cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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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

Views

Author

Paul Barry, Mar 19 2003

Keywords

Comments

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

Crossrefs

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.

Programs

Formula

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

Views

Author

William A. Tedeschi, Feb 29 2008

Keywords

Comments

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

Examples

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

Crossrefs

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.

Programs

Formula

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

Views

Author

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

Keywords

Comments

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

Crossrefs

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.

Programs

Formula

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

Extensions

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

Views

Author

N. J. A. Sloane, Jul 27 2018

Keywords

Comments

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)

Examples

			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)
		

Crossrefs

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.

Programs

  • Mathematica
    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 *)

Formula

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

A267682 a(n) = 2*a(n-1) - 2*a(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

Views

Author

Robert Price, Jan 19 2016

Keywords

Comments

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

References

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

Crossrefs

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.

Programs

  • Mathematica
    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 *)
  • PARI
    Vec((1-x+2*x^2+2*x^3)/((1-x)^3*(1+x)) + O(x^100)) \\ Colin Barker, Jan 19 2016
    
  • Python
    print([n*(n-1)+n//2+1 for n in range(51)]) # Karl V. Keller, Jr., Jul 14 2021

Formula

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

Extensions

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

A054555 Prime number spiral (clockwise, East spoke).

Original entry on oeis.org

2, 7, 47, 139, 283, 503, 797, 1151, 1579, 2089, 2689, 3361, 4099, 4967, 5861, 6883, 8011, 9199, 10457, 11903, 13313, 14887, 16547, 18269, 20161, 22091, 24083, 26297, 28573, 30941, 33347, 35899, 38593, 41299, 44111, 47149, 50131, 53327, 56597
Offset: 0

Views

Author

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

Keywords

Comments

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

Examples

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

Crossrefs

Programs

  • Mathematica
    Table[ Prime[4n^2 - 9n + 6], {n, 1, 40} ]

Formula

a(n) = A000040(A054556(n+1)). - R. J. Mathar, Aug 29 2018

Extensions

Edited by Robert G. Wilson v, Feb 25 2002

A168023 Noncomposite numbers in the northern ray of the Ulam spiral as oriented on the March 1964 cover of Scientific American.

Original entry on oeis.org

1, 61, 139, 1009, 1279, 2281, 3109, 3571, 4591, 6361, 8419, 13399, 14341, 17359, 19531, 23029, 35251, 39901, 44839, 46549, 51871, 55579, 61381, 73849, 76039, 102241, 110059, 135241, 153469, 156619
Offset: 1

Views

Author

Alonso del Arte, Nov 16 2009

Keywords

Crossrefs

Cf. A054556, all numbers of the form 4n^2 - 9n + 6. Noncomposites of the eastern ray are in A168022. Primes of the northeastern ray are in A073337. Noncomposites of the northwestern ray are in A168024. Noncomposites of the western ray are in A168025. Noncomposites of the southwestern ray are in A168026. Noncomposites of the southern ray are in A168027.

Programs

  • Mathematica
    Select[Table[4 n^2 - 9 n + 6, {n, 200}], Length[Divisors[ # ]] < 3 &]

Formula

Positive numbers of the form 4n^2 - 9n + 6 with no more than two divisors.

A266883 Numbers of the form m*(4*m+1)+1, where m = 0,-1,1,-2,2,-3,3,...

Original entry on oeis.org

1, 4, 6, 15, 19, 34, 40, 61, 69, 96, 106, 139, 151, 190, 204, 249, 265, 316, 334, 391, 411, 474, 496, 565, 589, 664, 690, 771, 799, 886, 916, 1009, 1041, 1140, 1174, 1279, 1315, 1426, 1464, 1581, 1621, 1744, 1786, 1915, 1959, 2094, 2140, 2281, 2329, 2476, 2526
Offset: 0

Views

Author

Bruno Berselli, Jan 05 2016

Keywords

Comments

Also, numbers m such that 16*m-15 is a square. Therefore, the terms 1 and 4 are the only squares in this sequence.
Conjecture: the sequence terms are the exponents in the expansion of Sum_{n >= 1} q^n * (Product_{k >= 2*n-1} 1 - q^k) = q + q^4 + q^6 + q^15 + q^19 + q^34 + .... Cf. A174114. - Peter Bala, May 10 2025

Crossrefs

Cf. A002061: m*(4*m+2)+1 for m = 0,0,-1,1,-2,2,-3,3, ...
Cf. A174114: m*(4*m+3)+1 for m = 0,-1,1,-2,2,-3,3,-4,4, ...
Cf. A054556: m*(4*m+1)+1 for nonpositive m.
Cf. A054567: m*(4*m+1)+1 for nonnegative m.
Cf. A074378: numbers m such that 16*m+1 is a square.

