A183575 a(n) = n - 1 + ceiling((n^2-2)/2); complement of A183574.

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

Offset: 1

Author

Clark Kimberling, Jan 05 2011

Keywords

• nonn
• easy

Comments

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

Links

• Colin Barker, Table of n, a(n) for n = 1..1000
• Eric Weisstein's World of Mathematics, Graph Circumference.
• Eric Weisstein's World of Mathematics, Stacked Book Graph.
• Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Crossrefs

Cf. A183574 (complement).

Programs

Mathematica

Table[Ceiling[n^2/2 - 1] + n - 1, {n, 20}] (* Eric W. Weisstein, May 18 2017 *)
Table[(2 n (n + 2) - 7 - (-1)^n)/4, {n, 20}] (* Eric W. Weisstein, May 18 2017 *)
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 *)

PARI

concat(0, Vec(2*x*(1 + x - x^2) / ((1 - x)^3*(1 + x)) + O(x^60))) \\ Colin Barker, Dec 07 2017

Formula

a(n) = n - 1 + ceiling(n^2/2-1).

a(n) = A000217(n-2) + A047215(n-1). - Wesley Ivan Hurt, Jul 15 2013

From Colin Barker, Dec 07 2017: (Start)

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

Extensions

Description corrected by Eric W. Weisstein, May 18 2017

a(1)=0 inserted by Amiram Eldar, Sep 16 2022