A002943 -id:A002943 - cp's OEIS frontend

A243424 Triangle T(n,k) read by rows of number of ways k domicules can be placed on an n X n square (n >= 0, 0 <= k <= floor(n^2/2)).

1, 1, 1, 6, 3, 1, 20, 110, 180, 58, 1, 42, 657, 4890, 18343, 33792, 27380, 7416, 280, 1, 72, 2172, 36028, 362643, 2307376, 9382388, 24121696, 37965171, 34431880, 16172160, 3219364, 170985, 1, 110, 5375, 154434, 2911226, 38049764, 355340561, 2408715568

Offset: 0

Alois P. Heinz, Jun 04 2014

A domicule is either a domino or it is formed by the union of two neighboring unit squares connected via their corners. In a tiling the connections of two domicules are allowed to cross each other.

The n-th row gives the coefficients of the matching-generating polynomial of the n X n king graph. - Eric W. Weisstein, Jun 20 2017

			T(2,1) = 6:
  +---+  +---+  +---+  +---+  +---+  +---+
  |o-o|  |   |  |o  |  |  o|  |o  |  |  o|
  |   |  |   |  ||  |  |  ||  | \ |  | / |
  |   |  |o-o|  |o  |  |  o|  |  o|  |o  |
  +---+  +---+  +---+  +---+  +---+  +---+
T(2,2) = 3:
  +---+  +---+  +---+
  |o-o|  |o o|  |o o|
  |   |  || ||  | X |
  |o-o|  |o o|  |o o|
  +---+  +---+  +---+
Triangle T(n,k) begins:
  1;
  1;
  1,  6,    3;
  1, 20,  110,   180,     58;
  1, 42,  657,  4890,  18343,   33792,   27380,     7416,      280;
  1, 72, 2172, 36028, 362643, 2307376, 9382388, 24121696, 37965171, ...
  ...

Alois P. Heinz, Rows n = 0..13, flattened
Eric Weisstein's World of Mathematics, King Graph
Eric Weisstein's World of Mathematics, Matching-Generating Polynomial

Columns k=0-5 give: A000012, A002943(n-1) for n>0, A243464, A243465, A243466, A243467.
Row sums give A220638.
T(n,floor(n^2/2)) gives A243510.
T(n,floor(n^2/4)) gives A243511.
Cf. A242861 (the same for dominoes), A239264.

b:= proc(n, l) option remember; local d, f, k;
      d:= nops(l)/2; f:=false;
      if n=0 then 1
    elif l[1..d]=[f$d] then b(n-1, [l[d+1..2*d][], true$d])
    else for k to d while not l[k] do od;
         expand(b(n, subsop(k=f, l))+
         `if`(k1 and l[k+d+1],
                            x*b(n, subsop(k=f, k+d+1=f, l)), 0)+
         `if`(k>1 and n>1 and l[k+d-1],
                            x*b(n, subsop(k=f, k+d-1=f, l)), 0)+
         `if`(n>1 and l[k+d], x*b(n, subsop(k=f, k+d=f, l)), 0)+
         `if`(k (p-> seq(coeff(p,x,i), i=0..degree(p)))(b(n, [true$(n*2)])):
seq(T(n), n=0..7);

b[n_, l_] := b[n, l] = Module[{d, f, k}, d = Length[l]/2; f = False; Which[ n == 0, 1, l[[1 ;; d]] == Table[f, d], b[n-1, Join[l[[d+1 ;; 2d]], Table[ True, d]]], True, For[k = 1, !l[[k]], k++]; Expand[b[n, ReplacePart[l, k -> f]] + If[k1 && l[[k+d+1]], x*b[n, ReplacePart[l, {k -> f, k + d + 1 -> f}]], 0] + If[k>1 && n>1 && l[[k + d - 1]], x*b[n, ReplacePart[ l, {k -> f, k + d - 1 -> f}]], 0] + If[n>1 && l[[k + d]], x*b[n, ReplacePart[l, {k -> f, k+d -> f}]], 0] + If[k f, k+1 -> f}]], 0]]]];
T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][
  b[n, Table[True, 2n]]];
Table[T[n], {n, 0, 7}] // Flatten (* Jean-François Alcover, Jun 09 2018, after Alois P. Heinz *)

A187586 T(n,k)=Number of n-step E, S, NW and NE-moving king's tours on a kXk board summed over all starting positions.

