A347559 Number of minimum dominating sets in the n-Moebius ladder.

9, 24, 10, 4, 14, 80, 18, 4, 22, 168, 26, 4, 30, 288, 34, 4, 38, 440, 42, 4, 46, 624, 50, 4, 54, 840, 58, 4, 62, 1088, 66, 4, 70, 1368, 74, 4, 78, 1680, 82, 4, 86, 2024, 90, 4, 94, 2400, 98, 4, 102, 2808, 106, 4, 110, 3248, 114, 4, 118, 3720, 122, 4, 126, 4224

Offset: 3

Author

Eric W. Weisstein, Sep 06 2021

Keywords

• nonn
• easy

Links

• Eric Weisstein's World of Mathematics, Minimum Dominating Set
• Eric Weisstein's World of Mathematics, Moebius Ladder
• Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

Crossrefs

Cf. A284663, A290337.

Programs

Mathematica

Table[Piecewise[{{9, n == 3}, {n (n + 2), Mod[n, 4] == 0}, {2 n, Mod[n, 2] == 1}, {4, Mod[n, 4] == 2}}, 0], {n, 3, 20}]
Join[{9}, LinearRecurrence[{0, 0, 0, 3, 0, 0, 0, -3, 0, 0, 0, 1}, {24, 10, 4, 14, 80, 18, 4, 22, 168, 26, 4, 30}, 20]]
CoefficientList[Series[(-9 - 24 x - 10 x^2 - 4 x^3 + 13 x^4 - 8 x^5 + 12 x^6 + 8 x^7 - 7 x^8 - 2 x^10 - 4 x^11 + 3 x^12)/((-1 + x)^3 (1 + x)^3 (1 + x^2)^3), {x, 0, 20}], x]

Formula

a(n) = n*(n+2) for n == 0 (mod 4).

a(n) = 2*n for n == 1 (mod 2) and n > 3.

a(n) = 4 for n == 2 (mod 4).

a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12) for n > 3.

G.f.: x^3*(-9 - 24*x - 10*x^2 - 4*x^3 + 13*x^4 - 8*x^5 + 12*x^6 + 8*x^7 - 7*x^8 - 2*x^10 - 4*x^11 + 3*x^12)/((-1 + x)^3*(1 + x)^3*(1 + x^2)^3).