A302653 Number of minimum total dominating sets in the n-cycle graph.
1, 1, 3, 4, 5, 9, 7, 4, 9, 25, 11, 4, 13, 49, 15, 4, 17, 81, 19, 4, 21, 121, 23, 4, 25, 169, 27, 4, 29, 225, 31, 4, 33, 289, 35, 4, 37, 361, 39, 4, 41, 441, 43, 4, 45, 529, 47, 4, 49, 625, 51, 4, 53, 729, 55, 4, 57, 841, 59, 4, 61, 961, 63, 4, 65, 1089, 67, 4, 69, 1225, 71, 4
Offset: 1
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
- Eric Weisstein's World of Mathematics, Cycle Graph.
- Eric Weisstein's World of Mathematics, Minimum Total Dominating Set.
- Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).
Programs
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Mathematica
Table[((-1)^n (n - 4)^2 + (n + 4)^2 - 2 (n - 4) (n + 4) cos(n Pi/2))/16, {n, 80}] Table[Piecewise[{{n, Mod[n, 2] == 1}, {4, Mod[n, 4] == 0}, {(n/2)^2, Mod[n, 4] == 2}}], {n, 80}] LinearRecurrence[{0, 0, 0, 3, 0, 0, 0, -3, 0, 0, 0, 1}, {1, 1, 3, 4, 5, 9, 7, 4, 9, 25, 11, 4}, 80]
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PARI
Vec(x*(1 + x + 3*x^2 + 4*x^3 + 2*x^4 + 6*x^5 - 2*x^6 - 8*x^7 - 3*x^8 + x^9 - x^10 + 4*x^11) / ((1 - x)^3*(1 + x)^3*(1 + x^2)^3) + O(x^70)) \\ Colin Barker, Dec 25 2019
Formula
a(n) = n for odd n.
a(n) = 4 for n mod 4 = 0.
a(n) = (n/2)^2 for n mod 4 = 2.
a(n) = ((-1)^n*(n - 4)^2 + (n + 4)^2 - 2*(n - 4)*(n + 4)*cos(n*Pi/2))/16.
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12).
G.f.: x*(1 + x + 3*x^2 + 4*x^3 + 2*x^4 + 6*x^5 - 2*x^6 - 8*x^7 - 3*x^8 + x^9 - x^10 + 4*x^11) / ((1 - x)^3*(1 + x)^3*(1 + x^2)^3). - Colin Barker, Dec 25 2019
Comments