A001107 -id:A001107 - cp's OEIS frontend

A049452 Pentagonal numbers with even index.

0, 5, 22, 51, 92, 145, 210, 287, 376, 477, 590, 715, 852, 1001, 1162, 1335, 1520, 1717, 1926, 2147, 2380, 2625, 2882, 3151, 3432, 3725, 4030, 4347, 4676, 5017, 5370, 5735, 6112, 6501, 6902, 7315, 7740, 8177, 8626, 9087, 9560, 10045, 10542

Offset: 0

Joe Keane (jgk(AT)jgk.org)

If Y is a 3-subset of an (2n+1)-set X then, for n>=4, a(n-1) is the number of 4-subsets of X having at least two elements in common with Y. - Milan Janjic, Dec 16 2007
Sequence found by reading the line (one of the diagonal axes) from 0, in the direction 0, 5,..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. - Omar E. Pol, Sep 08 2011
a(n) is the sum of 2*n consecutive integers starting from 2*n. - Bruno Berselli, Jan 16 2018

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

Cf. A000326, A033570, A049453, A001318, A033568, A185019.

See index to sequences with numbers of the form n*(d*n+10-d)/2 in A140090.

seq(n*(6*n-1),n=0..42); # Zerinvary Lajos, Jun 12 2007

Table[n(6n-1),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,5,22},50] (* Harvey P. Dale, Mar 07 2012 *)

a(n)=n*(6*n-1) \\ Charles R Greathouse IV, Sep 24 2015

a(n) = n*(6*n-1).
G.f.: x*(5+7*x)/(1-x)^3.
a(n) = C(6*n,2)/3. - Zerinvary Lajos, Jan 02 2007
a(n) = A001105(n) + A033991(n) = A033428(n) + A049450(n) = A022266(n) + A000326(n). - Zerinvary Lajos, Jun 12 2007
a(n) = 12*n + a(n-1) - 7 for n>0, a(0)=0. - Vincenzo Librandi, Aug 06 2010
a(n) = 4*A000217(n) + A001107(n). - Bruno Berselli, Feb 11 2011
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>2, a(0)=0, a(1)=5, a(2)=22. - Harvey P. Dale, Mar 07 2012
E.g.f.: (6*x^2 + 5*x)*exp(x). - G. C. Greubel, Jul 17 2017
From Amiram Eldar, Jul 03 2020: (Start)
Sum_{n>=1} 1/a(n) = 2*log(2) + 3*log(3)/2 - sqrt(3)*Pi/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi - log(2) - 2*sqrt(3)*arccoth(sqrt(3)). (End)

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

A011848 a(n) = floor(binomial(n, 2)/2).

Original entry on oeis.org

0, 0, 0, 1, 3, 5, 7, 10, 14, 18, 22, 27, 33, 39, 45, 52, 60, 68, 76, 85, 95, 105, 115, 126, 138, 150, 162, 175, 189, 203, 217, 232, 248, 264, 280, 297, 315, 333, 351, 370, 390, 410, 430, 451, 473, 495, 517, 540, 564, 588, 612, 637, 663, 689, 715, 742, 770, 798

