A002939 -id:A002939 - cp's OEIS frontend

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

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

A017089 a(n) = 8*n + 2.

Original entry on oeis.org

2, 10, 18, 26, 34, 42, 50, 58, 66, 74, 82, 90, 98, 106, 114, 122, 130, 138, 146, 154, 162, 170, 178, 186, 194, 202, 210, 218, 226, 234, 242, 250, 258, 266, 274, 282, 290, 298, 306, 314, 322, 330, 338, 346, 354, 362, 370, 378, 386, 394, 402, 410, 418, 426

Offset: 0

N. J. A. Sloane, Dec 11 1996

Apart from initial term(s), dimension of the space of weight 2n cusp forms for Gamma_0( 33 ).
Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0( 81 ).
First differences of A002939. - Aaron David Fairbanks, May 13 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
Tanya Khovanova, Recursive Sequences.
William A. Stein, Dimensions of the spaces S_k(Gamma_0(N)).
William A. Stein, Dimensions of the spaces S_k^{new}(Gamma_0(N)).
William A. Stein, The modular forms database.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A002939, A008589, A008590, A016813, A016993, A017005, A017017, A017077.

a017089 = (+ 2) . (* 8)
a017089_list = [2, 10 ..]  -- Reinhard Zumkeller, Jun 07 2015

[8*n+2: n in [0..60]]; // Vincenzo Librandi, May 28 2011

A017089:=n->8*n+2; seq(A017089(n), n=0..50); # Wesley Ivan Hurt, May 13 2014

Range[2, 1000, 8] (* Vladimir Joseph Stephan Orlovsky, May 27 2011 *)

a(n) = 8*n+2; \\ Michel Marcus, Sep 17 2015

a(n) = 8*n+2; a(n) = 2*a(n-1)-a(n-2). - Vincenzo Librandi, May 28 2011
Sum_{n>=0} (-1)^n/a(n) = (Pi + 2*log(cot(Pi/8)))/(8*sqrt(2)). - Amiram Eldar, Dec 11 2021
From Elmo R. Oliveira, Mar 17 2024: (Start)
G.f.: 2*(1+3*x)/(1-x)^2.
E.g.f.: 2*exp(x)*(1 + 4*x).
a(n) = 2*A016813(n) = A008590(n) + 2. (End)

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.

A033952 Write 1,2,... in a clockwise spiral on a square lattice, writing each digit at a separate lattice point, starting with 1 at the origin and 2 at x=0, y=-1; sequence gives the numbers on the positive x-axis.

Original entry on oeis.org

1, 8, 6, 2, 3, 6, 6, 0, 3, 1, 8, 0, 2, 7, 1, 3, 9, 3, 4, 3, 9, 6, 0, 6, 8, 9, 6, 1, 2, 4, 2, 1, 5, 9, 4, 1, 0, 7, 7, 1, 7, 8, 0, 2, 6, 3, 4, 2, 7, 1, 8, 2, 0, 2, 2, 3, 5, 6, 6, 3, 2, 4, 1, 4, 1, 5, 6, 4, 2, 9, 2, 5, 5, 6, 8, 5, 0, 6, 4, 6, 7, 9, 0, 7, 6, 6, 7, 7, 7, 6, 4, 8, 0, 9, 1, 9, 5, 5, 9, 1, 0, 4, 0, 0, 9

Offset: 1

Olivier Gorin (gorin(AT)roazhon.inra.fr)

Same as the South spoke of the Champernowne spiral (A244677).

			The spiral begins
.
  3---1---4---1---5
  |               |
  1   5---6---7   1
  |   |       |   |
  2   4   1   8   6
  |   |   |   |   |
  1   3---2   9   1
  |           |   |
  1---1---0---1   7
.

