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a(n) = 3^n - 2^(n+1) - 1.
1

%I #21 Mar 05 2024 10:12:37

%S -2,0,10,48,178,600,1930,6048,18658,57000,173050,523248,1577938,

%T 4750200,14283370,42915648,128878018,386896200,1161212890,3484687248,

%U 10456158898,31372671000,94126401610,282395982048,847221500578,2541731610600,7625329049530

%N a(n) = 3^n - 2^(n+1) - 1.

%C For even n >= 4, also the number of minimum vertex colorings of the n-Moebius ladder.

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/MinimumVertexColoring.html">Minimum Vertex Coloring</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_03">Index entries for linear recurrences with constant coefficients</a>, signature (6,-11,6).

%F a(n) = 6*(n-1) - 11*a(n-2) + 6*a(n-3).

%F G.f.: 2*x*(1-6*x+6*x^2)/((-1+x)*(-1+2*x)*(-1+3*x)).

%t Table[3^n - 2^(n + 1) - 1, {n, 20}]

%t LinearRecurrence[{6, -11, 6}, {-2, 0, 10}, 20]

%t CoefficientList[Series[2 (1 - 6 x + 6 x^2)/((-1 + x) (-1 + 2 x) (-1 + 3 x)), {x, 0, 20}], x]

%o (Python)

%o def A368956(n): return 3**n-(1<<n+1)-1 # _Chai Wah Wu_, Mar 05 2024

%K sign,easy

%O 1,1

%A _Eric W. Weisstein_, Mar 05 2024