Original entry on oeis.org

1, 4, 0, 9, 6, 0, 16, 20, 8, 0, 25, 42, 48, 5, 0, 36, 72, 120, 84, 0, 0, 49, 110, 224, 286, 106, 0, 0, 64, 156, 360, 604, 578, 104, 0, 0, 81, 210, 528, 1038, 1484, 1069, 78, 0, 0, 100, 272, 728, 1588, 2794, 3514, 1708, 34, 0, 0, 121, 342, 960, 2254, 4508, 7480, 7666, 2309, 13

Offset: 1

R. H. Hardin Mar 11 2011

Table starts
.1.4...9...16....25.....36.....49......64......81.....100.....121.....144
.0.6..20...42....72....110....156.....210.....272.....342.....420.....506
.0.8..48..120...224....360....528.....728.....960....1224....1520....1848
.0.5..84..286...604...1038...1588....2254....3036....3934....4948....6078
.0.0.106..578..1484...2794...4508....6626....9148...12074...15404...19138
.0.0.104.1069..3514...7480..12874...19696...27946...37624...48730...61264
.0.0..78.1708..7666..19104..35832...57592...84384..116208..153064..194952
.0.0..34.2309.15056..45718..95776..164135..250132..353767..475040..613951
.0.0..13.2792.27252.103108.246792..458018..732810.1069534.1468190.1928778
.0.0...0.3108.45960.219432.609070.1243461.2111652.3201436.4508924

			Some k=4 solutions for 4X4
..0..0..0..0....3..0..0..0....0..0..0..0....0..4..0..0....0..0..0..3
..0..0..0..0....4..2..0..0....4..2..0..0....3..0..0..0....0..0..2..4
..0..1..2..0....1..0..0..0....0..3..1..0....0..2..0..0....0..0..0..1
..0..0..3..4....0..0..0..0....0..0..0..0....1..0..0..0....0..0..0..0

R. H. Hardin, Table of n, a(n) for n = 1..200

Row 2 is A002943(n-1)

Row 3 is A152750(n-1)

Empirical: T(1,k) = k^2
Empirical: T(2,k) = 4*k^2 - 6*k + 2
Empirical: T(3,k) = 16*k^2 - 40*k + 24
Empirical: T(4,k) = 58*k^2 - 204*k + 174 for k>2
Empirical: T(5,k) = 202*k^2 - 912*k + 994 for k>3
Empirical: T(6,k) = 714*k^2 - 3888*k + 5104 for k>4
Empirical: T(7,k) = 2516*k^2 - 15980*k + 24408 for k>5
Empirical: T(8,k) = 8819*k^2 - 63926*k + 111127 for k>6
Empirical: T(9,k) = 30966*k^2 - 251630*k + 489234 for k>7
Empirical: T(10,k) = 108852*k^2 - 978404*k + 2100276 for k>8

A068377 Engel expansion of sinh(1).

Original entry on oeis.org

1, 6, 20, 42, 72, 110, 156, 210, 272, 342, 420, 506, 600, 702, 812, 930, 1056, 1190, 1332, 1482, 1640, 1806, 1980, 2162, 2352, 2550, 2756, 2970, 3192, 3422, 3660, 3906, 4160, 4422, 4692, 4970, 5256, 5550, 5852, 6162, 6480, 6806, 7140, 7482, 7832, 8190

Offset: 1

Benoit Cloitre, Mar 03 2002

This sequence is also the Pierce expansion of sin(1). - G. C. Greubel, Nov 14 2016

Simon Plouffe, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Engel Expansion.
Eric Weisstein's World of Mathematics, Hyperbolic Sine.
Eric Weisstein's World of Mathematics, Pierce Expansion.
Wikipedia, Engel Expansion.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A002943, A006784, A068379, A068380, A073742 (sinh(1)).

