A125202 -id:A125202 - cp's OEIS frontend

A081353 Diagonal of square maze arrangement of natural numbers A081349.

3, 5, 13, 19, 31, 41, 57, 71, 91, 109, 133, 155, 183, 209, 241, 271, 307, 341, 381, 419, 463, 505, 553, 599, 651, 701, 757, 811, 871, 929, 993, 1055, 1123, 1189, 1261, 1331, 1407, 1481, 1561, 1639, 1723, 1805, 1893, 1979, 2071, 2161, 2257, 2351, 2451, 2549

Offset: 0

Paul Barry, Mar 19 2003

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

Cf. A081349, A081350, A081351, A081352.
Cf. A109613, A137932.
Bisections are in A054554, A125202.

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

A081353:=n->(n+1)*(n+2)+(-1)^n: seq(A081353(n), n=0..100); # Wesley Ivan Hurt, Aug 09 2015

Table[(n + 1) (n + 2) + (-1)^n, {n, 0, 70}] (* Wesley Ivan Hurt, Aug 09 2015 *)
LinearRecurrence[{2,0,-2,1},{3,5,13,19},50] (* Harvey P. Dale, Aug 02 2021 *)

a(n) = (n+1)*(n+2)+(-1)^n = 2*binomial(n+2,2)+(-1)^n.
G.f.: (3-x)*(1+x^2)/((1-x)^3*(1+x)). [Colin Barker, Sep 03 2012]
From Wesley Ivan Hurt, Aug 09 2015: (Start)
a(n) = 2*a(n-1)-2*a(n-3)+a(n-4), n>4.
a(n) = n^2+3n+3 if n is even, otherwise n^2+3n+1.
a(n) = A137932(n+3) - A109613(n+1). (End)

A185669 a(n) = 4n^2 + 3n + 2.

Original entry on oeis.org

2, 9, 24, 47, 78, 117, 164, 219, 282, 353, 432, 519, 614, 717, 828, 947, 1074, 1209, 1352, 1503, 1662, 1829, 2004, 2187, 2378, 2577, 2784, 2999, 3222, 3453, 3692, 3939, 4194, 4457, 4728, 5007, 5294, 5589, 5892, 6203, 6522, 6849, 7184, 7527, 7878, 8237, 8604, 8979, 9362, 9753, 10152, 10559, 10974, 11397, 11828

Offset: 0

Paul Curtz, Feb 09 2011

Natural numbers A000027 written clockwise as a square spiral:
.
  43--44--45--46--47--48--49
   |
  42  21--22--23--24--25--26
   |   |                   |
  41  20   7---8---9--10  27
   |   |   |           |   |
  40  19   6   1---2  11  28
   |   |   |       |   |   |
  39  18   5---4---3  12  29
   |   |               |   |
  38  17--16--15--14--13  30
   |                       |
  37--36--35--34--33--32--31
.
Walking in straight lines away from the center:
1,  2, 11, ... = A054552(n)     = 1 -3*n+4*n^2,
1,  8, 23, ... = A033951(n)     = 1 +3*n+4*n^2,
1,  3, 13, ... = A054554(n+1)   = 1 -2*n-4*n^2,
1,  7, 21, ... = A054559(n+1)   = 1 +2*n+4*n^2,
1,  4, 15, ... = A054556(n+1)   = 1   -n+4*n^2,
1,  6, 19, ... = A054567(n+1)   = 1   +n+4*n^2,
1,  5, 17, ... = A053755(n)     = 1     +4*n^2,
1,  9, 25, ... = A016754(n)     = 1 +4*n+4*n^2 = (1+2*n)^2,
2,  8, 22, ... = 2*A084849(n)   = 2 +2*n+4*n^2,
2, 12, 30, ... = A002939(n+1)   = 2 +6*n+4*n^2,
2,  9, 24, ... = a(n)           = 2 +3*n+4*n^2,
2, 10, 26, ... = A069894(n)     = 2 +4*n+4*n^2,
3, 11, 27, ... = A164897(n)     = 3 +4*n+4*n^2,
3, 12, 29, ... = A054552(n+1)+1 = 3 +5*n+4*n^2,
3, 14, 33, ... = A033991(n+1)   = 3 +7*n+4*n^2,
3, 15, 35, ... = A000466(n+1)   = 3 +8*n+4*n^2,
4, 14, 32, ... = 2*A130883(n+1) = 4 +6*n+4*n^2,
4, 16, 36, ... = A016742(n+1)   = 4 +8*n+4*n^2 = (2+2*n)^2,
5, 18, 39, ... = A007742(n+1)   = 5 +9*n+4*n^2,
5, 19, 41, ... = A125202(n+2)   = 5+10*n+4*n^2.