Andrey Zabolotskiy, Table of n, a(n) for n = 1..10000

Sequences from spirals: A001107, A002939, A007742, A033951, A033952, A033953, A033954, A033989, A033990, A033991, A002943, A033996, A033988.

almostNatural[n_, b_] := Block[{m = 0, d = n, i = 1, l, p}, While[m <= d, l = m; m = (b - 1) i*b^(i - 1) + l; i++]; i--; p = Mod[d - l, i]; q = Floor[(d - l)/i] + b^(i - 1); If[p != 0, IntegerDigits[q, b][[p]], Mod[q - 1, b]]];
f[n_] := 4n^2 - 5n + 2; Array[ almostNatural[ f@#, 10] &, 105] (* Robert G. Wilson v, Aug 08 2014 *)

More terms from Andrew J. Gacek (andrew(AT)dgi.net)

Edited by Charles R Greathouse IV, Nov 01 2009

A296030 Pairs of coordinates for successive integers in the square spiral (counterclockwise).

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 1, -1, 1, -1, 0, -1, -1, 0, -1, 1, -1, 2, -1, 2, 0, 2, 1, 2, 2, 1, 2, 0, 2, -1, 2, -2, 2, -2, 1, -2, 0, -2, -1, -2, -2, -1, -2, 0, -2, 1, -2, 2, -2, 3, -2, 3, -1, 3, 0, 3, 1, 3, 2, 3, 3, 2, 3, 1, 3, 0, 3, -1, 3, -2, 3, -3, 3, -3, 2

Offset: 1

Benjamin Mintz, Dec 03 2017

The spiral is also called the Ulam spiral, cf. A174344, A274923 (x and y coordinates). - M. F. Hasler, Oct 20 2019
The n-th positive integer occupies the point whose x- and y-coordinates are represented in the sequence by a(2n-1) and a(2n), respectively. - Robert G. Wilson v, Dec 03 2017
From Robert G. Wilson v, Dec 05 2017: (Start)
The cover of the March 1964 issue of Scientific American (see link) depicts the Ulam Spiral with a heavy black line separating the numbers from their non-sequential neighbors. The pairs of coordinates for the points on this line, assuming it starts at the origin, form this sequence, negated.
The first number which has an abscissa value of k beginning at 0: 1, 2, 10, 26, 50, 82, 122, 170, 226, 290, 362, 442, 530, 626, 730, 842, 962, ...; g.f.: -(x^3 +7x^2 -x +1)/(x-1)^3;
The first number which has an abscissa value of -k beginning at 0: 1, 5, 17, 37, 65, 101, 145, 197, 257, 325, 401, 485, 577, 677, 785, 901, ...; g.f.: -(5x^2 +2x +1)/(x-1)^3;
The first number which has an ordinate value of k beginning at 0: 1, 3, 13, 31, 57, 91, 133, 183, 241, 307, 381, 463, 553, 651, 757, 871, 993, ...; g.f.: -(7x^2+1)/(x-1)^3;
The first number which has an ordinate value of -k beginning at 0: 1, 7, 21, 43, 73, 111, 157, 211, 273, 343, 421, 507, 601, 703, 813, 931, ...; g.f.: -(3x^2+4x+1)/(x-1)^3;
The union of the four sequences above is A033638.
(End)
Sequences A174344, A268038 and A274923 start with the integer 0 at the origin (0,0). One might then prefer offset 0 as to have (a(2n), a(2n+1)) as coordinates of the integer n. - M. F. Hasler, Oct 20 2019
This sequence can be read as an infinite table with 2 columns, where row n gives the x- and y-coordinate of the n-th point on the spiral. If the point at the origin has number 0, then the points with coordinates (n,n), (-n,n), (n,-n) and (n,-n) have numbers given by A002939(n) = 2n(2n-1): (0, 2, 12, 30, ...), A016742(n) = 4n^2: (0, 4, 16, 36, ...), A002943(n) = 2n(2n+1): (0, 6, 20, 42, ...) and A033996(n) = 4n(n+1): (0, 8, 24, 48, ...), respectively. - M. F. Hasler, Nov 02 2019

			The integer 1 occupies the initial position, so its coordinates are {0,0}; therefore a(1)=0 and a(2)=0.
The integer 2 occupies the position immediately to the right of 1, so its coordinates are {1,0}.
The integer 3 occupies the position immediately above 2, so its coordinates are {1,1}; etc.

S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 935.

Benjamin Mintz, Table of n, a(n) for n = 1..100000
BackIssues.com, Scientific American March 1964 back issue
Scientific American, March 1964 cover
Wikipedia, Ulam Spiral.

Cf. A033638, A063826, A174344, A180714, A268038, A274923.

