^3)) \\ Altug Alkan, Mar 26 2016

a(n) = (n + 1)*(n + 2) - (-1)^n = 2*C(n+2, 2) - (-1)^n.
G.f.: (1 +5*x -3*x^2 +x^3) / ((1+x)*(1-x)^3). [Bruno Berselli, Aug 01 2010]
a(n) -2*a(n-1) +2*a(n-3) -a(n-4) = 0 with n>3. [Bruno Berselli, Aug 01 2010]
a(n) = 3*A000982(n + 2) - A000982(n + 3). - Miko Labalan, Mar 26 2016
a(n) = A116940(n) + A236283(n + 1). - Miko Labalan, Dec 04 2016
a(n) = (2*n^2 + 6*n - 2*(-1)^n + (-1)^(2*n) + 3)/2. - Kritsada Moomuang, Oct 24 2019

A036826 a(n) = A036800(n)/2.

Original entry on oeis.org

0, 1, 9, 45, 173, 573, 1725, 4861, 13053, 33789, 84989, 208893, 503805, 1196029, 2801661, 6488061, 14876669, 33816573, 76283901, 170917885, 380633085, 843055101, 1858076669, 4076863485, 8908701693, 19394461693, 42077257725, 90999619581, 196226318333

Offset: 0

N. J. A. Sloane

Binomial transform of A054569 (with leading 0). Partial sums of A014477 (with leading 0). - Paul Barry, Jun 11 2003

This sequence is related to A000337 by a(n) = n*A000337(n) - Sum_{i=0..n-1} A000337(i). - Bruno Berselli, Mar 06 2012

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-18,20,-8).

Cf. A000337, A014477, A036800, A054569, A209359.

m:=28; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!((1+2*x)/((1-x)*(1-2*x)^3))); // Bruno Berselli, Mar 06 2012

A036826:= n-> 2^n*(3-2*n+n^2) -3; seq(A036826(n), n=0..30); # G. C. Greubel, Mar 31 2021

LinearRecurrence[{7,-18,20,-8}, {0,1,9,45}, 29] (* Bruno Berselli, Mar 06 2012 *)

for(n=0, 28, print1(2^n*(n^2-2*n+3)-3", ")); \\ Bruno Berselli, Mar 06 2012

[2^n*(3-2*n+n^2) -3 for n in (0..30)] # G. C. Greubel, Mar 31 2021

From Paul Barry, Jun 11 2003: (Start)
G.f.: x*(1+2*x)/((1-x)*(1-2*x)^3).
a(n) = 2^n*(n^2-2*n+3) - 3.
a(n) = Sum_{k=0..n} k^2*2^(k-1). (End)
a(n) = 7*a(n-1) -18*a(n-2) +20*a(n-3) -8*a(n-4). - Harvey P. Dale, Mar 04 2015
E.g.f.: -3*exp(x) + (3 -2*x +4*x^2)*exp(2*x). - G. C. Greubel, Mar 31 2021

A054568 Prime number spiral (clockwise, Southwest spoke).

Original entry on oeis.org

2, 17, 73, 191, 367, 607, 919, 1297, 1753, 2309, 2909, 3623, 4421, 5303, 6257, 7307, 8447, 9643, 10979, 12421, 13883, 15467, 17203, 19037, 20921, 22853, 24953, 27091, 29389, 31859, 34267, 36847, 39551, 42307, 45233, 48311, 51407, 54547, 57781

Offset: 0

Enoch Haga and G. L. Honaker, Jr., Apr 10 2000

			Begin a prime number spiral at shell 0 (prime 2), proceed clockwise, Southwest.
From _Omar E. Pol_, Feb 19 2022: (Start)
The spiral with four terms in every spoke looks like this:
.
  227  101--103--107--109--113--127
   |     |                       |
  223   97   29---31---37---41  131
   |     |    |              |   |
  211   89   23    3----5   43  137
   |     |    |    |    |    |   |
  199   83   19    2    7   47  139
   |     |    |         |    |   |
  197   79   17---13---11   53  149
   |     |                   |   |
  193   73---71---67---61---59  151
   |                             |
  191--181--179--173--167--163--157
.
(End)

Ivan Panchenko, Table of n, a(n) for n = 0..1000

Cf. A054569, A054551, A054553, A054555, A054564, A054566, A054570, A053999.

Table[ Prime[4n^2 - 6n + 3], {n, 1, 40} ]

8-spoke wheel overlays prime number spiral; hub is 2 in shell 0; 8 spokes radiate from this hub; this is Southwest, clockwise.

a(n) = A000040(A054569(n+1)). - R. J. Mathar, Aug 29 2018

Edited by Robert G. Wilson v, Feb 25 2002

cp's OEIS Frontend

A008527 Coordination sequence for body-centered tetragonal lattice.

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A080335 Diagonal in square spiral or maze arrangement of natural numbers.

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A137932 Terms in an n X n spiral that do not lie on its principal diagonals.

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A156859 The main column of a version of the square spiral.

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

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A267682 a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n > 3, with initial terms 1, 1, 4, 8.

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A082040 a(n) = 9*n^2 + 3*n + 1.

A267682 a(n) = 2a(n-1) - 2a(n-3) + a(n-4) for n > 3, with initial terms 1, 1, 4, 8.

A082040 a(n) = 9n^2 + 3n + 1.