%I #44 Oct 03 2019 04:08:49

%S -1,1,-4,4,-13,16,-43,61,-145,226,-496,823,-1714,2965,-5965,10609,

%T -20860,37792,-73189,134236,-257359,475897,-906313,1685050,-3194836,

%U 5961463,-11269558,21079225,-39770137,74507233,-140389636,263291836,-495676141,930265144

%N a(n) = 3*a(n-2) - a(n-3) with a(0)=-1, a(1)=1, a(2)=-4.

%C Ramanujan-type sequence number 2 for argument 2Pi/9 is connected with the sequence A214699 (see also sequences A006053, A214683) - all have "similar" trigonometric description, for example in the case of a(n) the following formula hold true: 9^(1/3)*a(n) = (c(1)/c(4))^(1/3)*c(1)^n + (c(2)/c(1))^(1/3)*c(2)^n + (c(4)/c(2))^(1/3)*c(4)^n = -( (c(2)/c(1))^(1/3)*c(1)^(n+1) + (c(4)/c(2))^(1/3)*c(2)^(n+1) + (c(1)/c(4))^(1/3)*c(4)^(n+1) ), where c(j) := 2*Cos(2Pi*j/9) - for the proof see Witula et al.'s papers.

%C From a(0),A214699(0),a(2) and c(1)+c(2)+c(4)=0 we deduce

%C x^3 - 9^(1/3)*x - 1 = (x - (c(1)/c(2))^(1/3))*(x - (c(2)/c(4))^(1/3))*(x - (c(4)/c(1))^(1/3)), and

%C x^3 - 7*9^(1/3)*x - 1 = (x - (c(1)/c(2))^(1/3)*c(1)^2)*(x - (c(2)/c(4))^(1/3)*c(2)^2)*(x - (c(4)/c(1))^(1/3)*c(4)^2). We note that applying the Newton-Girard formulas to these polynomials two new sequences of real numbers can be discussed: X(n) := (c(1)/c(2))^(n/3) + (c(2)/c(4))^(n/3) + (c(4)/c(1))^(n/3), and Y(n) := ((c(1)/c(2))^(1/3)*c(1)^2)^n + ((c(2)/c(4))^(1/3)*c(2)^2)^n + ((c(4)/c(1))^(1/3)*c(4)^2)^n, where X(n)=9^(1/3)*X(n-2)+X(n-3), X(0)=3, X(1)=0, X(2)=2*9^(1/3), Y(n)=7*9^(1/3)Y(n-2)+Y(n-3), Y(0)=3, Y(1)=0, Y(2)=14*9^(1/3). It could be obtained the following decompositions: X(n) = ax(n) + 9^(1/3)*bx(n) + 81^(1/3)*cx(n), ax(0)=3, bx(0)=cx(0)=ax(1)=bx(1)=cx(1)=ax(2)=bx(2)=0, cx(2)=2, ax(n)=ax(n-3)+9*cx(n-2), bx(n)=bx(n-3)+ax(n-2), cx(n)=cx(n-3)+bx(n-2), and Y(n) = ay(n) + 9^(1/3)*by(n) + 81^(1/3)*cy(n), ay(0)=3, by(0)=cy(0)=ay(1)=by(1)=cy(1)=ay(2)=cy(2)=0, by(2)=14, ay(n)=ay(n-3)+63*cy(n-2), by(n)=by(n-3)+7*ay(n-2), cy(n)=cy(n-3)+7*by(n-2). All these new sequence of positive integers ax(n),bx(n),...,cy(n) will be presented separately as A214778, A214951, A214954. - _Roman Witula_, Sep 27 2012

%C We note that all sums a(n+1) + a(n) are divisible by 3, which easily from recurrence formula for a(n) follows. Then it can be deduced the formula a(n+1) + a(n) = -A214699(n). - _Roman Witula_, Oct 06 2012

%D 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.

%H Roman Witula, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL13/Witula/witula30.html">Full Description of Ramanujan Cubic Polynomials</a>, Journal of Integer Sequences, Vol. 13 (2010), Article 10.5.7.

%H Roman Witula, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL13/Witula2/witula40r.html">Ramanujan Cubic Polynomials of the Second Kind</a>, Journal of Integer Sequences, Vol. 13 (2010), Article 10.7.5.

%H Roman Witula, <a href="https://doi.org/10.1515/dema-2013-0418">Ramanujan Type Trigonometric Formulae</a>, Demonstratio Math. 45 (2012) 779-796.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (0,3,-1).

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

%e From a(0)=-1 and A214699(0)=0 we obtain (c(1)/c(4))^(2/3) + (c(2)/c(1))^(2/3) + (c(4)/c(2))^(2/3) = 3*3^(1/3), whereas from a(1)=-1 and A214699(1)=3*3^(1/3) we get (c(1)/c(4))^(2/3)*2c(2) + (c(2)/c(1))^(2/3)*2c(4) + (c(4)/c(2))^(2/3)*2c(1) = 3*3^(1/3).

%t LinearRecurrence[{0, 3, -1}, {-1, 1, -4}, 40] (* _T. D. Noe_, Jul 30 2012 *)

%Y Cf. A214699, A214778, A214951, A214954, A217053, A217052, A217069.

%K sign,easy

%O 0,3

%A _Roman Witula_, Jul 28 2012