A327480 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.
0, 0, 2, 4, 8, 12, 22, 28, 40, 48, 64, 76, 94, 108, 130, 148, 172, 192, 220, 244, 274, 300, 334, 364, 400, 432, 472, 508, 550, 588, 634, 676, 724, 768, 820, 868, 922, 972, 1030, 1084, 1144, 1200, 1264, 1324, 1390, 1452, 1522, 1588, 1660, 1728, 1804, 1876, 1954
Offset: 0
Examples
Illustrations for n = 2..7: __ __ __ __ __ |__|__ |__|__|__| |__|__|__| |__| __|__|__ __|__|__ __ |__| |__| |__| | | |__ __| a(2) = 2 a(3) = 4 a(4) = 8 __ __ __ __ __ __ __ __ __ __ __ |__|__|__ __ __| |__|__|__| |__|__ |__|__|__| |__|__|__| __|__|__ __ __|__|__ __|__| __|__|__ __|__|__ | | |__|__|__| |__| |__|__|__| |__| |__|__|__| |__| | | __|__|__ __ __|__|__ __ __|__|__ __ |__| |__| |__| |__|__|__| |__|__ |__|__|__| |__|__|__| |__| |__| __|__|__ __|__|__ |__| |__| |__| |__| a(5) = 12 a(6) = 22 a(7) = 28
Links
- Stefano Spezia, Table of n, a(n) for n = 0..10000
- Index entries for linear recurrences with constant coefficients, signature (2,-1,0,1,-2,1).
Programs
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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]];
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Maple
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): seq(factorial(n)*coeff(ser, x, n), n = 0 .. 52)
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Mathematica
Join[{0,0,2,4,8},Table[(1/8)*(21-12n+6n^2+11*(-1)^n-4*Sin[n*Pi/2]),{n,5,52}]]
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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)))
Formula
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)).
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)).
a(n) = 2*a(n-1) - a(n-2) + a(n-4) - 2*a(n-5) + a(n-6) for n > 10.
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. [corrected by Jason Yuen, Dec 17 2024]
Limit_{n->oo} a(n)/A000290(n) = 3/4.
Comments