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 A214699 a(n) = 3*a(n-2) - a(n-3) with a(0)=0, a(1)=3, a(2)=0. 18
 0, 3, 0, 9, -3, 27, -18, 84, -81, 270, -327, 891, -1251, 3000, -4644, 10251, -16932, 35397, -61047, 123123, -218538, 430416, -778737, 1509786, -2766627, 5308095, -9809667, 18690912, -34737096, 65882403, -122902200, 232384305, -434589003, 820055115, -1536151314 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS The Ramanujan-type sequence number 1 for the argument 2*Pi/9 defined by the following identity: 3^(1/3)*a(n) = (c(1)/c(2))^(1/3)*c(1)^n + (c(2)/c(4))^(1/3)*c(2)^n + (c(4)/c(1))^(1/3)*c(4)^n = -( (c(1)/c(2))^(1/3)*c(2)^(n+1) + (c(2)/c(4))^(1/3)*c(4)^(n+1) + (c(4)/c(1))^(1/3)*c(1)^(n+1) ), where c(j) := 2*Cos(2*Pi*j/9). The definitions other Ramanujan-type sequences for the argument 2*Pi/9 in one's from Crossrefs are given. We note that all a(n) are divisible by 3. REFERENCES R. Witula, E. Hetmaniok, D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, Proceedings of the Fifteenth International Conference on Fibonacci Numbers and Their Applications, Eger, Hungary, 2012 LINKS Roman Witula, Full Description of Ramanujan Cubic Polynomials, Journal of Integer Sequences, Vol. 13 (2010), Article 10.5.7. Roman Witula, Ramanujan Cubic Polynomials of the Second Kind, Journal of Integer Sequences, Vol. 13 (2010), Article 10.7.5. Roman Witula, Ramanujan Type Trigonometric Formulae, Demonstratio Math. 45 (2012) 779-796. Index entries for linear recurrences with constant coefficients, signature (0,3,-1). FORMULA G.f.: 3*x/(1 - 3*x^2 + x^3). a(n+1) = 3*A052931(n)*(-1)^n, which from recurrence relations for a(n) and A052931 can easily be proved inductively. - Roman Witula, Oct 06 2012 a(n) = -A214779(n+1)-A214779(n). - Roman Witula, Oct 06 2012 EXAMPLE We have a(2)=a(1)+a(4)=a(4)+a(7)+a(8)=-a(3)+a(5)+a(6)=0, which implies (c(1)/c(2))^(1/3)*c(1)^2 + (c(2)/c(4))^(1/3)*c(2)^2 + (c(4)/c(1))^(1/3)*c(4)^2 = (c(1)/c(2))^(1/3)*(c(1) + c(1)^4) + (c(2)/c(4))^(1/3)*(c(2) + c(2)^4) + (c(4)/c(1))^(1/3)*(c(4) + c(4)^4) = (c(1)/c(2))^(1/3)*(c(1)^4 + c(1)^7 + c(1)^8) + (c(2)/c(4))^(1/3)*(c(2)^4 + c(2)^7 + c(2)^8) + (c(4)/c(1))^(1/3)*(c(4)^4 + c(4)^7 + c(4)^8) = 0. Moreover we have 3000*3^(1/3) = (c(1)/c(2))^(1/3)*c(1)^13 + (c(2)/c(4))^(1/3)*c(2)^13 + (c(4)/c(1))^(1/3)*c(4)^13. - Roman Witula, Oct 06 2012 MATHEMATICA LinearRecurrence[{0, 3, -1}, {0, 3, 0}, 30] CROSSREFS Cf. A006053, A214683. Cf. A214779, A214778, A214951, A214954, A217053, A217052, A217069. Sequence in context: A159760 A021101 A154202 * A317825 A002346 A021327 Adjacent sequences:  A214696 A214697 A214698 * A214700 A214701 A214702 KEYWORD sign,easy AUTHOR Roman Witula, Jul 26 2012 STATUS approved

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Last modified July 14 03:19 EDT 2020. Contains 335716 sequences. (Running on oeis4.)