Programs

  • Magma
    [n*(n+1)+1-((2*n+1)*(-1)^n-1)/4: n in [0..50]];
    
  • Magma
    I:=[1,4,6,15,19]; [n le 5 select I[n] else Self(n-1) + 2*Self(n-2) -2*Self(n-3)-Self(n-4)+Self(n-5): n in [1..60]]; // Vincenzo Librandi, Jan 06 2016
  • Mathematica
    Table[n (n + 1) + 1 - ((2 n + 1) (-1)^n - 1)/4, {n, 0, 50}]
    LinearRecurrence[{1, 2, -2, -1, 1}, {1, 4, 6, 15, 19}, 60] (* Vincenzo Librandi, Jan 06 2016 *)
  • PARI
    vector(50, n, n--; n*(n+1)+1-((2*n+1)*(-1)^n-1)/4)
    
  • PARI
    Vec((1+3*x+3*x^3+x^4)/((1+x)^2*(1-x)^3) + O(x^100)) \\ Altug Alkan, Jan 06 2016
    
  • Python
    [n*(n+1)+1-((2*n+1)*(-1)**n-1)/4 for n in range(60)]
    
  • Sage
    [n*(n+1)+1-((2*n+1)*(-1)^n-1)/4 for n in range(50)]
    

Formula

O.g.f.: (1 + 3*x + 3*x^3 + x^4)/((1 + x)^2*(1 - x)^3).
E.g.f.: (5 + 8*x + 4*x^2)*exp(x)/4 -(1 - 2*x)*exp(-x)/4.
a(n) = a(-n-1) = n*(n + 1) + 1 - ((2*n + 1)*(-1)^n - 1)/4 = (2*n + 1)*floor((n + 1)/2) + 1.
a(n) = A002061(n+1) + A001057(n) = A074378(n)+1.
a(n+1) + a(n+2) = A049486(n+3).

A365673 Array A(n, k) read by ascending antidiagonals. Polygonal number weighted generalized Catalan sequences.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 4, 1, 1, 1, 4, 15, 8, 1, 1, 1, 5, 34, 105, 16, 1, 1, 1, 6, 61, 496, 945, 32, 1, 1, 1, 7, 96, 1385, 11056, 10395, 64, 1, 1, 1, 8, 139, 2976, 50521, 349504, 135135, 128, 1, 1, 1, 9, 190, 5473, 151416, 2702765, 14873104, 2027025, 256, 1
Offset: 0

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Author

Peter Luschny, Sep 30 2023

Keywords

Comments

Using polygonal numbers as weights, a recursion for triangles is defined, whose main diagonals represents a family of sequences, which include, among others, the powers of 2, the double factorial of odd numbers, the reduced tangent numbers, and the Euler numbers.
Apart from the edge cases k = 0 and k = n the recursion is T(n, k) = w(n, k) * T(n, k - 1) + T(n - 1, k). T(n, 0) = 1 and T(n, n) = T(n, n-1) if n > 0.
The weights w(n, k) identical to 1 yield the recursion of the Catalan triangle A009766 (with main diagonal the Catalan numbers). Here the polygonal numbers are used as weights in the form w(n, k) = p(s, n - k + 1), where the parameter s is the number of sides of the polygon and p(s, n) = ((s-2) * n^2 - (s-4) * n) / 2, see A317302.

Examples

			Array A(n, k) starts:                            (polygon|diagonal|triangle)
[0] 1, 1, 1,   1,     1,       1,         1, ...  A258837  A000012
[1] 1, 1, 2,   4,     8,      16,        32, ...  A080956  A011782
[2] 1, 1, 3,  15,   105,     945,     10395, ...  A001477  A001147  A001498
[3] 1, 1, 4,  34,   496,   11056,    349504, ...  A000217  A002105  A365674
[4] 1, 1, 5,  61,  1385,   50521,   2702765, ...  A000290  A000364  A060058
[5] 1, 1, 6,  96,  2976,  151416,  11449296, ...  A000326  A126151  A366138
[6] 1, 1, 7, 139,  5473,  357721,  34988647, ...  A000384  A126156  A365672
[7] 1, 1, 8, 190,  9080,  725320,  87067520, ...  A000566  A366150  A366149
[8] 1, 1, 9, 249, 14001, 1322001, 188106489, ...  A000567
           A054556                         A366137
		

Crossrefs

Cf. A009766, A366137 (central diagonal), A317302 (table of polygonal numbers).

