A016742 -id:A016742 - cp's OEIS frontend

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

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

A139098 a(n) = 8*n^2.

Original entry on oeis.org

0, 8, 32, 72, 128, 200, 288, 392, 512, 648, 800, 968, 1152, 1352, 1568, 1800, 2048, 2312, 2592, 2888, 3200, 3528, 3872, 4232, 4608, 5000, 5408, 5832, 6272, 6728, 7200, 7688, 8192, 8712, 9248, 9800, 10368, 10952, 11552, 12168, 12800, 13448, 14112, 14792, 15488, 16200

Offset: 0

Omar E. Pol, Apr 25 2008

Opposite numbers to the centered 16-gonal numbers (A069129) in the square spiral whose vertices are the triangular numbers (A000217).
8 times the squares. - Omar E. Pol, Dec 09 2008
a(n-1) is the molecular topological index of the n-wheel graph W_n. - Eric W. Weisstein, Jul 11 2011
An n X n pandiagonal magic square has a(n) orientations. - Kausthub Gudipati, Sep 15 2011
Area of a square with diagonal 4n. - Wesley Ivan Hurt, Jun 19 2014
Sum of all the parts in the partitions of 4n into exactly two parts. - Wesley Ivan Hurt, Jul 23 2014
Equivalently: integers k such that k$ / (k/2-1)! and k$ / (k/2)! are both squares when A000178 (k) = k$ = 1!*2!*...*k! is the superfactorial of k (see A348692 for further information). - Bernard Schott, Dec 02 2021

Vincenzo Librandi, Table of n, a(n) for n = 0..800
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Eric Weisstein's World of Mathematics, Molecular Topological Index.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Subsequence of A008586 and of A349081.
Cf. A000217, A000290, A000178, A001105, A016742, A016766, A033582, A069129, A245058.
Cf. A008590, A348692.

[8*n^2: n in [0..50]]; // Vincenzo Librandi, Apr 26 2011

A139098:=n->8*n^2; seq(A139098(n), n=0..50); # Wesley Ivan Hurt, Jun 19 2014

8 Range[0, 50]^2 (* Wesley Ivan Hurt, Jun 19 2014 *)
LinearRecurrence[{3,-3,1},{0,8,32},50] (* Harvey P. Dale, Oct 05 2023 *)

a(n)=8*n^2 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 8*A000290(n) = 4*A001105(n) = 2*A016742(n). - Omar E. Pol, Dec 13 2008
G.f.: -8*x*(1+x)/(x-1)^3. - R. J. Mathar, Nov 27 2015
From Amiram Eldar, Feb 03 2021: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/48 (A245058).
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/96.
Product_{n>=1} (1 + 1/a(n)) = sqrt(8)*sinh(Pi/sqrt(8))/Pi.
Product_{n>=1} (1 - 1/a(n)) = sqrt(8)*sin(Pi/sqrt(8))/Pi. (End)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Wesley Ivan Hurt, Dec 03 2021
From Elmo R. Oliveira, Dec 01 2024: (Start)
E.g.f.: 8*x*(1 + x)*exp(x).
a(n) = n*A008590(n) = A001105(2*n). (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.

A016814 a(n) = (4*n + 1)^2.

Original entry on oeis.org

1, 25, 81, 169, 289, 441, 625, 841, 1089, 1369, 1681, 2025, 2401, 2809, 3249, 3721, 4225, 4761, 5329, 5929, 6561, 7225, 7921, 8649, 9409, 10201, 11025, 11881, 12769, 13689, 14641, 15625, 16641, 17689, 18769, 19881, 21025, 22201, 23409, 24649, 25921, 27225, 28561, 29929

Offset: 0

N. J. A. Sloane

A bisection of A016754. Sequence arises from reading the line from 1, in the direction 1, 25, ..., in the square spiral whose vertices are the squares A000290. - Omar E. Pol, May 24 2008

G. C. Greubel, Table of n, a(n) for n = 0..10000 (terms 0..200 from Ivan Panchenko).
Leo Tavares, Illustration: Square trapeziums
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Sequences of the form (m*n+1)^2: A000012 (m=0), A000290 (m=1), A016754 (m=2), A016778 (m-3), this sequence (m=4), A016862 (m=5), A016922 (m=6), A016994 (m=7), A017078 (m=8), A017174 (m=9), A017282 (m=10), A017402 (m=11), A017534 (m=12), A134934 (m=14).
Cf. A001539, A006752, A016742, A016802, A016813.
Cf. A016826, A016838, A017126, A068467, A087197.
Cf. A272399, A014105.