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

[2+3*n+4*n^2: n in [0..80]];  // Vincenzo Librandi, Feb 09 2011

Table[4n^2 + 3n + 2, {n,0,50}] (* G. C. Greubel, Jul 09 2017 *)
LinearRecurrence[{3,-3,1},{2,9,24},60] (* Harvey P. Dale, Aug 11 2021 *)

a(n)=4*n^2+3*n+2 \\ Charles R Greathouse IV, Apr 14 2014

a(n) = a(n-1) + 8*n - 1.
a(n) = 2*a(n-1) - a(n-2) + 8.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: (2 +3*x +3*x^2)/(1-x)^3 . - R. J. Mathar, Feb 11 2011
a(n) = A033954(n) + 2. - Bruno Berselli, Apr 10 2011
E.g.f.: (4*x^2 + 7*x + 2)*exp(x). - G. C. Greubel, Jul 09 2017

A271725 T(n,k) is an array read by rows, with n > 0 and k=1..4, where row n gives four prime numbers in increasing order with locations in right angles of each concentric square drawn on a distorted version of the Ulam spiral.

Original entry on oeis.org

3, 7, 17, 19, 13, 23, 37, 41, 307, 359, 401, 419, 13807, 14159, 14401, 14519, 41413, 42023, 42437, 42641, 6317683, 6325223, 6330257, 6332771, 22958473, 22972847, 22982437, 22987229, 39081253, 39100007, 39112517, 39118769, 110617807, 110649359, 110670401, 110680919

Offset: 1

Michel Lagneau, Apr 13 2016

nonn

See the illustration for more information.
Conjecture: there is an infinity of concentric squares having a prime number in each right angle. The number 5 is the center of all the squares.
It seems that the drawing of an infinite number of concentric squares having a prime number in each corner is impossible in an Ulam spiral. But with a slight distortion of this space, the problem becomes possible.
The illustration (see the link) shows the new version of a spiral with two remarkable orthogonal diagonals containing four classes of prime numbers given by the sequences A125202, A121326, A028871 and A073337 supported by four line segments. These intersect at a single point represented by the prime number 5.
The sequence of the corresponding length of the sides is {s(k)} = {2, 4, 18, 118, 204, 2514, 4792, 6252, 10518, 14032, 16752, 17598, ...}
The primes are defined by the polynomials: [4*m^2-10*m+7, (2*m-1)^2-2, 4*m^2+1, 4*(m+1)^2-6*(m+1)+1]. The sequence of the corresponding m is {b(k)} = {2, 3, 10, 60, 103, 1258, 2397, 3127, 5260, 7017, 8377, 8800, 10375, 11518, 11523, 12498, 15415, 15888, ...} with the relation b(k) = 1 + s(k)/2.
The array begins:
      3,     7,    17,    19;
     13,    23,    37,    41;
    307,   359,   401,   419;
  13807, 14159, 14401, 14519;
  41413, 42023, 42437, 42641;
  ...
Construction of the spiral (see the illustration in the link):
                .  .  .  .  .  .  .  .  .  .  .  .
               .  42  41  40  39  38  37   .  .  .
                                       |
               .  43  20  19  18  17  36  35  .  .
                                   |
               .  .   21   6   5  16  15  34  .  .
                               |
               .  .   22   7   4   3  14  33  .  .
               .  .   23   8   1   2  13  32  .  .
               .  .   24   9  10  11  12  31  .  .
               .  .   25  26  27  28  29  30  .  .
                   .  .  .  .  .  .  .  .  .  .  .
The first squares of center 5 having a prime number in each vertex are:
    19  18  17      41  40  39  38  37
     6   5  16      20  19  18  17  36
     7   4  3       21   6   5  16  15   . . . .
                    22   7   4   3  14
                    23   8   1   2  13