Join[{1}, Table[(2 n - 2) (2 n - 1), {n, 2, 50}]] (* Bruno Berselli, Aug 04 2015 *)
LinearRecurrence[{3,-3,1}, {1,6,20,42}, 25] (* G. C. Greubel, Oct 27 2016; a(1)=1 by Georg Fischer, Apr 02 2019*)
Rest@ CoefficientList[Series[x (1 + 3 x + 5 x^2 - x^3)/(1 - x)^3, {x, 0, 46}], x] (* Michael De Vlieger, Oct 28 2016 *)
PierceExp[A_, n_] := Join[Array[1 &, Floor[A]], First@Transpose@ NestList[{Floor[1/Expand[1 - #[[1]] #[[2]]]], Expand[1 - #[[1]] #[[2]]]} &, {Floor[1/(A - Floor[A])], A - Floor[A]}, n - 1]]; PierceExp[N[Sin[1] , 7!], 50] (* G. C. Greubel, Nov 14 2016 *)

A068377(n)=(n+n--)*n*2+!n \\ M. F. Hasler, Jul 19 2015

A068377 = lambda n: rising_factorial(n*2,2) if n>0 else 1
print([A068377(n) for n in (0..45)]) # Peter Luschny, Aug 04 2015

a(n) = (2*n-2)*(2*n-1) = A002943(n-1) = 2*A000217(2n-2) for n>1. [Corrected and extended by M. F. Hasler, Jul 19 2015]
From Colin Barker, Apr 13 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>4.
G.f.: x*(1 + 3*x + 5*x^2 - x^3)/(1-x)^3. (End)
E.g.f.: -2 + x + 2*(1 - x + 2*x^2)*exp(x). - G. C. Greubel, Oct 27 2016
From Amiram Eldar, May 05 2025: (Start)
Sum_{n>=1} 1/a(n) = 2 - log(2).
Sum_{n>=1} (-1)^(n+1)/a(n) = 2 - Pi/4 - log(2)/2. (End)

A173511 a(n) = 4*n^2 + floor(n/2).

Original entry on oeis.org

0, 4, 17, 37, 66, 102, 147, 199, 260, 328, 405, 489, 582, 682, 791, 907, 1032, 1164, 1305, 1453, 1610, 1774, 1947, 2127, 2316, 2512, 2717, 2929, 3150, 3378, 3615, 3859, 4112, 4372, 4641, 4917, 5202, 5494, 5795, 6103, 6420, 6744, 7077, 7417, 7766, 8122

Offset: 0

Reinhard Zumkeller, Feb 20 2010

			a(6) = 147; 4(6)^2 + floor(6/3) = 144 + 3 = 147.

Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A002943, A007494, A007742, A110654, A157474, A173512.

A173511:=n->4*n^2 + floor(n/2); seq(A173511(k), k=0..100); # Wesley Ivan Hurt, Nov 01 2013

Table[4n^2 + Floor[n/2], {n,0,100}] (* Wesley Ivan Hurt, Nov 01 2013 *)
LinearRecurrence[{2,0,-2,1},{0,4,17,37},50] (* Harvey P. Dale, Nov 23 2019 *)

a(n) = 4*n^2 + n\2 \\ Charles R Greathouse IV, Jun 11 2015

def A173511(n): return (n**2<<2)+(n>>1) # Chai Wah Wu, Jan 18 2023

a(n) = floor((2*n + 1/8)^2).
a(n+1) - a(n) = A173512(n).
a(n) = A002943(n) - A007494(n) = A007742(n) - A110654(n).
a(2*n) = A157474(n) for n>0.
From - R. J. Mathar, Feb 21 2010: (Start)
a(n)= 2*a(n-1) -2*a(n-3) +a(n-4).
G.f.: -x*(4+9*x+3*x^2)/((1+x)*(x-1)^3). (End)
E.g.f.: (x*(8*x + 9)*cosh(x) + (8*x^2 + 9*x - 1)*sinh(x))/2. - Stefano Spezia, Apr 24 2024

A113689 Number of semiprimes in clumps of size > 1 through n^2 in the semiprime spiral.

