|
%I #29 Oct 03 2019 04:09:22
%S 2,5,106,550,6531,44999,435973,3384404,30252969,246877464,2135653370,
%T 17793576423,151867661753,1276243154087,10832435479322,91356359187721,
%U 773637352766062,6534137016412674,55281085635664595,467187197014742851,3951025667301212597,33398969150217473532
%N a(n) = 3*a(n-1) + 46*a(n-2) + a(n-3) with a(0)=2, a(1)=5, a(2)=106.
%C The Ramanujan-type sequence number 8 for the argument 2Pi/7 (see also A214683, A215112, A006053, A006054, A215076, A215100, A120757, A215560, A215569 for the numbers: 1, 1a, 2, 2a, 3-7 respectively). The sequence a(n) is one of the three special sequences (the remaining two are A215560 and A215569) connected with the following recurrence relation:
%C (c(1)^4/c(2))^(n/3) + (c(2)^4/c(4))^(n/3) + (c(4)^4/c(1))^(n/3) = at(n) + bt(n)*7^(1/3) + ct(n)*49^(1/3), where c(j):=2*cos(2*Pi*j/7), and the sequences at(n), bt(n), and ct(n) are defined in comments to A215560 (see also A215569). It follows that a(n)=ct(3*n+2), at(3*n+2)=bt(3*n+2)=0, which implies the first formula below.
%C We note that if a(n), a(n+1) and a(n+2) are all odd for some n in N then a(n+3) is even, a(n+4) is odd, a(n+5) and a(n+6) are both even, and the numbers a(n+7), a(n+8), a(n+9) are all odd again. In consequence, this situation hold for every n of the form 7*k+4, k=0,1,..., in the other words cyclical through all sequence a(n), n=4,5,... (from n=1 whenever we start from odd-even-even sequence).
%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 G. C. Greubel, <a href="/A215572/b215572.txt">Table of n, a(n) for n = 0..1000</a>
%H Roman Witula, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL12/Witula/witula17.html">Ramanujan Type Trigonometric Formulas: The General Form for the Argument 2*Pi/7</a>, Journal of Integer Sequences, Vol. 12 (2009), Article 09.8.5.
%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 (3,46,1).
%F 49^(1/3)*a(n) = (c(1)^4/c(2))^(n+2/3) + (c(2)^4/c(4))^(n+2/3) + (c(4)^4/c(1))^(n+2/3) = (c(1)*(c(1)/c(2))^(1/3))^(3*n+2) + (c(2)*(c(2)/c(4))^(1/3))^(3*n+2) + (c(4)*(c(4)/c(1))^(1/3))^(3*n+2).
%F G.f.: (2-x-x^2)/(1-3*x-46*x^2-x^3).
%e From 4*a(1)+5*a(2)=a(3) we obtain 4*((c(1)^4/c(2))^(5/3) + (c(2)^4/c(4))^(5/3) + (c(4)^4/c(1))^(5/3)) + 5*((c(1)^4/c(2))^(8/3) + (c(2)^4/c(4))^(8/3) + (c(4)^4/c(1))^(8/3)) = (4 + 5*c(1)^4/c(2))*((c(1)^4/c(2))^(5/3) + (4 + 5*c(2)^4/c(4))*((c(2)^4/c(4))^(5/3) + (4 + 5*c(4)^4/c(1))*((c(4)^4/c(1))^(5/3) = (c(1)^4/c(2))^(11/3) + (c(2)^4/c(4))^(11/3) + (c(4)^4/c(1))^(11/3) = 550*49^(1/3).
%t LinearRecurrence[{3,46,1}, {2,5,106}, 50]
%t CoefficientList[Series[(2 - x - x^2)/(1 - 3*x - 46*x^2 - x^3), {x,0,50}], x] (* _G. C. Greubel_, Apr 16 2017 *)
%o (PARI) Vec((2-x-x^2)/(1-3*x-46*x^2-x^3) + O(x^40)) \\ _Michel Marcus_, Apr 20 2016
%Y Cf. A214683, A215112, A006053, A006054, A215076, A215100, A120757, A215560, A215569.
%K nonn,easy
%O 0,1
%A _Roman Witula_, Aug 16 2012
%E More terms from _Michel Marcus_, Apr 20 2016
|
