OFFSET
1,1
COMMENTS
Sequence extrapolated to n=1 using recurrence. - Andrew Howroyd, Apr 17 2018
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..200
Eric Weisstein's World of Mathematics, Prism Graph
Eric Weisstein's World of Mathematics, Total Dominating Set
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,5,0,0,0,0,0,-10,0,0,0,0,0,10,0,0,0,0,0,-5,0,0,0,0,0,1).
FORMULA
From Andrew Howroyd, Apr 17 2018: (Start)
a(n) = 5*a(n-6) - 10*a(n-12) + 10*a(n-18) - 5*a(n-24) + a(n-30) for n > 30.
a(6*k) = 9, a(6*k+1) = 2*(6*k+1), a(6*k+2) = (6*k+2)^2, a(6*k+3) = 3, a(6*k+4) = ((2*k + 3)*(3*k + 2))^2, a(6*k+5) = (4*k + 5)*(6*k + 5). (End)
MATHEMATICA
Table[(432 + 132 n + 85 n^2 + 10 n^3 + n^4 + (216 - 132 n + 37 n^2 + 10 n^3 + n^4) (-1)^n +(432 + 132 n - 37 n^2 - 10 n^3 - n^4) Cos[n Pi/3] + (864 - 132 n - 85 n^2 - 10 n^3 - n^4) Cos[2 n Pi/3] +Sqrt[3] (12 n - 13 n^2 - 10 n^3 - n^4) Sin[n Pi/3] + Sqrt[3] (12 n - 35 n^2 + 10 n^3 + n^4) Sin[2 n Pi/3])/216, {n, 200}]
Table[Piecewise[{{9, Mod[n, 6] == 0}, {2 n, Mod[n, 6] == 1}, {n^2, Mod[n, 6] == 2}, {3, Mod[n, 6] == 3}, {n^2 (n + 5)^2/36, Mod[n, 6] == 4}, {n (2 n + 5)/3, Mod[n, 6] == 5}}], {n, 200}]
LinearRecurrence[{0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, -10, 0, 0, 0, 0, 0, 10, 0, 0, 0, 0, 0, -5, 0, 0, 0, 0, 0, 1}, {2, 4, 3, 36, 25, 9, 14, 64, 3, 625, 99, 9, 26, 196, 3, 3136, 221, 9, 38, 400, 3, 9801, 391, 9, 50, 676, 3, 23716, 609, 9}, 200]
Rest @ CoefficientList[Series[(9 x^6)/(1 - x^6) - (3 x^3)/(-1 + x^6) + (2 x (1 + 5 x^6))/(-1 + x^6)^2 + (x^5 (-25 - 24 x^6 + x^12))/(-1 + x^6)^3 - (4 x^2 (1 + 13 x^6 + 4 x^12))/(-1 + x^6)^3 - (x^4 (36 + 445 x^6 + 371 x^12 + 11 x^18 + x^24))/(-1 + x^6)^5, {x, 0, 200}], x]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Eric W. Weisstein, Apr 17 2018
EXTENSIONS
a(1)-a(2) and terms a(15) and beyond from Andrew Howroyd, Apr 17 2018
STATUS
approved