Cf. Diagonal rays (+-n,+-n): A002939 (2n(2n-1): 0, 2, 12, 30, ...: NE), A016742 (4n^2: 0, 4, 16, 36, ...: NW), A002943 (2n(2n+1): 0, 6, 20, 42, ...: SW) and A033996 (4n(n+1): 0, 8, 24, 48, ...: SE).

f[n_] := Block[{k = Ceiling[(Sqrt[n] - 1)/2], m, t}, t = 2k +1; m = t^2; t--; If[n >= m - t, {k -(m - n), -k}, m -= t; If[n >= m - t, {-k, -k +(m - n)}, m -= t; If[n >= m - t, {-k +(m - n), k}, {k, k -(m - n - t)}]]]]; Array[f, 40] // Flatten (* Robert G. Wilson v, Dec 04 2017 *)
f[n_] := Block[{k = Mod[ Floor[ Sqrt[4 If[OddQ@ n, (n + 1)/2 - 2, (n/2 - 2)] + 1]], 4]}, f[n - 2] + If[OddQ@ n, Sin[k*Pi/2], -Cos[k*Pi/2]]]; f[1] = f[2] = 0; Array[f, 90] (* Robert G. Wilson v, Dec 14 2017 *)
f[n_] := With[{t = Round@ Sqrt@ n}, 1/2*(-1)^t*({1, -1}(Abs[t^2 - n] - t) + t^2 - n - Mod[t, 2])]; Table[f@ n, {n, 0, 95}] // Flatten (* Mikk Heidemaa May 23 2020, after Stephen Wolfram *)

apply( {coords(n)=my(m=sqrtint(n), k=m\/2); if(m <= n -= 4*k^2, [n-3*k,-k], n >= 0, [-k,k-n], n >= -m, [-k-n,k], [k,3*k+n])}, [0..99]) \\ Use concat(%) to remove brackets '[', ']'. This function gives the coordinates of n on the spiral starting with 0 at (0,0), as shown in Examples for A174344, A274923, ..., so (a(2n-1),a(2n)) = coords(n-1). To start with 1 at (0,0), change n to n-=1 in sqrtint(). The inverse function is pos(x,y) given e.g. in A316328. - M. F. Hasler, Oct 20 2019

from math import ceil, sqrt
def get_coordinate(n):
    k=ceil((sqrt(n)-1)/2)
    t=2*k+1
    m=t**2
    t=t-1
    if n >= m - t:
        return k - (m-n), -k
    else:
        m -= t
    if n >= m - t:
        return -k, -k+(m-n)
    else:
        m -= t
    if n >= m-t:
        return -k+(m-n), k
    else:
        return k, k-(m-n-t)

a(2*n-1) = A174344(n).
a(2*n) = A274923(n) = -A268038(n).
abs(a(n+2) - a(n)) < 2.
a(2*n-1)+a(2*n) = A180714(n).
f(n) = floor(-n/4)*ceiling(-3*n/4 - 1/4) mod 2 + ceiling(n/8) (gives the pairs of coordinates for integers in the diagonal rays). - Mikk Heidemaa, May 07 2020

A017485 a(n) = 11*n + 8.

Original entry on oeis.org

8, 19, 30, 41, 52, 63, 74, 85, 96, 107, 118, 129, 140, 151, 162, 173, 184, 195, 206, 217, 228, 239, 250, 261, 272, 283, 294, 305, 316, 327, 338, 349, 360, 371, 382, 393, 404, 415, 426, 437, 448, 459, 470, 481, 492, 503, 514, 525, 536, 547, 558, 569, 580, 591, 602, 613

Offset: 0

N. J. A. Sloane, Dec 11 1996

a(n) = A125199(n+1,3) for n>1. - Reinhard Zumkeller, Nov 24 2006

G. C. Greubel, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A002939, A016789, A017041, A125202.

Powers of the form (11*n+8)^m: this sequence (m=1), A017486 (m=2), A017487 (m=3), A017488 (m=4), A017489 (m=5), A017490 (m=6), A017491 (m=7), A017492 (m=8), A017493 (m=9), A017494 (m=10), A017495 (m=11), A017496 (m=12).

