login
a(n) = 3*a(n-1) + 24*a(n-2) + a(n-3), with a(0)=0, a(1)=2, and a(2)=7.
5

%I #12 Jun 13 2015 00:54:21

%S 0,2,7,69,377,2794,17499,119930,782560,5243499,34631867,230522137,

%T 1527974718,10151087309,67355177296,447219602022,2968334148479,

%U 19705628071261,130804123379301,868315777996646,5763951923164423,38262238564792074,253989877628319020

%N a(n) = 3*a(n-1) + 24*a(n-2) + a(n-3), with a(0)=0, a(1)=2, and a(2)=7.

%C The Ramanujan sequence number 11 for the argument 2Pi/9 defined by the relation a(n)*9^(1/3) = (3^(n-1))*(((-1)^(n-1))*(c(1) - 1/3)^(n + 1/3) + (((-1)^(n-1))*(c(2) - 1/3)^(n + 1/3) + (1/3 - c(4))^(n + 1/3)), where c(j) := 2*cos(2*Pi*j/9). The sequences A217052 and A217053 are conjugate with the sequence a(n). For more information on these connections - see Comments in A217053.

%C The 3-valuation of the sequence a(n) is equal to (0,2,1).

%D R. Witula, E. Hetmaniok, and D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, Proceedings of the 15th International Conference on Fibonacci Numbers and Their Applications, Eger, Hungary, 2012, in review.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,24,1).

%F G.f.: x*(2+x)/(1-3*x-24*x^2-x^3).

%e We have a(4)-5*a(3)=32, 8*a(4)-a(5)=222, a(9)-a(6)=5226000. Furthermore from a(0)=0 we get (c(1) - 1/3)^( 1/3) + (c(2) - 1/3)^(1/3) = (1/3 - c(4))^(1/3), while from a(3)=69 we obtain 23*9^(-1/6) = (c(1) - 1/3)^(10/3) + (c(2) - 1/3)^(10/3) + (1/3 - c(4))^(10/3).

%t LinearRecurrence[{3,24,1}, {0,2,7}, 30]

%o (PARI) Vec((2+x)/(1-3*x-24*x^2-x^3)+O(x^99)) \\ _Charles R Greathouse IV_, Sep 27 2012

%Y Cf. A217052, A217053, A214699, A214779, A214778, A214951, A214954, A215560, A215569, A215572.

%K nonn,easy

%O 0,2

%A _Roman Witula_, Sep 26 2012