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Circumference of the (n,n)-knight graph.
1

%I #20 Jan 14 2024 16:35:58

%S 8,14,24,36,48,64,80,100,120,144,168,196,224,256,288,324,360,400,440,

%T 484,528,576,624,676,728,784,840,900,960,1024,1088,1156,1224,1296,

%U 1368,1444,1520,1600,1680,1764,1848,1936,2024,2116,2208,2304,2400,2500,2600,2704,2808,2916,3024,3136,3248,3364

%N Circumference of the (n,n)-knight graph.

%C Colin Barker's conjectures confirmed using first Mathematica program. - _Ray Chandler_, Jan 14 2024

%H Ray Chandler, <a href="/A248427/b248427.txt">Table of n, a(n) for n = 3..1000</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/GraphCircumference.html">Graph Circumference</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/KnightGraph.html">Knight Graph</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-2,1).

%F From _Colin Barker_, Oct 07 2014: (Start)

%F a(n) = (-1+(-1)^n+2*n^2)/2 for n>4.

%F a(n) = 2*a(n-1)-2*a(n-3)+a(n-4) for n>8.

%F G.f.: -2*x^3*(x^5-2*x^4+2*x^3-2*x^2-x+4) / ((x-1)^3*(x+1)).

%F (End)

%t Table[Piecewise[{{14, n == 4}, {n^2, Mod[n, 2] == 0}, {n^2 - 1, Mod[n, 2] == 1}}], {n, 3, 50}]

%t CoefficientList[Series[-2x^3(x^5-2x^4+2x^3-2x^2-x+4)/((x-1)^3(x+1)),{x,0,50}],x] (* or *) LinearRecurrence[{2,0,-2,1},{8,14,24,36,48,64},50] (* _Harvey P. Dale_, Jan 05 2024 *)

%K nonn

%O 3,1

%A _Eric W. Weisstein_, Oct 06 2014