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Number of Eulerian cycles in the n-antiprism graph.
2

%I #22 Jan 08 2020 09:56:04

%S 4,44,372,2932,22484,170196,1279828,9590612,71736660,536055124,

%T 4003591508,29892900180,223162389844,1665861735764,12434781197652,

%U 92816950121812,692805066118484,5171207088198996,38598573880071508,288104312443589972,2150442403051689300

%N Number of Eulerian cycles in the n-antiprism graph.

%C Sequence extrapolated to n=1 and n=2 using the recurrence. - _Andrew Howroyd_, Jan 11 2018

%H Andrew Howroyd, <a href="/A297384/b297384.txt">Table of n, a(n) for n = 1..200</a>

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

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

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (13, -48, 52, -16).

%F From _Andrew Howroyd_, Jan 11 2018: (Start)

%F a(n) = 13*a(n-1) - 48*a(n-2) + 52*a(n-3) - 16*a(n-4).

%F G.f.: 4*x*(1 - 2*x - 2*x^2)/((1 - 8*x + 4*x^2)*(1 - 4*x)*(1 - x)).

%F (End)

%F From _Eric W. Weisstein_, Jan 12 2018: (Start)

%F a(n) = 2^n*((2 - sqrt(3))^n + (2 + sqrt(3))^n) - 2/3*(2 + 4^n).

%F a(n) = 2^n*A003500(n) - 2/3*(2 + 4^n).

%F a(n) = A003500(n) - A039301(n-1).

%F (End)

%t Table[2^n ((2 - Sqrt[3])^n + (2 + Sqrt[3])^n) - 2/3 (2 + 4^n), {n, 20}] // Expand

%t Table[2^n LucasL[2 n, Sqrt[2]] - 2/3 (2 + 4^n), {n, 20}] // Round

%t LinearRecurrence[{13, -48, 52, -16}, {4, 44, 372, 2932}, 20]

%t CoefficientList[Series[-4 (-1 + 2 x + 2 x^2)/(1 - 13 x + 48 x^2 - 52 x^3 + 16 x^4), {x, 0, 20}], x]

%o (PARI) Vec(4*(1 - 2*x - 2*x^2)/((1 - 8*x + 4*x^2)*(1 - 4*x)*(1 - x)) + O(x^30)) \\ _Andrew Howroyd_, Jan 11 2018

%K nonn,easy

%O 1,1

%A _Eric W. Weisstein_, Dec 29 2017