

A217053


a(n) = 3*a(n1) + 24*a(n2) + a(n3), with a(0) = 2, a(1) = 5, and a(2) = 62.


5



2, 5, 62, 308, 2417, 14705, 102431, 662630, 4460939, 29388368, 195890270, 1297452581, 8623112591, 57204089987, 379864424726, 2521114546457, 16737293922782, 111098495308040, 737511654617345, 4895636145167777, 32498286641627651, 215727639063526946
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,1


COMMENTS

The Ramanujan type sequence number 9 for the argument 2Pi/9 defined by the following relation a(n)*3^(n + 1/3) = ((1))^n)*(c(1)  1/3)^(n + 2/3) + ((1)^n)*(c(2)  1/3)^(n + 2/3) + (1/3  c(4))^(n + 2/3), where c(j) := 2*cos(2*Pi*j/9). We note that c(4) = cos(Pi/9). The conjugate with a(n) are the sequences A217052 and A217069.
In Witula's et al.'s paper the following sequence is discussed: S(n) := sum{j=0,1,2} (1/3  c(2^j))^(n/3). It is proved that S(n+3) = (3^(1/3))*S(n+1) + S(n)/3, n=0,1,... Moreover the following decomposition holds true S(n) = x(n) + y(n)*3^(1/3) + z(n)*9^(1/3), which implies the system of recurrence relations: x(n+3) = 3*z(n+1) + x(n)/3, y(n+3) = x(n+1) + y(n)/3, z(n+3) = y(n+1) + z(n)/3, x(0)=3, y(0)=z(0)=x(1)=y(1)=z(1)=x(2)=z(2)=0, y(2)=2. Then it can be generated the relations X'(n+9)  3*X'(n+6)  24*X'(n+3)  X'(n) = 0, where X'(n) = X(n)*3^n for every X=x,y,z and n=0,1,..., from which we obtain that x(3*n+1)=x(3*n+2)=y(3*n)=y(3*n+1)=z(3*n)=z(3*n+2)=0 and a(n) = y(3*n+2)*3^n, A217052(n) = x(3*n)*3^(n1), and A217069(n) = z(3*n+1)*3^(n1).
Each a(n)+1 is divisible by 3 but which a(n)+1 are divisible by 9  it is a question?


REFERENCES

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.


LINKS

Table of n, a(n) for n=0..21.
Index entries for linear recurrences with constant coefficients, signature (3,24,1).


FORMULA

G.f.: (2xx^2)/(13*x24*x^2x^3).


EXAMPLE

We have 2*3^(1/3) = (c(1)  1/3)^(2/3) + (c(2)  1/3)^(2/3) + (1/3  c(4))^(2/3), and 5*3^(2/3) = (c(1)  1/3)^(5/3)  (c(2)  1/3)^(5/3) + (1/3  c(4))^(5/3).
Moreover we have 12*a(1) + a(0) = a(2), 5*a(2) = a(3) + a(0).


MATHEMATICA

LinearRecurrence[{3, 24, 1}, {2, 5, 62}, 30]


PROG

(PARI) Vec((2xx^2)/(13*x24*x^2x^3)+O(x^99)) \\ Charles R Greathouse IV, Oct 01 2012


CROSSREFS

Cf. A217052, A217069, A214699, A214778, A214779, A214951, A214954, A215560, A215569, A215572.
Sequence in context: A093484 A250195 A041069 * A134590 A012978 A012949
Adjacent sequences: A217050 A217051 A217052 * A217054 A217055 A217056


KEYWORD

nonn,easy


AUTHOR

Roman Witula, Sep 25 2012


STATUS

approved



