A303054 Number of minimum total dominating sets in the n-ladder graph.

1, 4, 1, 16, 9, 1, 64, 16, 1, 169, 25, 1, 361, 36, 1, 676, 49, 1, 1156, 64, 1, 1849, 81, 1, 2809, 100, 1, 4096, 121, 1, 5776, 144, 1, 7921, 169, 1, 10609, 196, 1, 13924, 225, 1, 17956, 256, 1, 22801, 289, 1, 28561, 324, 1, 35344, 361, 1, 43264, 400, 1, 52441

Offset: 1

Author

Eric W. Weisstein, Apr 17 2018

Keywords

• nonn
• easy

Comments

Each vertex can dominate up to three others. A ladder with a length that is an exact multiple of three can be dominated in only one way with 2n/3 vertices. - Andrew Howroyd, Apr 21 2018

Examples

			From _Andrew Howroyd_, Apr 21 2018: (Start)
a(9) = 1 because there is only one arrangement of 6 vertices that is totally dominating and no set with fewer vertices can be totally dominating:
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(End)
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..200
• Eric Weisstein's World of Mathematics, Ladder Graph.
• Eric Weisstein's World of Mathematics, Minimum Total Dominating Set.
• Index entries for linear recurrences with constant coefficients, signature (0,0,5,0,0,-10,0,0,10,0,0,-5,0,0,1).

Crossrefs

Row 2 of A303293.

Cf. A302402, A303072.

Programs

Mathematica

Table[Piecewise[{{1, Mod[n, 3] == 0}, {((n^2 + 13 n + 4)/18)^2, Mod[n, 3] == 1}, {((n + 4)/3)^2, Mod[n, 3] == 2}}], {n, 58}] (* Eric W. Weisstein, Apr 23 2018 and Michael De Vlieger, Apr 21 2018 *)
Table[(916 + 392 n + 213 n^2 + 26 n^3 + n^4 - (-56 + 392 n + 213 n^2 + 26 n^3 + n^4) Cos[2 n Pi/3] + Sqrt[3] (-20 + 7 n + n^2) (28 + 19 n + n^2) Sin[2 n Pi/3])/972, {n, 20}] (* Eric W. Weisstein, Apr 23 2018 *)
LinearRecurrence[{0, 0, 5, 0, 0, -10, 0, 0, 10, 0, 0, -5, 0, 0, 1}, {1, 4, 1, 16, 9, 1, 64, 16, 1, 169, 25, 1, 361, 36, 1}, 20] (* Eric W. Weisstein, Apr 23 2018 *)
CoefficientList[Series[(-1 - 4 x - x^2 - 11 x^3 + 11 x^4 + 4 x^5 + 6 x^6 - 11 x^7 - 6 x^8 + x^9 + 5 x^10 + 4 x^11 - x^12 - x^13 - x^14)/(-1 + x^3)^5, {x, 0, 20}], x] (* Eric W. Weisstein, Apr 23 2018 *)

PARI

a(n)={if(n%3==0, 1, if(n%3==1, (n^2 + 13*n + 4)/18, (n + 4)/3))^2} \\ Andrew Howroyd, Apr 21 2018

PARI

Vec(x*(1 + 4*x + x^2 + 11*x^3 - 11*x^4 - 4*x^5 - 6*x^6 + 11*x^7 + 6*x^8 - x^9 - 5*x^10 - 4*x^11 + x^12 + x^13 + x^14) / ((1 - x)^5*(1 + x + x^2)^5) + O(x^60)) \\ Colin Barker, Apr 23 2018

Formula

a(n) = 1 for n mod 3 = 0

= ((n^2 + 13*n + 4)/18)^2 for n mod 3 = 1

= ((n + 4)/3)^2 for n mod 3 = 2.

G.f.: x*(-1 - 4*x - x^2 - 11*x^3 + 11*x^4 + 4*x^5 + 6*x^6 - 11*x^7 - 6*x^8 + x^9 + 5*x^10 + 4*x^11 - x^12 - x^13 - x^14)/(-1 + x^3)^5.

a(n) = 5*a(n-3) - 10*a(n-6) + 10*a(n-9) - 5*a(n-12) + a(n-15) for n>15. - Colin Barker, Apr 23 2018

Extensions

Terms a(14) and beyond from Andrew Howroyd, Apr 21 2018