Original entry on oeis.org

0, 0, 2, 6, 9, 13, 17, 21, 23, 31, 37, 45, 54, 59, 72, 77, 83, 93, 104, 116, 125, 140, 150, 164, 180, 188, 203, 219, 236, 255, 272, 287, 301, 317, 334, 354, 378, 403, 419, 430, 450, 475, 498, 521, 542, 560, 588, 608, 626, 652, 677, 698

Offset: 1

Jonathan Vos Post, Nov 05 2005

Write the integers 1, 2, 3, 4, ... in a counterclockwise square spiral. Analogous to Ulam coloring in the primes in the spiral and discovering unexpectedly many connected diagonals, we construct a semiprime spiral by coloring in all semiprimes (A001358). Each integer has 8 adjacent integers in the spiral, horizontally, vertically and diagonally. Curious extended clumps coagulate, slightly denser towards the origin, of semiprimes connected by adjacency. This sequence, A113689, gives an enumeration of the number of semiprimes in clumps of size > 1 through n^2, not looking past the square boundary. A113688 gives isolated semiprimes in the semiprime spiral, namely those semiprimes none of whose adjacent integers in the spiral are semiprimes.

			a(3) = 2 because there is one visible clump through 3^2 = 9, {4,6}, which two semiprimes are diagonally connected.
a(4) = 6 because there are 6 semiprimes in the 2 visible clumps through 4^2 = 16, {4, 6, 14, 15}, {9, 10}.
a(5) = 9 because there are 9 semiprimes in the 3 visible clumps through 5^2 = 25, {4, 6, 14, 15}, {9, 10, 25}, {21, 22}.
......................
... 17 16 15 14 13 ...
... 18  5  4  3 12 ...
... 19  6  1  2 11 ...
... 20  7  8  9 10 ...
... 21 22 23 24 25 ...
......................

S. M. Ellerstein, The square spiral, J. Recreational Mathematics 29 (#3, 1998) 188; 30 (#4, 1999-2000), 246-250.

M. Stein and S. M. Ulam, An Observation on the Distribution of Primes, Amer. Math. Monthly 74, 43-44, 1967.
M. Stein, S. M. Ulam and M. B. Wells, A Visual Display of Some Properties of the Distribution of Primes, Amer. Math. Monthly 71, 516-520, 1964.
Eric Weisstein's World of Mathematics, Prime Spiral.
Eric Weisstein's World of Mathematics, Semiprime

Cf. A001107, A001358, A002939, A002943, A004526, A005620, A007742, A033951-A033954, A033988, A033989-A033991, A033996, A063826, A113688.

Corrected and extended by Alois P. Heinz, Jan 02 2011

A181995 a(n) = if n mod 2 = 1 then n*(n - 1) else (n - 1)^2 + (n - 2)/2.

