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a(n) is the maximum number of squares of unit area that can be removed from an n X n square while still obtaining a connected figure without holes and of the longest perimeter.
3

%I #38 Jun 25 2023 04:24:20

%S 0,0,2,4,8,12,22,28,40,48,64,76,94,108,130,148,172,192,220,244,274,

%T 300,334,364,400,432,472,508,550,588,634,676,724,768,820,868,922,972,

%U 1030,1084,1144,1200,1264,1324,1390,1452,1522,1588,1660,1728,1804,1876,1954

%N a(n) is the maximum number of squares of unit area that can be removed from an n X n square while still obtaining a connected figure without holes and of the longest perimeter.

%C a(n) is equal to h_1(n) + h_2(n) as defined in A309038.

%H Stefano Spezia, <a href="/A327480/b327480.txt">Table of n, a(n) for n = 0..10000</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 O.g.f.: 2*x^2*(1 + x^2 + 2*x^4 - 2*x^5 + 2*x^6 - 2*x^7 + x^8)/((1 - x)^3*(1 + x)*(1 + x^2)).

%F E.g.f.: (1/24)*exp(-x)*(33 + 9*exp(2*x)*(7 - 2*x + 2*x^2) - 2*exp(x)*(48 + 12*x^2 + x^4) - 12*exp(x)*sin(x)).

%F a(n) = 2*a(n-1) - a(n-2) + a(n-4) - 2*a(n-5) + a(n-6) for n > 10.

%F a(n) = (1/8)*(21 - 12*n + 6*n^2 + 11*(-1)^n - 4*A056594(n+1) for n > 4, a(0) = 0, a(1) = 0, a(2) = 2, a(3) = 4, a(4) = 8.

%F Limit_{n->oo} a(n)/A000290(n) = 3/4.

%e Illustrations for n = 2..7:

%e __ __ __ __ __

%e |__|__ |__|__|__| |__|__|__|

%e |__| __|__|__ __|__|__ __

%e |__| |__| |__| | |

%e |__ __|

%e a(2) = 2 a(3) = 4 a(4) = 8

%e __ __ __ __ __ __ __ __ __ __ __

%e |__|__|__ __ __| |__|__|__| |__|__ |__|__|__| |__|__|__|

%e __|__|__ __ __|__|__ __|__| __|__|__ __|__|__

%e | | |__|__|__| |__| |__|__|__| |__| |__|__|__| |__|

%e | | __|__|__ __ __|__|__ __ __|__|__ __

%e |__| |__| |__| |__|__|__| |__|__ |__|__|__| |__|__|__|

%e |__| |__| __|__|__ __|__|__

%e |__| |__| |__| |__|

%e a(5) = 12 a(6) = 22 a(7) = 28

%p gf := (1/24)*exp(-x)*(33+9*exp(2*x)*(2*x^2-2*x+7)-2*exp(x)*(x^4+12*x^2+48)-12*exp(x)*sin(x)); ser := series(gf, x, 53):

%p seq(factorial(n)*coeff(ser, x, n), n = 0 .. 52)

%t Join[{0,0,2,4,8},Table[(1/8)*(21-12n+6n^2+11*(-1)^n-4*Sin[n*Pi/2]),{n,5,52}]]

%o (Magma) I:=[0, 0, 2, 4, 8, 12, 22, 28, 40, 48, 64]; [n le 11 select I[n] else 2*Self(n-1)-Self(n-2)+Self(n-4)-2*Self(n-5)+Self(n-6): n in [1..53]];

%o (PARI) concat([0, 0], Vec(2*x^2*(1+x^2+2*x^4-2*x^5+2*x^6-2*x^7+x^8)/((1-x)^3*(1+x)*(1+x^2))+O(x^53)))

%Y Cf. A000290, A056594, A309038, A326118, A327479.

%K nonn,easy

%O 0,3

%A _Stefano Spezia_, Sep 16 2019