Programs

  • Maple
    poly := (s, n) -> ((s - 2) * n^2 - (s - 4) * n) / 2:
    T := proc(s, n, k) option remember; if k = 0 then 1 else if k = n then T(s, n, k-1) else poly(s, n - k + 1) * T(s, n, k - 1) + T(s, n - 1, k) fi fi end:
    for n from 0 to 8 do A := (n, k) -> T(n, k, k): seq(A(n, k), k = 0..9) od;
    # Alternative, using continued fractions:
    A := proc(p, L) local CF, poly, k, m, P, ser;
       poly := (s, n) -> ((s - 2)*n^2 - (s - 4)*n)/2;
       CF := 1 + x;
       for k from 1 to L do
           m := L - k + 1;
           P := poly(p, m);
           CF := 1/(1 - P*x*CF)
       od;
       ser := series(CF, x, L);
       seq(coeff(ser, x, m), m = 0..L-1)
    end:
    for p from 0 to 8 do lprint(A(p, 8)) od;
  • Mathematica
    poly[s_, n_] := ((s - 2) * n^2 - (s - 4) * n) / 2;
    T[s_, n_, k_] := T[s, n, k] = If[k == 0, 1, If[k == n, T[s, n, k - 1], poly[s, n - k + 1] * T[s, n, k - 1] + T[s, n - 1, k]]];
    A[n_, k_] := T[n, k, k];
    Table[A[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 27 2023, from first Maple program *)
  • PARI
    A(p, n) = {
           my(CF = 1 + x,
               poly(s, n) = ((s - 2)*n^2 - (s - 4)*n)/2,
               m, P
           );
           for(k = 1, n,
               m = n - k + 1;
               P = poly(p, m);
               CF = 1/(1 - P*x*CF)
            );
            Vec(CF + O(x^(n)))
    }
    for(p = 0, 8, print(A(p, 8)))
    \\  Michel Marcus and Peter Luschny, Oct 02 2023
  • Python
    from functools import cache
    @cache
    def T(s, n, k):
        if k == 0: return 1
        if k == n: return T(s, n, k - 1)
        p = (n - k + 1) * ((s - 2) * (n - k + 1) - (s - 4)) // 2
        return p * T(s, n, k - 1) + T(s, n - 1, k)
    def A(n, k): return T(n, k, k)
    for n in range(9): print([A(n, k) for k in range(9)])
    

A143861 Ulam's spiral (NNE spoke).

Original entry on oeis.org

1, 14, 59, 136, 245, 386, 559, 764, 1001, 1270, 1571, 1904, 2269, 2666, 3095, 3556, 4049, 4574, 5131, 5720, 6341, 6994, 7679, 8396, 9145, 9926, 10739, 11584, 12461, 13370, 14311, 15284, 16289, 17326, 18395, 19496, 20629, 21794, 22991, 24220
Offset: 1

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Author

Keywords

Comments

Stanislaw M. Ulam was doodling during the presentation of a "long and very boring paper" at a scientific meeting in 1963. The spiral is its result. Note that conforming to trigonometric conventions, the spiral begins on the abscissa and rotates counterclockwise. Other spirals, orientations, direction of rotation and initial values exist, even in the OEIS.
Also sequence found by reading the segment (1, 14) together with the line from 14, in the direction 14, 59, ..., in the square spiral whose vertices are the generalized decagonal numbers A074377. - Omar E. Pol, Nov 05 2012

References

  • Chris K. Caldwell & G. L. Honaker, Jr., Prime Curios! The Dictionary of Prime Number Trivia, CreateSpace, Sept 2009, pp. 2-3.

Crossrefs

Programs

  • GAP
    List([1..40], n-> ((32*n-35)^2 +55)/64); # G. C. Greubel, Nov 09 2019
  • Magma
    [((32*n-35)^2 +55)/64: n in [1..40]]; // G. C. Greubel, Nov 09 2019
    
  • Maple
    seq( ((32*n-35)^2 +55)/64, n=1..40); # G. C. Greubel, Nov 09 2019
  • Mathematica
    (* From Robert G. Wilson v, Oct 29 2011 *)
    f[n_]:= 16n^2 -35n +20; Array[f, 40]
    LinearRecurrence[{3,-3,1}, {1,14,59}, 40]
    FoldList[#1 + #2 &, 1, 32Range@ 10 - 19] (* End *)
    ((32*Range[40] -35)^2 +55)/64 (* G. C. Greubel, Nov 09 2019 *)
  • PARI
    a(n)=16*n^2-35*n+20 \\ Charles R Greathouse IV, Oct 29 2011
    
  • Sage
    [((32*n-35)^2 +55)/64 for n in (1..40)] # G. C. Greubel, Nov 09 2019
    

Formula

a(n) = 16*n^2 - 35*n + 20. - R. J. Mathar, Sep 08 2008
G.f.: x*(1 + 11*x + 20*x^2)/(1-x)^3. - Colin Barker, Aug 03 2012
E.g.f.: -20 + (20 - 19*x + 16*x^2)*exp(x). - G. C. Greubel, Nov 09 2019
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