%I #77 Aug 10 2024 19:25:49
%S 0,1,3,5,8,12,17,22,28,35,43,51,60,70,81,92,104,117,131,145,160,176,
%T 193,210,228,247,267,287,308,330,353,376,400,425,451,477,504,532,561,
%U 590,620,651,683,715,748,782,817,852,888,925,963
%N a(n) = floor((3*n + 2)^2/24 + 1/3).
%C 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.
%C After 0, subsequence of A025767.
%C 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
%H Harvey P. Dale, <a href="/A282513/b282513.txt">Table of n, a(n) for n = 0..1000</a>
%H Luce ETIENNE, <a href="/A282513/a282513.pdf">Permutations</a>
%H Dmitry Kamenetsky and Rob Pratt, <a href="https://puzzling.stackexchange.com/questions/90128/10x10-grid-with-no-unpainted-hexominoes/">10x10 grid with no unpainted hexominoes</a>, Puzzling StackExchange, October 2019.
%H Rob Pratt, <a href="/A282513/a282513.png">Optimal solution for n=11</a>.
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,0,1,-2,1).
%F G.f.: x*(1 + x + x^3)/((1 + x)*(1 + x^2)*(1 - x)^3).
%F a(n) = 2*a(n-1) - a(n-2) + a(n-4) - 2*a(n-5) + a(n-6) for n>5.
%F a(n) = floor((3*n + 2)^2/24 + 2/3).
%F 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:
%F a(2*k) = (6*k^2 + 4*k + 1 - (-1)^k)/4,
%F a(2*k+1) = (k + 1)*(3*k + 2)/2.
%F a(n) = (6*n^2 + 8*n + 3 + cos(n*Pi) - 4*cos(n*Pi/2))/16.
%F 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).
%F a(n) = A130519(n+3)+A130519(n+2)+A130519(n). - _R. J. Mathar_, Jun 23 2021
%e Rectangular array with four columns:
%e . 0, 1, 3, 5;
%e . 8, 12, 17, 22;
%e . 28, 35, 43, 51;
%e . 60, 70, 81, 92;
%e . 104, 117, 131, 145, etc.
%e From _Rob Pratt_, Aug 30 2020: (Start)
%e For n = 3, painting only 2 cells would leave an unpainted hexomino, but painting the following 3 cells avoids all unpainted hexominoes:
%e . . .
%e . . X
%e X X .
%e (End)
%t 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 *)
%t LinearRecurrence[{2,-1,0,1,-2,1},{0,1,3,5,8,12},60] (* _Harvey P. Dale_, Aug 10 2024 *)
%o (PARI) a(n)=(3*n^2 + 4*n + 4)\8 \\ _Charles R Greathouse IV_, Feb 17 2017
%o (Magma) [(3*n^2+4*n+4) div 8: n in [0..50]]; // _Bruno Berselli_, Feb 17 2017
%Y Cf. A033436: floor((3*n)^2/24 + 1/3).
%Y Cf. A000326, A000567, A025767, A033570, A033580, A049452, A064412, A126587, A222017, A269064, A274221.
%Y Cf. A130519.
%Y Minimum number of painted cells in other n-ominoes: A337501, A337502, A337503.
%K nonn,easy
%O 0,3
%A _Luce ETIENNE_, Feb 17 2017
%E Corrected and extended by _Bruno Berselli_, Feb 17 2017