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A327479
a(n) is the minimum number of squares of unit area that must be removed from an n X n square to obtain a connected figure without holes and of the longest perimeter.
2
0, 0, 0, 4, 6, 12, 16, 28, 32, 44, 52, 68, 76, 92, 104, 124, 136, 156, 172, 196, 212, 236, 256, 284, 304, 332, 356, 388, 412, 444, 472, 508, 536, 572, 604, 644, 676, 716, 752, 796, 832, 876, 916, 964, 1004, 1052, 1096, 1148, 1192, 1244, 1292, 1348, 1396, 1452, 1504
OFFSET
0,4
COMMENTS
a(n) is equal to h_1(n) as defined in A309038.
All the terms are even numbers (A005843).
FORMULA
O.g.f.: 2*x^3*(-2 + x - 2*x^2 + x^3 - 2*x^4 + 3*x^5 - 2*x^6 + x^7)/((-1 + x)^3*(1 + x + x^2 + x^3)).
E.g.f.: 8 + 4*x + 2*x^2 + x^4/12 + (1/4)*(-7*exp(-x) + exp(x)*(-25 + 6*x + 2*x^2) - 4*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) = (1/4)*(- 25 + 2*n*(2 + n) - 7*cos(n*Pi) - 4*sin(n*Pi/2)) for n > 4, a(0) = 0, a(1) = 0, a(2) = 0, a(3) = 4, a(4) = 6.
Lim_{n->inf} a(n)/A000290(n) = 1/2.
EXAMPLE
Illustrations for n = 3..8:
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a(3) = 4 a(4) = 6 a(5) = 12
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a(6) = 16 a(7) = 28 a(8) = 32
MAPLE
gf := 8+4*x+2*x^2+(1/12)*x^4+1/4*(-7*exp(-x)+exp(x)*(2*x^2+6*x-25)-4*sin(x)):
ser := series(gf, x, 55): seq(factorial(n)*coeff(ser, x, n), n = 0..54);
MATHEMATICA
Join[{0, 0, 0, 4, 6}, Table[(1/4)*(-25+2n*(2+n)-7*Cos[n*Pi]-4*Sin[n*Pi/2]), {n, 5, 54}]]
PROG
(Magma) I:=[0, 0, 0, 4, 6, 12, 16, 28, 32, 44, 52]; [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..55]];
(PARI) concat([0, 0, 0], Vec(2*x^3*(-2+x-2*x^2+x^3-2*x^4+3*x^5-2*x^6+x^7)/((-1+x)^3*(1+x+x^2+x^3))+O(x^55)))
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Stefano Spezia, Sep 16 2019
STATUS
approved