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 A217053 a(n) = 3*a(n-1) + 24*a(n-2) + a(n-3), 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^(n-1), and A217069(n) = z(3*n+1)*3^(n-1). 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 Index entries for linear recurrences with constant coefficients, signature (3,24,1). FORMULA G.f.: (2-x-x^2)/(1-3*x-24*x^2-x^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((2-x-x^2)/(1-3*x-24*x^2-x^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

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Last modified December 16 06:18 EST 2019. Contains 330016 sequences. (Running on oeis4.)