A282513 a(n) = floor((3*n + 2)^2/24 + 1/3).

0, 1, 3, 5, 8, 12, 17, 22, 28, 35, 43, 51, 60, 70, 81, 92, 104, 117, 131, 145, 160, 176, 193, 210, 228, 247, 267, 287, 308, 330, 353, 376, 400, 425, 451, 477, 504, 532, 561, 590, 620, 651, 683, 715, 748, 782, 817, 852, 888, 925, 963

Offset: 0

Author

Luce ETIENNE, Feb 17 2017

Keywords

• nonn
• easy

Comments

List of quadruples: 2*n*(3*n+1), (2*n+1)*(3*n+1), 6*n^2+8*n+3, (n+1)*(6*n+5). These terms belong to the sequences A033580, A033570, A126587 and A049452, respectively. See links for all the permutations.

After 0, subsequence of A025767.

It seems that a(n) is the smallest number of cells that need to be painted in a (n+1) X (n+1) grid, such that it has no unpainted hexominoes (see link to Kamenetsky and Pratt). - Rob Pratt, Dmitry Kamenetsky, Aug 30 2020

Examples

			Rectangular array with four columns:
.   0,   1,   3,   5;
.   8,  12,  17,  22;
.  28,  35,  43,  51;
.  60,  70,  81,  92;
. 104, 117, 131, 145, etc.
From _Rob Pratt_, Aug 30 2020: (Start)
For n = 3, painting only 2 cells would leave an unpainted hexomino, but painting the following 3 cells avoids all unpainted hexominoes:
    . . .
    . . X
    X X .
(End)
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..1000
• Luce ETIENNE, Permutations
• Dmitry Kamenetsky and Rob Pratt, 10x10 grid with no unpainted hexominoes, Puzzling StackExchange, October 2019.
• Rob Pratt, Optimal solution for n=11.
• Index entries for linear recurrences with constant coefficients, signature (2,-1,0,1,-2,1).

Crossrefs

Cf. A033436: floor((3*n)^2/24 + 1/3).

Cf. A000326, A000567, A025767, A033570, A033580, A049452, A064412, A126587, A222017, A269064, A274221.

Cf. A130519.

Minimum number of painted cells in other n-ominoes: A337501, A337502, A337503.

Programs

Magma

[(3*n^2+4*n+4) div 8: n in [0..50]]; // Bruno Berselli, Feb 17 2017

Mathematica

Table[Floor[(3 n + 2)^2/24 + 1/3], {n, 0, 50}] (* or *) CoefficientList[Series[x (1 + x + x^3)/((1 + x) (1 + x^2) (1 - x)^3), {x, 0, 50}], x] (* or *) Table[(6 n^2 + 8 n + 3 + Cos[n Pi] - 4 Cos[n Pi/2])/16, {n, 0, 50}] (* or *) Table[(3 n + 2)^2/24 + 1/3 + (-6 + (1 + (-1)^n) (1 + 2 I^((n + 1) (n + 2))))/16, {n, 0, 50}] (* Michael De Vlieger, Feb 17 2017 *)
LinearRecurrence[{2,-1,0,1,-2,1},{0,1,3,5,8,12},60] (* Harvey P. Dale, Aug 10 2024 *)

PARI

a(n)=(3*n^2 + 4*n + 4)\8 \\ Charles R Greathouse IV, Feb 17 2017

Formula

G.f.: x*(1 + x + x^3)/((1 + x)*(1 + x^2)*(1 - x)^3).

a(n) = 2*a(n-1) - a(n-2) + a(n-4) - 2*a(n-5) + a(n-6) for n>5.

a(n) = floor((3*n + 2)^2/24 + 2/3).

a(n) = (6*n^2 + 8*n + 3 + (-1)^n - 2*((-1)^((2*n - 1 + (-1)^n)/4) + (-1)^((2*n + 1 - (-1)^n)/4)))/16. Therefore:

a(2*k) = (6*k^2 + 4*k + 1 - (-1)^k)/4,

a(2*k+1) = (k + 1)*(3*k + 2)/2.

a(n) = (6*n^2 + 8*n + 3 + cos(n*Pi) - 4*cos(n*Pi/2))/16.

a(n) = (3*n + 2)^2/24 + 1/3 + (-6 + (1 + (-1)^n)*(1 + 2*i^((n+1)*(n+2))))/16, where i=sqrt(-1).

a(n) = A130519(n+3)+A130519(n+2)+A130519(n). - R. J. Mathar, Jun 23 2021

Extensions

Corrected and extended by Bruno Berselli, Feb 17 2017