Michel Lagneau, Illustration

Cf. A028871, A073337, A121326, A125202, A200975.

for n from 1 to 10000 do :
  x1:=4*n^2-10*n+7:x2:=(2*n-1)^2-2:
  x3:=4*(n+1)^2-6*(n+1)+1:x4:=4*n^2+1:
   if isprime(x1) and isprime(x2) and isprime(x3) and isprime(x4)
    then
     printf("%d %d %d %d %d \n",n,x1,x2,x4,x3):
    else
    fi:
od:

A329451 Maximum number of pieces that can be captured during one move on an n X n board according to the international draughts capture rules.

Original entry on oeis.org

0, 0, 0, 1, 1, 4, 5, 9, 10, 16, 19, 25, 28, 36, 41, 49, 54, 64, 71, 81, 88, 100, 109, 121, 130, 144, 155, 169, 180, 196, 209, 225, 238, 256, 271, 289, 304, 324, 341, 361, 378, 400, 419, 441, 460, 484, 505, 529, 550, 576, 599, 625, 648, 676, 701, 729, 754, 784

Offset: 0

Stéphane Rézel, Nov 14 2019

nonn

Captures are made diagonally, forward and backward. Kings have the long-range capturing capability. During the multiple capture, a piece may pass over the same empty square several times, but no opposing piece can be jumped twice. Captured pieces can only be lifted from the board after the end of the multiple capture.

			It is possible to capture in a single move 19 opposing pieces on a 10 X 10 board, but not one more, so a(10) = 19.

Stéphane Rézel, Table of n, a(n) for n = 0..1000
Fabien Gigante, Solution to Problem J122 La grande rafle de la dame, Diophante, 2009 (in French).
Stéphane Rézel, Solution to Problem J128 La méga-rafle de la dame, Diophante, 2020 (in French).
World Draughts Federation, Official rules for international draughts.
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,1,-2,1).

Cf. A000290, A059193, A125202, A000982 (active squares).

a(n) = if(n<5, floor(n/3), (n^2 - 2*n + if(n%2, 1, 2*(n%4) - 8))/4)

a(2*t+1) = t^2 = A000290(t).
a(4*t+6) = 4*t^2 + 10*t + 5 = A125202(t+2).
a(4*t+8) = 4*t^2 + 14*t + 10 = A059193(t+2).
a(0) = a(2) = 0; a(4) = 1.
Recurrence: For t >= 1, a(2*t+1) = a(2*t-1) + 2*t - 1;
For t >= 1, a(4*t+3) = a(4*t+2) + 2*t + 2; a(4*t+2) = a(4*t+1) + 2*t - 1;
For t >= 2, a(4*t+1) = a(4*t) + 2*t + 2; a(4*t) = a(4*t-1) + 2*t - 3.
From Colin Barker, Nov 14 2019: (Start)
G.f.: x^3*(1 - x + 3*x^2 - 2*x^3 + 2*x^4 - 2*x^5 + 2*x^6 - x^7) / ((1 - x)^3*(1 + x)*(1 + x^2)).
a(n) = 2*a(n-1) - a(n-2) + a(n-4) - 2*a(n-5) + a(n-6) for n>10.
(End)

cp's OEIS Frontend

A081353 Diagonal of square maze arrangement of natural numbers A081349.

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A185669 a(n) = 4*n^2 + 3*n + 2.

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Magma

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A271725 T(n,k) is an array read by rows, with n > 0 and k=1..4, where row n gives four prime numbers in increasing order with locations in right angles of each concentric square drawn on a distorted version of the Ulam spiral.

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Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

A329451 Maximum number of pieces that can be captured during one move on an n X n board according to the international draughts capture rules.

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Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A185669 a(n) = 4n^2 + 3n + 2.