[(4*n+1)^2: n in [0..40]]; // G. C. Greubel, Dec 28 2022

A016814:=n->(4*n+1)^2; seq(A016814(k), k=0..100); # Wesley Ivan Hurt, Nov 02 2013

(4*Range[0,40] +1)^2 (* or *) LinearRecurrence[{3,-3,1}, {1,25,81}, 40] (* Harvey P. Dale, Nov 20 2012 *)
Accumulate[32Range[0, 47] - 8] + 9 (* Alonso del Arte, Aug 19 2017 *)

a(n)=(4*n+1)^2 \\ Charles R Greathouse IV, Oct 07 2015

[(4*n+1)^2 for n in range(41)] # G. C. Greubel, Dec 28 2022

a(n) = a(n-1) + 32*n - 8, n > 0. - Vincenzo Librandi, Dec 15 2010
From George F. Johnson, Sep 28 2012: (Start)
G.f.: (1 + 22*x + 9*x^2)/(1 - x)^3.
a(n+1) = a(n) + 16 + 8*sqrt(a(n)).
a(n+1) = 2*a(n) - a(n-1) + 32 = 3*a(n) - 3*a(n-1) + a(n-2).
a(n-1)*a(n+1) = (a(n) - 16)^2 ; a(n+1) - a(n-1) = 16*sqrt(a(n)).
a(n) = A016754(2*n) = (A016813(n))^2. (End)
Sum_{n>=0} 1/a(n) = G/2 + Pi^2/16, where G is the Catalan constant (A006752). - Amiram Eldar, Jun 28 2020
Product_{n>=1} (1 - 1/a(n)) = 2*Gamma(5/4)^2/sqrt(Pi) = 2 * A068467^2 * A087197. - Amiram Eldar, Feb 01 2021
From G. C. Greubel, Dec 28 2022: (Start)
a(2*n) = A017078(n).
a(2*n+1) = A017126(n).
E.g.f.: (1 + 24*x + 16*x^2)*exp(x). (End)
a(n) = A272399(n+1) - A014105(n). - Leo Tavares, Dec 24 2023

A016802 a(n) = (4*n)^2.

Original entry on oeis.org

0, 16, 64, 144, 256, 400, 576, 784, 1024, 1296, 1600, 1936, 2304, 2704, 3136, 3600, 4096, 4624, 5184, 5776, 6400, 7056, 7744, 8464, 9216, 10000, 10816, 11664, 12544, 13456, 14400, 15376, 16384, 17424, 18496, 19600, 20736, 21904, 23104, 24336, 25600, 26896, 28224

Offset: 0

N. J. A. Sloane

A bisection of A016742. Sequence arises from reading the line from 0, in the direction 0, 16, ... in the square spiral whose vertices are the squares A000290. - Omar E. Pol, May 24 2008

Also, sequence found by reading the line from 0, in the direction 0, 16, ... in the square spiral whose vertices are the generalized decagonal numbers A074377. - Omar E. Pol, Sep 10 2011

Karl-Heinz Hofmann, Table of n, a(n) for n = 0..10000 (first 200 terms from Ivan Panchenko).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000290, A001105, A001539, A016742, A016754, A016814, A016826, A016838, A112628.

Cf. A008598, A074377, A139098, A195146.

a(n) = (4*n)^2; \\ Michel Marcus, Mar 04 2014

def A016802(n): return (4*n)**2 # Karl-Heinz Hofmann, Sep 11 2024

a(n) = 16*n^2 = 16*A000290(n). - Omar E. Pol, Dec 11 2008
a(n) = 8*A001105(n) = 4*A016742(n) = 2*A139098(n). - Omar E. Pol, Dec 13 2008
a(n) = a(n-1) + 16*(2*n-1) (with a(0)=0). - Vincenzo Librandi, Nov 20 2010
From Amiram Eldar, Jan 25 2021: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/96.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/192.
Product_{n>=1} (1 + 1/a(n)) = sinh(Pi/4)/(Pi/4).
Product_{n>=1} (1 - 1/a(n)) = sin(Pi/4)/(Pi/4) = 2*sqrt(2)/Pi (A112628). (End)
From Elmo R. Oliveira, Nov 30 2024: (Start)
G.f.: 16*x*(1 + x)/(1-x)^3.
E.g.f.: 16*x*(1 + x)*exp(x).
a(n) = n*A008598(n) = A195146(2*n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A022264 a(n) = n(7n - 1)/2.