Original entry on oeis.org

0, 0, 1, 6, 10, 20, 27, 42, 52, 72, 85, 110, 126, 156, 175, 210, 232, 272, 297, 342, 370, 420, 451, 506, 540, 600, 637, 702, 742, 812, 855, 930, 976, 1056, 1105, 1190, 1242, 1332, 1387, 1482, 1540, 1640, 1701, 1806, 1870, 1980, 2047, 2162, 2232, 2352, 2425, 2550, 2626, 2756, 2835, 2970, 3052, 3192, 3277, 3422, 3510, 3660, 3751, 3906, 4000, 4160, 4257, 4422

Offset: 0

N. J. A. Sloane, Apr 05 2012

Decagonal numbers (A001107) and twice second hexagonal numbers (A002943) interleaved. - Omar E. Pol, Aug 03 2012
Similar to A074377. Members of this family are A093005, A210977, A006578, A210978, this sequence, A210981, A210982. - Omar E. Pol, Aug 09 2012
Number of kites whose vertices are the vertices a regular 2n-gon. - Halil Ibrahim Kanpak, Nov 08 2018

H. L. Abbott, D. Hanson, N. Sauer, Intersection theorems for systems of sets, J. Combinatorial Theory Ser. A 12 (1972), 381--389.MR0297579 (45 #6633).
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A001107, A002943, A006578, A074377, A093005.

Cf. A210977, A210978, A210981, A210982.

[n*(4*n - 5 - (-1)^n)/4 : n in [0..80]]; // Wesley Ivan Hurt, Apr 11 2016

f:=n->if n mod 2 = 1 then n*(n-1) else (n-1)^2+(n-2)/2; fi;
[seq(f(n),n=0..130)];

Table[n*(4*n - 5 - (-1)^n)/4, {n, 0, 80}] (* Wesley Ivan Hurt, Apr 11 2016 *)

a(n)=n*(4*n-5-(-1)^n)/4 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: -x^2*(1 + 5*x + 2*x^2)/((1 + x)^2*(x - 1)^3). - R. J. Mathar, Apr 06 2012
a(n) = n*(4*n - 5 - (-1)^n)/4. - Luce ETIENNE, Oct 04 2014
From Wesley Ivan Hurt, Apr 11 2016: (Start)
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n>4.
a(n) = Sum_{i=floor((n-1)/2)..floor(3*(n-1)/2)} i. (End)
E.g.f.: x^2*cosh(x) - x*(1 - 2*x)*sinh(x)/2. - Franck Maminirina Ramaharo, Nov 08 2018

A133280 Triangle formed by: 1 even, 2 odd, 3 even, 4 odd, ... starting with zero.

Original entry on oeis.org

0, 1, 3, 4, 6, 8, 9, 11, 13, 15, 16, 18, 20, 22, 24, 25, 27, 29, 31, 33, 35, 36, 38, 40, 42, 44, 46, 48, 49, 51, 53, 55, 57, 59, 61, 63, 64, 66, 68, 70, 72, 74, 76, 78, 80, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 100, 102, 104, 106, 108, 110, 112, 114, 116, 118, 120

Offset: 0

Omar E. Pol, Aug 27 2008

This sequence is related to the Connell sequence (A001614).
First member of every row is a square (A000290).
A127366(T(n,k)) mod 2 = 0 or equal parity of T(n,k) and A000196(T(n,k)); complement of A195437. - Reinhard Zumkeller, Oct 12 2011
Written as a square array the main diagonal gives A002943. - Omar E. Pol, Aug 13 2013
Last member of every row is one less than a square (A005563). - Harvey P. Dale, Oct 02 2013

			Written as a triangle the sequence begins:
    0;
    1,   3;
    4,   6,   8;
    9,  11,  13,  15;
   16,  18,  20,  22,  24;
   25,  27,  29,  31,  33,  35;
   36,  38,  40,  42,  44,  46,  48;
   49,  51,  53,  55,  57,  59,  61,  63;
   64,  66,  68,  70,  72,  74,  76,  78,  80;
   81,  83,  85,  87,  89,  91,  93,  95,  97,  99;
  100, 102, 104, 106, 108, 110, 112, 114, 116, 118, 120;

