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If n mod 3 = 0 then a(n) = 3^(n/3) + 12*n, if n mod 3 = 1 then a(n) = 4*3^((n-4)/3) + 12*n + 51, otherwise a(n) = 2*3^((n-2)/3) + 12*n - 36.
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%I #25 Oct 13 2022 10:58:11

%S 39,103,30,81,147,78,135,207,150,225,315,294,423,567,654,945,1251,

%T 1662,2439,3231,4614,6849,9099,13398,20007,26631,39678,59409,79155,

%U 118446,177543,236655,354678,531873,709083,1063302,1594791,2126295,3189102,4783473,6377859

%N If n mod 3 = 0 then a(n) = 3^(n/3) + 12*n, if n mod 3 = 1 then a(n) = 4*3^((n-4)/3) + 12*n + 51, otherwise a(n) = 2*3^((n-2)/3) + 12*n - 36.

%H Seiichi Manyama, <a href="/A276401/b276401.txt">Table of n, a(n) for n = 3..6287</a>

%H Natasha Morrison and Alex Scott, <a href="https://people.maths.ox.ac.uk/scott/Papers/maxinduced.pdf">Maximizing the number of induced cycles in a graph</a>, Preprint, 2016. See Cor. 1.3.

%H Natasha Morrison and Alex Scott, <a href="http://dx.doi.org/10.1016/j.jctb.2017.03.007">Maximising the number of induced cycles in a graph</a>, Journal of Combinatorial Theory Series B 126 (2017), 24-61.

%F From _Chai Wah Wu_, Sep 11 2016: (Start)

%F a(n) = a(n-1) + 4*a(n-3) - 4*a(n-4) - 3*a(n-6) + 3*a(n-7) for n > 9.

%F a(n) = 4*a(n-3) - 3*a(n-6) - 72 for n > 8.

%F G.f.: x^3*(30*x^6 - 223*x^5 + 190*x^4 + 105*x^3 + 73*x^2 - 64*x - 39)/((x - 1)^2*(3*x^3 - 1)*(x^2 + x + 1)). (End)

%p f:=n-> if n mod 3 = 0 then 3^(n/3)+12*n

%p elif n mod 3 = 1 then 4*3^((n-4)/3)+12*n+51

%p else 2*3^((n-2)/3)+12*n-36; fi;

%p [seq(f(n),n=3..40)];

%t CoefficientList[Series[(30 x^6 -223 x^5 + 190 x^4 + 105 x^3 + 73 x^2 - 64 x - 39) / ((x-1)^2 (3 x^3-1) (x^2+x+1)), {x, 0, 33}], x] (* _Vincenzo Librandi_, Sep 12 2016 *)

%t Table[Which[Mod[n,3]==0,3^(n/3)+12n,Mod[n,3]==1,4*3^((n-4)/3)+12n+51,True,2*3^((n-2)/3)+12n-36],{n,3,40}] (* _Harvey P. Dale_, Nov 16 2020 *)

%Y Cf. A352669.

%K nonn,easy

%O 3,1

%A _N. J. A. Sloane_, Sep 11 2016