Offset: 0

N. J. A. Sloane, Dec 11 1996

Column sums of an array of the odd numbers repeatedly shifted 4 places to the right:
1 3 5 7  9 11 13 15 17...
         1  3  5  7  9...
                     1...
.........................
-------------------------
1 3 5 7 10 14 18 22 27...
Floor of the area under the polygon connecting the lattice points (n, floor(n/2)) from 0..n. - Wesley Ivan Hurt, Jun 09 2014
Beginning with a(4)=3, the sequence might be called the "off-axis" Ulam-Spiral numbers because they are the numbers in ascending order on the horizontal and vertical spokes (heading outward) starting with the first turning points on the spiral (i.e., 3, 5, 7 and 10). That is, starting with: 3 (upward); 5 (leftward); 7 (downward) and 10 (rightward). These are A033991 (starting at a(1)), A007742 (starting at a(1)), A033954 (starting at a(1)) and A001107 (starting at a(2)), respectively. These quadri-sections are summarized in the formulas of Sep 26 2015. - Bob Selcoe, Oct 05 2015
Conjecture: For n = 2, a(n) is the greatest k such that A123663(k) < A000217(n - 2). - Peter Kagey, Nov 18 2016
a(n) is also the matching number of the n-triangular graph, (n-1)-triangular honeycomb queen graph, (n-1)-triangular honeycomb bishop graphs, and (for n > 7) (n-1)-triangular honeycomb obtuse knight graphs. - Eric W. Weisstein, Jun 02 2017 and Apr 03 2018
After 0, 0, 0, add 1, then add 2 three times, then add 3, then add 4 three times, then add 5, etc.; i.e., first differences are A004524 = (0, 0, 0, 1, 2, 2, 2, 3, 4, 4, 4, 5, ...). - M. F. Hasler, May 09 2018
Let s(0) = s(1) = 1, s(-1) = s(2) = x, and s(n+2)*s(n-2) = s(n+1)*s(n-1) + s(n)^2 for all n in Z. Then s(n) = p(n) / x^e(n) is a Laurent polynomial in x with p(n) a polynomial with nonnegative integer coefficients of degree a(n) for all n in Z. If x = 1, then s(n) = p(n) = A006720(n+1). - Michael Somos, Mar 22 2023

			G.f. = x^3 + 3*x^4 + 5*x^5 + 7*x^6 + 10*x^7 + 14*x^8 + 18*x^9 + 22*x^10 + ...
p(0) = p(1) = 1, p(2) = 1 + x, p(3) = 1 + x + x^3, p(4) = 1 + 2*x + 2*x^2 + x^3 + x^5. - _Michael Somos_, Mar 22 2023

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Kyu-Hwan Lee and Se-jin Oh, Catalan triangle numbers and binomial coefficients, arXiv:1601.06685 [math.CO], 2016.
Eric Weisstein's World of Mathematics, Matching Number.
Eric Weisstein's World of Mathematics, Triangular Graph.
Index entries for linear recurrences with constant coefficients, signature (3,-4,4,-3,1).

A column of triangle A011857.
First differences are in A004524.
Cf. A007318, A033991, A007742, A033954, A001107, A006720, A035608 (bisection), A156859 (bisection).

List([0..60],n->Int(Binomial(n,2)/2)); # Muniru A Asiru, Apr 05 2018

a011848 n = if n < 2 then 0 else flip div 2 $ a007318 n 2
-- Reinhard Zumkeller, Mar 04 2015

[ Floor(n*(n-1)/4) : n in [0..50] ]; // Wesley Ivan Hurt, Jun 09 2014

seq(floor(binomial(n,2)/2), n=0..57); # Zerinvary Lajos, Jan 12 2009

Table[Floor[n (n - 1)/4], {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Jun 28 2011 *)
CoefficientList[Series[x^3/((1 + x^2) (1 - x)^3), {x, 0, 70}], x] (* Vincenzo Librandi, Jun 21 2013 *)
LinearRecurrence[{3, -4, 4, -4, 1}, {0, 0, 1, 3, 5}, {0, 20}] (* Eric W. Weisstein, Jun 02 2017 *)
Table[Floor[Binomial[n, 2]/2], {n, 0, 20}] (* Eric W. Weisstein, Jun 02 2017 *)
Table[1/4 (-1 + (-1 + n) n + Cos[n Pi/2] + Sin[n Pi/2]), {n, 0, 20}] (* Eric W. Weisstein, Jun 02 2017 *)
Floor[Binomial[Range[0, 20], 2]/2] (* Eric W. Weisstein, Apr 03 2018 *)