Reinhard Zumkeller, Rows n=0..100 of triangle, flattened

Column 1 is A000290. Right border gives A005563.
Cf. A001614.
Cf. A045991 (row sums). - R. J. Mathar, Jul 20 2009

a133280 n k = a133280_tabl !! n !! k
a133280_tabl = f 0 1 [0..] where
   f m j xs = (filter ((== m) . (`mod` 2)) ys) : f (1 - m) (j + 2) xs'
     where (ys,xs') = splitAt j xs
b133280 = bFile' "A133280" (concat $ take 101 a133280_tabl) 0
-- Reinhard Zumkeller, Oct 12 2011

Flatten[Table[Range[(n-1)^2,n^2-1,2],{n,20}]] (* Harvey P. Dale, Oct 02 2013 *)

T(n,k) = n^2 + 2*k;
for(n=0,10,for(k=0,n,print1(T(n,k),", "))); \\ Joerg Arndt, Aug 13 2013

from math import isqrt
def A133280(n): return (m:=(n<<1)+1)-((isqrt(m+1<<2)+1)>>1) # Chai Wah Wu, Aug 01 2022

a(n) = A005408(n) - A002024(n+1). - Ivan N. Ianakiev, Aug 13 2013

T(n,k) = n^2 + 2*k. - Joerg Arndt, Aug 13 2013

A161896 Integers n for which k = (9^n - 3 * 3^n - 4n) / (2n * (2n + 1)) is an integer.

Original entry on oeis.org

5, 11, 23, 29, 41, 53, 83, 89, 113, 131, 173, 179, 191, 233, 239, 251, 281, 293, 359, 419, 431, 443, 491, 509, 593, 641, 653, 659, 683, 719, 743, 761, 809, 911, 953, 1013, 1019, 1031, 1049, 1103, 1223, 1229, 1289, 1409, 1439, 1451, 1481, 1499, 1511, 1541, 1559

Offset: 1

Reikku Kulon, Jun 21 2009

Near superset of the Sophie Germain primes (A005384), excluding 2 and 3: 2n + 1 is prime. Nearly all members of this sequence are also prime, but four members less than 10000 are composite: 1541 = 23 * 67, 2465 = 5 * 17 * 29, 3281 = 17 * 193, and 4961 = 11^2 * 41.
The congruence of n modulo 4 is evenly distributed between 1 and 3. n is congruent to 5 (mod 6) for all n less than two billion.
This sequence has roughly twice the density of the sequence (A158034) corresponding to the Diophantine equation
f = (4^n - 2^n + 8n^2 - 2) / (2n * (2n + 1)),
and contains most members of that sequence. Those it does not contain are composite and often congruent to 3 (mod 6).
Composite terms appear to predominantly belong to A262051. - Bill McEachen, Aug 29 2024

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000

Cf. A161897 A000040, A002515, A005384, A158034, A158035, A158036, A145918, A002943.

a161896 n = a161896_list !! (n-1)
a161896_list = [x | x <- [1..],
                    (9^x - 3*3^x - 4*x) `mod` (2*x*(2*x + 1)) == 0]
-- Reinhard Zumkeller, Jan 12 2014

is(n)=my(m=2*n*(2*n+1),t=Mod(3,m)^n); t^2-3*t==4*n \\ Charles R Greathouse IV, Nov 25 2014

A302488 Total domination number of the n X n grid graph.

Original entry on oeis.org

0, 1, 2, 3, 6, 9, 12, 15, 20, 25, 30, 35, 42, 49, 56, 63, 72, 81, 90, 99, 110, 121, 132, 143, 156, 169, 182, 195, 210, 225, 240, 255, 272, 289, 306, 323, 342, 361, 380, 399, 420, 441, 462, 483, 506, 529, 552, 575, 600, 625, 650, 675, 702, 729, 756, 783, 812, 841, 870, 899, 930

Offset: 0

Eric W. Weisstein, Apr 08 2018

Extended to a(0) and a(1) using the formula/recurrence. The total domination number of the 1 X 1 grid graph is undefined.

Andrew Howroyd, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Grid Graph.
Eric Weisstein's World of Mathematics, Total Domination Number.
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,1,-2,1).

