A020878 - OEIS

%I #32 Feb 16 2025 08:32:34

%S 2,3,3,6,7,13,18,31,47,78,123,201,322,523,843,1366,2207,3573,5778,

%T 9351,15127,24478,39603,64081,103682,167763,271443,439206,710647,

%U 1149853,1860498,3010351,4870847,7881198,12752043,20633241,33385282,54018523,87403803

%N Number of maximum matchings in the n-Moebius ladder M_n.

%D J. P. McSorley, Counting structures in the Moebius ladder, Discrete Math., 184 (1998), 137-164.

%H Vincenzo Librandi, <a href="/A020878/b020878.txt">Table of n, a(n) for n = 0..1000</a>

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Matching.html">Matching</a>

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/MaximumIndependentEdgeSet.html">Maximum Independent Edge Set</a>

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/MoebiusLadder.html">Moebius Ladder</a>

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

%F If n mod 2 = 0 then L(n) else L(n)+2, where L() are the Lucas numbers.

%F a(n) = A001350(n) + 2.

%F G.f.: (2 + x - 4*x^2 - x^3) / ((1 - x)*(1 + x)*(1 - x - x^2)). - _Colin Barker_, Jan 23 2012

%F From _Colin Barker_, Jul 12 2017: (Start)

%F a(n) = ((1 - sqrt(5))/2)^n + ((1 + sqrt(5))/2)^n for n even.

%F a(n) = ((1 - sqrt(5))/2)^n + ((1 + sqrt(5))/2)^n + 2 for n odd.

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

%F (End)

%t CoefficientList[Series[(2+x-4*x^2-x^3)/((1+x)*(1-x)*(1-x-x^2)),{x,0,40}],x] (* _Vincenzo Librandi_, Apr 20 2012 *)

%t Table[1 - (-1)^n + LucasL[n], {n, 20}] (* _Eric W. Weisstein_, Dec 31 2017 *)

%t LinearRecurrence[{1, 2, -1, -1}, {3, 3, 6, 7}, 20] (* _Eric W. Weisstein_, Dec 31 2017 *)

%o (Magma) I:=[2, 3, 3, 6]; [n le 4 select I[n] else Self(n-1)+2*Self(n-2)-Self(n-3)-Self(n-4): n in [1..40]]; // _Vincenzo Librandi_, Apr 20 2012

%o (PARI) Vec((2 + x - 4*x^2 - x^3) / ((1 - x)*(1 + x)*(1 - x - x^2)) + O(x^50)) \\ _Colin Barker_, Jul 12 2017

%K nonn,easy

%O 0,1

%A _N. J. A. Sloane_