PARI
```
a(n) = binomial(n, 2)\2;
```

vector(100, n, n--; floor(n*(n-1)/4)) \\ Altug Alkan, Sep 30 2015

def a(n): return n*(n-1)//4 # Christoph B. Kassir, Oct 07 2022

[floor(binomial(n,2)/2) for n in range(0,58)] # Zerinvary Lajos, Dec 01 2009

G.f.: x^3*(1-x^2)/((1-x)^3*(1-x^4)).
G.f.: x^3/((1+x^2)*(1-x)^3). - Jon Perry, Mar 31 2004
a(n) = +3*a(n-1) -4*a(n-2) +4*a(n-3) -3*a(n-4) +a(n-5). - R. J. Mathar, Apr 15 2010
a(n) = floor((n/(1+e^(1/n)))^2). - Richard R. Forberg, Jun 19 2013
a(n) = floor(n*(n-1)/4). - T. D. Noe, Jun 20 2013
a(n) = (1/4) * ( n^2 - n - 1 + (-1)^floor(n/2) ). - Ralf Stephan, Aug 11 2013
a(n) = A054925(n) - A133872(n+2). - Wesley Ivan Hurt, Jun 09 2014
a(4*n) = A033991(n). a(4*n+1) = A007742(n). a(4*n+2) = A033954(n). a(4*n+3) = A001107(n+1). - Bob Selcoe, Sep 26 2015
E.g.f.: (sin(x) + cos(x) + (x^2 - 1)*exp(x))/4. - Ilya Gutkovskiy, Nov 18 2016
A054925(n) = a(-n). A035608(n) = a(2*n+1). Wesley Ivan Hurt, Jun 09 2014
A156859(n) = a(2*n+2). - Michael Somos, Nov 18 2016
Euler transform of length 4 sequence [ 3, -1, 0, 1]. - Michael Somos, Nov 18 2016
From Amiram Eldar, Mar 18 2022: (Start)
Sum_{n>=3} 1/a(n) = 40/9 - 2*Pi/3.
Sum_{n>=3} (-1)^(n+1)/a(n) = 32/9 - 4*log(2). (End)
0 = a(n+2)*(a(n)*(a(n) -6*a(n+1) +4*a(n+2)) +a(n+1)*(8*a(n+1) -10*a(n+2)) + 3*a(n+2)^2) +a(n+3)*(a(n)*(+a(n) -2*a(n+1)) +a(n+2)*(2*a(n+1) -a(n+2))) for all n in Z. - Michael Somos, Mar 22 2023
2*a(n) + 2*a(n-2) = (n-1)*(n-2). - R. J. Mathar, Feb 12 2024

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

A158405 Triangle T(n,m) = 1+2*m of odd numbers read along rows, 0<=m

Original entry on oeis.org

1, 1, 3, 1, 3, 5, 1, 3, 5, 7, 1, 3, 5, 7, 9, 1, 3, 5, 7, 9, 11, 1, 3, 5, 7, 9, 11, 13, 1, 3, 5, 7, 9, 11, 13, 15, 1, 3, 5, 7, 9, 11, 13, 15, 17, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23

Offset: 1

Paul Curtz, Mar 18 2009

Row sums are n^2 = A000290(n).
The triangle sums, see A180662 for their definitions, link this triangle of odd numbers with seventeen different sequences, see the crossrefs. The knight sums Kn14 - Kn110 have been added. - Johannes W. Meijer, Sep 22 2010
A208057 is the eigentriangle of A158405 such that as infinite lower triangular matrices, A158405 * A208057 shifts the latter, deleting the right border of 1's. - Gary W. Adamson, Feb 22 2012
T(n,k) = A099375(n-1,n-k), 1<=k<=n. [Reinhard Zumkeller, Mar 31 2012]

			The triangle contains the first n odd numbers in row n:
  1;
  1,3;
  1,3,5;
  1,3,5,7;
From _Seiichi Manyama_, Dec 02 2017: (Start)
    |       a(n)        |                               | A000290(n)
   -----------------------------------------------------------------
   0|                                                      (=  0)
   1|                 1 = 1/3 * ( 3)                       (=  1)
   2|             1 + 3 = 1/3 * ( 5 +  7)                  (=  4)
   3|         1 + 3 + 5 = 1/3 * ( 7 +  9 + 11)             (=  9)
   4|     1 + 3 + 5 + 7 = 1/3 * ( 9 + 11 + 13 + 15)        (= 16)
   5| 1 + 3 + 5 + 7 + 9 = 1/3 * (11 + 13 + 15 + 17 + 19)   (= 25)
(End)