Main diagonal of A300358.
The four quadrasections are A002943, A016754, A002939(n+1), A000466(n+1).
Bisections are A002378 and A085046.
Cf. A303142.

R:=RealField(); [Round(((-1)^n + 2*n*(n + 2) + 4*Sin(n*Pi(R)/2) - 1)/8): n in [0..30]]; // G. C. Greubel, Apr 09 2018

Table[(-1 + (-1)^n + 2 n (2 + n) + 4 Sin[n Pi/2])/8, {n, 0, 20}]
LinearRecurrence[{2, -1, 0, 1, -2, 1}, {0, 1, 2, 3, 6, 9}, 20]
CoefficientList[Series[x (-1 - 2 x^3 + x^4)/((-1 + x)^3 (1 + x + x^2 + x^3)), {x, 0, 20}], x]

for(n=0,30, print1(round(((-1)^n + 2*n*(n + 2) + 4*sin(n*Pi/2) - 1)/8), ", ")) \\ G. C. Greubel, Apr 09 2018

a(n)=my(m=n\4); (2*m+1)*(2*m + n%4) \\ Andrew Howroyd, Aug 17 2025

a(n) = ((-1)^n + 2*n*(n + 2) + 4*sin(n*Pi/2) - 1)/8.
a(n) = 2*a(n-1) - a(n-2) + a(n-4) - 2*a(n-5) + a(n-6).
G.f.: x*(1 + 2*x^3 - x^4)/((1 - x)^3*(1 + x + x^2 + x^3)).
a(4*m + r) = (2*m + 1)*(2*m + r) for 0 <= r < 4. - Charles Kusniec, Aug 16 2025
From Amiram Eldar, Aug 26 2025: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/8 + 3/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/8 - 1/2. (End)

a(0)=0 prepended and offset corrected by Andrew Howroyd, Aug 17 2025

A181018 Maximum number of 1's in an n X n binary matrix with no three 1's adjacent in a line along a row, column or diagonally.

Original entry on oeis.org

1, 4, 6, 9, 16, 20, 26, 36, 42, 52, 64, 74, 86, 100, 114, 130

Offset: 1

R. H. Hardin, Sep 30 2010

Diagonal of A181019.
Three or more "1"s may be adjacent in an L-shape or step shape (cf. bottom of first example) or 2 X 2 square (top right of 2nd example) or similar. One possible (not always optimal) solution is therefore to fill the square with 2 X 2 squares of "1"s, separated by rows of "0"s: this yields the lower bound (n - floor(n/3))^2 = ceiling(2n/3)^2 given in FORMULA. I conjecture that this is optimal for n = 2 (mod 3) and that a(n) ~ (2n/3)^2. For n = 3k, the array can be filled with 2k(2k+1) "1"s by repeating the optimal solution for n = 3 on the diagonal, and filling the rest with 2 X 2 blocks separated by rows of "0"s, cf. the 4th example for 6 X 6. - M. F. Hasler, Jul 17 2015 [Conjecture proved to be wrong, see below. - M. F. Hasler, Jan 19 2016]
74 <= a(12) <= 77. - Manfred Scheucher, Jul 23 2015
You can repeat a 4 X 2 block [1100; 0011] infinitely in both directions and then crop the needed square. That gives ceiling(n^2/2). It eventually surpasses the solutions we've found so far: at 17*17 the pattern above gives 12*12=144 but this one ceiling(17*17/2)=145. The credit for finding this goes to Jaakko Himberg. - Juhani Heino, Aug 11 2015