Original entry on oeis.org

0, 3, 13, 30, 54, 85, 123, 168, 220, 279, 345, 418, 498, 585, 679, 780, 888, 1003, 1125, 1254, 1390, 1533, 1683, 1840, 2004, 2175, 2353, 2538, 2730, 2929, 3135, 3348, 3568, 3795, 4029, 4270, 4518, 4773, 5035, 5304, 5580, 5863, 6153, 6450, 6754, 7065, 7383

Offset: 0

N. J. A. Sloane

Sequence found by reading the line from 0, in the direction 0, 13, ..., and the parallel line from 3, in the direction 3, 30, ..., in the square spiral whose edges have length A195019 and whose vertices are the numbers A195020. - Omar E. Pol, Sep 09 2011

G. C. Greubel, Table of n, a(n) for n = 0..1000
Leo Tavares, Illustration: Crysta-gons
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A022265.
Cf. sequences listed in A254963.
Cf. similar sequences listed in A022288.
Cf. A016742, A049450, A174738, A195019, A195020.
Cf. A005449, A000384.

[n*(7*n-1)/2 : n in [0..50]]; // Wesley Ivan Hurt, Dec 04 2016

[seq(binomial(7*n,2)/7, n=0..37)]; # Zerinvary Lajos, Jan 02 2007

Table[n (7*n - 1)/2, {n, 0, 40}] (* Zerinvary Lajos, Jul 10 2009 *)

a(n)=n*(7*n-1)/2 \\ Charles R Greathouse IV, Mar 08 2013

a(n) = C(7*n,2)/7, n >= 0. - Zerinvary Lajos, Jan 02 2007
a(n) = A049450(n) + A000217(n). - Reinhard Zumkeller, Oct 09 2008
a(n) = 7*n + a(n-1) - 4 for n > 0, a(0)=0. - Vincenzo Librandi, Aug 04 2010
a(n) = (2*n)^2 - n*(n+1)/2 = A016742(n) - A000217(n). - Philippe Deléham, Mar 08 2013
a(n) = A174738(7*n+2). - Philippe Deléham, Mar 26 2013
G.f.: x*(3 + 4*x)/(1 - x)^3. - R. J. Mathar, Aug 04 2016
a(n) = A000217(4*n-1) - A000217(3*n-1). - Bruno Berselli, Oct 17 2016
a(n) = (1/5) * Sum_{i=n..(6*n-1)} i. - Wesley Ivan Hurt, Dec 04 2016
E.g.f.: (1/2)*x*(7*x + 6)*exp(x). - G. C. Greubel, Aug 19 2017
a(n) = A005449(n) + A000384(n). See Crysta-gons illustration. - Leo Tavares, Nov 21 2021

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

A055808 a(n) and floor(a(n)/4) are both squares; i.e., squares that remain squares when written in base 4 and last digit is removed.

Original entry on oeis.org

0, 1, 4, 16, 36, 64, 100, 144, 196, 256, 324, 400, 484, 576, 676, 784, 900, 1024, 1156, 1296, 1444, 1600, 1764, 1936, 2116, 2304, 2500, 2704, 2916, 3136, 3364, 3600, 3844, 4096, 4356, 4624, 4900, 5184, 5476, 5776, 6084, 6400, 6724, 7056, 7396, 7744, 8100

Offset: 0

Henry Bottomley, Jul 14 2000

Let A(x) = (1 + k*x + (k-1)*x^2). Then the coefficients of (A(x))^2 sum to 4*k^2, where k = (n - 1). Examples: if k = 3 we have (1 + 3*x + 2*x^2)^2 = (1 + 6*x + 13x^2 + 12*x^3 + 4*x^4), and ( 1 + 6 + 13 + 12 + 4) = 36. If k = 4 we have (1 + 4*x + 3*x^2)^2 = (1 + 8*x + 22*x^2 + 24*x^3 + 9*x^4), and (1 + 8 + 22 + 24 + 9) = 64 = a(5). - Gary W. Adamson, Aug 02 2015

For n>0, a(n) are the Engel expansion of A197036. - Benedict W. J. Irwin, Dec 15 2016

			36 is in the sequence because 36 = 6^2 = 210 base 3 and 21 base 4 = 9 = 3^2.

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A023110. Essentially A016742 with one addition.