List([0..60], n-> 11*n+8); # G. C. Greubel, Sep 21 2019

[11*n+8 : n in [0..60]]; // Wesley Ivan Hurt, May 21 2014

A017485:=n->11*n+8; seq(A017485(n), n=0..60); # Wesley Ivan Hurt, May 21 2014

Range[8, 1000, 11] (* Vladimir Joseph Stephan Orlovsky, May 29 2011 *)
LinearRecurrence[{2,-1},{8,19},60] (* Harvey P. Dale, May 10 2021 *)

Vec((8+3*x)/(1-x)^2 + O(x^60)) \\ Colin Barker, Oct 05 2014

[11*n+8 for n in (0..60)] # G. C. Greubel, Sep 22 2019

a(n) = 22*n + 5 - a(n-1), with n>0, a(0)=8. - Vincenzo Librandi, Dec 24 2010
From Colin Barker, Oct 05 2014: (Start)
a(n) = 2*a(n-1) - a(n-2).
G.f.: (8 + 3*x)/(1-x)^2. (End)
E.g.f.: (8 + 11*x)*exp(x). - G. C. Greubel, Sep 21 2019

A226488 a(n) = n(13n - 9)/2.

Original entry on oeis.org

0, 2, 17, 45, 86, 140, 207, 287, 380, 486, 605, 737, 882, 1040, 1211, 1395, 1592, 1802, 2025, 2261, 2510, 2772, 3047, 3335, 3636, 3950, 4277, 4617, 4970, 5336, 5715, 6107, 6512, 6930, 7361, 7805, 8262, 8732, 9215, 9711, 10220, 10742, 11277, 11825, 12386, 12960

Offset: 0

Bruno Berselli, Jun 09 2013

Sum of n-th octagonal number and n-th 9-gonal (nonagonal) number.

Sum of reciprocals of a(n), for n>0: 0.629618994194109711163742089971688...

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000567, A001106, A153080 (first differences).

Cf. numbers of the form n*(n*k-k+4)/2 listed in A005843 (k=0), A000096 (k=1), A002378 (k=2), A005449 (k=3), A001105 (k=4), A005476 (k=5), A049450 (k=6), A218471 (k=7), A002939 (k=8), A062708 (k=9), A135706 (k=10), A180223 (k=11), A139267 (n=12), this sequence (k=13), A139268 (k=14), A226489 (k=15), A139271 (k=16), A180232 (k=17), A152995 (k=18), A226490 (k=19), A152965 (k=20), A226491 (k=21), A152997 (k=22).

List([0..50], n-> n*(13*n-9)/2); # G. C. Greubel, Aug 30 2019

Magma
```
[n*(13*n-9)/2: n in [0..50]];
```

I:=[0,2,17]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2) +Self(n-3): n in [1..50]]; // Vincenzo Librandi, Aug 18 2013

A226488:=n->n*(13*n - 9)/2; seq(A226488(n), n=0..50); # Wesley Ivan Hurt, Feb 25 2014

Table[n(13n-9)/2, {n, 0, 50}]
LinearRecurrence[{3, -3, 1}, {0, 2, 17}, 50] (* Harvey P. Dale, Jun 19 2013 *)
CoefficientList[Series[x(2+11x)/(1-x)^3, {x, 0, 45}], x] (* Vincenzo Librandi, Aug 18 2013 *)

a(n)=n*(13*n-9)/2 \\ Charles R Greathouse IV, Sep 24 2015

[n*(13*n-9)/2 for n in (0..50)] # G. C. Greubel, Aug 30 2019

G.f.: x*(2+11*x)/(1-x)^3.
a(n) + a(-n) = A152742(n).
a(0)=0, a(1)=2, a(2)=17; for n>2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Jun 19 2013
E.g.f.: x*(4 + 13*x)*exp(x)/2. - G. C. Greubel, Aug 30 2019
a(n) = A000567(n) + A001106(n). - Michel Marcus, Aug 31 2019

cp's OEIS Frontend

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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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A017089 a(n) = 8*n + 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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A033952 Write 1,2,... in a clockwise spiral on a square lattice, writing each digit at a separate lattice point, starting with 1 at the origin and 2 at x=0, y=-1; sequence gives the numbers on the positive x-axis.

Views

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

A226488 a(n) = n(13n - 9)/2.