			Some solutions for 6 X 6:
  0 1 1 0 1 1    0 1 1 0 1 1    0 1 1 0 1 1    0 1 1 0 1 1
  1 0 1 0 0 1    1 0 1 0 1 1    1 0 1 0 0 1    1 0 1 0 1 1
  1 1 0 0 1 0    1 1 0 0 0 0    1 1 0 0 1 0    1 1 0 0 0 0
  0 0 0 0 1 1    0 0 0 0 1 1    0 0 0 0 1 1    0 0 0 0 1 1
  1 0 1 1 0 1    1 0 1 1 0 1    1 1 0 1 0 1    1 1 0 1 0 1
  1 1 0 1 1 0    1 1 0 1 1 0    1 1 0 1 1 0    1 1 0 1 1 0
A solution with 73 ones for 12 X 12 (I replaced "0" with "." for readability):
  1 1 . 1 1 . 1 1 . 1 . 1
  1 1 . . 1 1 . 1 1 . 1 1
  . . . 1 . . . . . . 1 .
  1 1 . 1 . 1 . 1 1 . . 1
  . 1 1 . . 1 1 . . 1 1 .
  1 . . . 1 . 1 . 1 . . 1
  1 1 . . 1 1 . . 1 . 1 .
  . 1 . 1 . 1 . 1 . . 1 1
  1 . . 1 1 . . 1 1 . . 1
  . 1 . . . . 1 . 1 . 1 .
  1 1 . 1 1 . 1 1 . . 1 1
  1 . 1 . 1 1 . 1 . 1 . 1
- _Manfred Scheucher_, Jul 23 2015
An optimal solution with 74 ones (denoted by O) for 12 X 12 (also symmetric):
  O . O . O . O O . O O .
  O O . O O . . . O O . O
  . O . O . O O . . . O O
  O . . . O O . O O . O .
  . O O . . . O . . . . O
  O O . O O . O . O O . .
  . . O O . O . O O . O O
  O . . . . O . . . O O .
  . O . O O . O O . . . O
  O O . . . O O . O . O .
  O . O O . . . O O . O O
  . O O . O O . O . O . O - _Giovanni Resta_, Jul 29 2015

Manfred Scheucher, Python Script
Peter J. Taylor, Java program to compute terms

Cf. A000769, A181019, A219760, A225623.

See Taylor link
(MATLAB with CPLEX)
function v = A181018(n)
%
Grid = [1:n]' * ones(1,n) + n*ones(n,1)*[0:n-1];
f = -ones(n^2,1);
A = sparse(4*(n-2)*(n-1),n^2);
count = 0;
for i =1:n
  for j = 1:n-2
    count = count+1;
    A(count, [Grid(i,j),Grid(i,j+1),Grid(i,j+2)]) = 1;
  end
end
for i = 1:n-2
  for j = 1:n
    count = count+1;
    A(count, [Grid(i,j),Grid(i+1,j),Grid(i+2,j)]) = 1;
  end
end
for i = 1:n-2
  for j = 1:n-2
    count = count+2;
    A(count-1,[Grid(i,j+2),Grid(i+1,j+1),Grid(i+2,j)]) = 1;
    A(count, [Grid(i,j),Grid(i+1,j+1),Grid(i+2,j+2)]) = 1;
  end
end
b = 2*ones(4*(n-2)*(n-1),1);
[x,v,exitflag,output] = cplexbilp(f,A,b);
end;
for n = 1:11
  A(n) = A181018(n);
end
A % Robert Israel, Jan 14 2016

a(n) >= ceiling(2n/3)^2; a(3k) >= A002943(k) = 2k(2k+1). - M. F. Hasler, Jul 17 2015; revised by Juhani Heino, Aug 11 2015

a(n) >= ceiling(n^2/2). - Juhani Heino, Aug 11 2015

a(11)-a(12) from M. F. Hasler, Jul 20 2015
a(12) deleted by Manfred Scheucher, Jul 23 2015
a(12) from Giovanni Resta, Jul 29 2015
PARI code (which implemented a conjectured formula shown to underestimate) deleted by Peter J. Taylor, Jan 06 2016
a(13)-a(15) from Peter J. Taylor, Jan 09 2016
a(16) from Peter J. Taylor, Jan 14 2016

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