[Floor((2*n^2)/(1 + n))^2: n in [0..60]]; // Vincenzo Librandi, Aug 03 2015

Join[{0, 1}, LinearRecurrence[{3, -3, 1}, {4, 16, 36}, 50]] (* Vincenzo Librandi, Aug 03 2015 *)

concat(0, Vec(x*(x^3-7*x^2-x-1)/(x-1)^3 + O(x^100))) \\ Colin Barker, Sep 15 2014

is_ok(n)=issquare(n) && issquare(floor(n/4));
first(m)=my(v=vector(m),r=0);for(i=1,m,while(!is_ok(r),r++);v[i]=r;r++;);v; /* Anders Hellström, Aug 08 2015 */

a(n) = A004275(n)^2. - M. F. Hasler, Jan 16 2012

a(n) = 4*(-1+n)^2 for n>1; a(n) = 3*a(n-1)-3*a(n-2)+a(n-3) for n>4; G.f.: x*(x^3-7*x^2-x-1) / (x-1)^3. - Colin Barker, Sep 15 2014

A114949 a(n) = n^2 + 6.

Original entry on oeis.org

6, 7, 10, 15, 22, 31, 42, 55, 70, 87, 106, 127, 150, 175, 202, 231, 262, 295, 330, 367, 406, 447, 490, 535, 582, 631, 682, 735, 790, 847, 906, 967, 1030, 1095, 1162, 1231, 1302, 1375, 1450, 1527, 1606, 1687, 1770, 1855, 1942, 2031, 2122, 2215, 2310, 2407, 2506

Offset: 0

Cino Hilliard, Feb 21 2006

2/a(n) = R(n)/r, n >= 0, with R(n) the n-th radius of the counterclockwise Pappus chain of the arbelos with semicircle radii r, r1 = 2r/3, r2 = r - r1 = r/3. See the MathWorld link for such a Pappus chain. The clockwise chain companion has circle radii R'(n)/r = 2/A222465(n), n >= 0. - Wolfdieter Lang, Mar 01 2013

			The arbelos chain defined in a comment above has circle radii [1/3, 2/7, 1/5, 2/15, 1/11, 2/31, 1/21, 2/55, 1/35, 2/87, 1/53,...], for n >= 0. - _Wolfdieter Lang_, Mar 01 2013

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Pappus chain.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000290, A002522, A016742, A059100, A082044, A087475, A114964 (see comment), A117619, A117950, A117951, A222465.

A114949:=n->n^2+6: seq(A114949(n), n=0..100); # Wesley Ivan Hurt, Apr 28 2017

Range[0, 49]^2 + 6 (* Alonso del Arte, Jan 30 2013 *)

From R. J. Mathar, May 17 2009: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: -(6 - 11*x + 7*x^2)/(x - 1)^3. (End)
a(n) = 2*n + a(n - 1) - 1, with n > 0, a(0)=6. - Vincenzo Librandi, Nov 13 2010
a(n) = A000290(n) + 6. - Omar E. Pol, Mar 02 2013
a(n) = ((n-2)^3 + (n-1)^3 + n^3 + (n+1)^3 + (n+2)^3)/(5*n) for n>=1. - Bruno Berselli, May 12 2014
For n >= 1, a(n) = (A016742(n) + A082044(n) - 1)/A000290(n). - Bruce J. Nicholson, Apr 19 2017
From Amiram Eldar, Nov 02 2020: (Start)
Sum_{n>=0} 1/a(n) = (1 + sqrt(6)*Pi*coth(sqrt(6)*Pi))/12.
Sum_{n>=0} (-1)^n/a(n) = (1 + sqrt(6)*Pi*cosech(sqrt(6)*Pi))/12. (End)
From Amiram Eldar, Feb 05 2024: (Start)
Product_{n>=0} (1 - 1/a(n)) = sqrt(5/6)*sinh(sqrt(5)*Pi)/sinh(sqrt(6)*Pi).
Product_{n>=0} (1 + 1/a(n)) = sqrt(7/6)*sinh(sqrt(7)*Pi)/sinh(sqrt(6)*Pi). (End)
E.g.f.: exp(x)*(6 + x + x^2). - Elmo R. Oliveira, Jan 17 2025

cp's OEIS Frontend

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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A139098 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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A016814 a(n) = (4*n + 1)^2.

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A016802 a(n) = (4*n)^2.

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A022264 a(n) = n*(7*n - 1)/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.

A022264 a(n) = n(7n - 1)/2.