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A183575
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a(n) = n - 1 + ceiling((n^2-2)/2); complement of A183574.
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4
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0, 2, 6, 10, 16, 22, 30, 38, 48, 58, 70, 82, 96, 110, 126, 142, 160, 178, 198, 218, 240, 262, 286, 310, 336, 362, 390, 418, 448, 478, 510, 542, 576, 610, 646, 682, 720, 758, 798, 838, 880, 922, 966, 1010, 1056, 1102, 1150, 1198, 1248, 1298, 1350, 1402, 1456, 1510, 1566, 1622, 1680, 1738, 1798, 1858, 1920, 1982, 2046, 2110, 2176, 2242, 2310, 2378, 2448, 2518
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OFFSET
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1,2
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COMMENTS
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Agrees with the circumference of the n X n stacked book graph for n = 2 up to at least n = 8. - Eric W. Weisstein, Dec 05 2017
It seems that a(n-1) is the maximal length of an optimal solution path required to solve any n X n maze. Here the maze has a single start point, a single end point, and any number of walls that cannot be traversed. The maze is 4-connected, so the allowed moves are: up, down, left and right. For odd n, the hardest mazes have walls located in a spiral, start point in one corner and end point in the center. - Dmitry Kamenetsky, Mar 06 2018
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LINKS
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FORMULA
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a(n) = n - 1 + ceiling(n^2/2-1).
G.f.: 2*x^2*(1 + x - x^2) / ((1 - x)^3*(1 + x)).
a(n) = (n^2 + 2*n - 4)/2 for n even.
a(n) = (n^2 + 2*n - 3)/2 for n odd.
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n > 4.
(End)
Sum_{n>=2} 1/a(n) = 7/8 + tan(sqrt(5)*Pi/2)*Pi/(2*sqrt(5)). - Amiram Eldar, Sep 16 2022
E.g.f.: (4 + (x^2 + 3*x - 4)*cosh(x) + (x^2 + 3*x - 3)*sinh(x))/2. - Stefano Spezia, Sep 05 2023
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MATHEMATICA
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Table[If[Mod[n, 2] == 0, n^2 + 2 n - 4, (n + 3) (n - 1)]/2, {n, 20}] (* Eric W. Weisstein, May 18 2017 *)
LinearRecurrence[{2, 0, -2, 1}, {0, 2, 6, 10}, 80] (* Harvey P. Dale, Feb 19 2021 *)
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PROG
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(PARI) concat(0, Vec(2*x*(1 + x - x^2) / ((1 - x)^3*(1 + x)) + O(x^60))) \\ Colin Barker, Dec 07 2017
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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