Seiichi Manyama, Rows n = 1..140, flattened
Daniel Erman, The Josephus Problem, Numberphile video (2016)
Boris Putievskiy, Transformations [of] Integer Sequences And Pairing Functions arXiv:1212.2732 [math.CO], 2012.
Index entries for sequences related to the Josephus Problem

Cf. A129326, A099375, A005408.
Triangle sums (see the comments): A000290 (Row1; Kn11 & Kn4 & Ca1 & Ca4 & Gi1 & Gi4); A000027 (Row2); A005563 (Kn12); A028347 (Kn13); A028560 (Kn14); A028566 (Kn15); A098603 (Kn16); A098847 (Kn17); A098848 (Kn18); A098849 (Kn19); A098850 (Kn110); A000217 (Kn21. Kn22, Kn23, Fi2, Ze2); A000384 (Kn3, Fi1, Ze3); A000212 (Ca2 & Ze4); A000567 (Ca3, Ze1); A011848 (Gi2); A001107 (Gi3). - Johannes W. Meijer, Sep 22 2010
Cf. A063656, A063657, A208057, A292610, A292611.

a158405 n k = a158405_row n !! (k-1)
a158405_row n = a158405_tabl !! (n-1)
a158405_tabl = map reverse a099375_tabl
-- Reinhard Zumkeller, Mar 31 2012

Table[2 Range[1, n] - 1, {n, 12}] // Flatten (* Michael De Vlieger, Oct 01 2015 *)

a(n) = 2*(n-floor((-1+sqrt(8*n-7))/2)*(floor((-1+sqrt(8*n-7))/2)+1)/2)-1;
vector(100, n, a(n)) \\ Altug Alkan, Oct 01 2015

a(n) = 2*i-1, where i = n-t(t+1)/2, t = floor((-1+sqrt(8*n-7))/2). - Boris Putievskiy, Feb 03 2013

a(n) = 2*A002262(n-1) + 1. - Eric Werley, Sep 30 2015

Edited by R. J. Mathar, Oct 06 2009

A139271 a(n) = 2n(4*n-3).

Original entry on oeis.org

0, 2, 20, 54, 104, 170, 252, 350, 464, 594, 740, 902, 1080, 1274, 1484, 1710, 1952, 2210, 2484, 2774, 3080, 3402, 3740, 4094, 4464, 4850, 5252, 5670, 6104, 6554, 7020, 7502, 8000, 8514, 9044, 9590, 10152, 10730, 11324, 11934, 12560, 13202, 13860, 14534, 15224

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 2, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A033585 in the same spiral.
Twice decagonal numbers (or twice 10-gonal numbers). - Omar E. Pol, May 15 2008
a(n) is the number of walks in a cubic lattice of n dimensions that reach the point of origin for the first time after 4 steps. - Shel Kaphan, Mar 20 2023

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A139272, A139273, A139274, A139275, A139276, A139278, A139279, A139280, A139281, A139282.
Cf. A001107.
Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488 (this sequence is the case k=16). - Bruno Berselli, Jun 10 2013
Row n=2 of A361397.

Table[8n^2-6n,{n,0,40}] (* or *) LinearRecurrence[{3,-3,1},{0,2,20},50] (* Harvey P. Dale, Sep 26 2016 *)

a(n)=2*n*(4*n-3) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 8*n^2 - 6*n.
Sequences of the form a(n) = 8*n^2 + c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n) = 3a(n-1) - 3a(n-2) + a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = A001107(n)*2. - Omar E. Pol, May 15 2008
a(n) = 16*n + a(n-1) - 14 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: (2*x)*(7*x+1)/(1-x)^3.
E.g.f.: (8*x^2 + 2*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = Pi/12 + log(2)/2. - Amiram Eldar, Mar 28 2023

Corrected by Harvey P. Dale, Sep 26 2016

cp's OEIS Frontend

A049452 Pentagonal numbers with even index.

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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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A011848 a(n) = floor(binomial(n, 2)/2).

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

A139271 a(n) = 2n(4*n-3).