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%I #80 Feb 15 2024 06:31:00
%S 2,4,13,52,221,949,4056,17186,72163,300482,1241981,5100758,20833813,
%T 84695026,342920942,1383646433,5566235714,22334785486,89420529809,
%U 357319721889,1425447435997,5678246483273,22590565547134,89775857333032,356428030609222,1413891596961194,5604509198580578
%N a(n) = 13*a(n-1) - 65*a(n-2) + 156*a(n-3) - 182*a(n-4) + 91*a(n-5) - 13*a(n-6), with a(1)..a(6) as shown.
%C a(n) is equal to the rational part (with respect to the field Q(sqrt(13))) of the product sqrt(2*(13-3*sqrt(13))/13)*X(2*n-1), where X(n) = sqrt((13 + 3*sqrt(13))/2)*X(n-1) - sqrt(13)*X(n-2) + sqrt((13 - 3*sqrt(13))/2)*X(n-3), with X(0)=3, X(1)=sqrt((13 + 3*sqrt(13))/2), and X(2)=(13 - sqrt(13))/2.
%C The Berndt-type sequence number 6 for the argument 2*Pi/13 defined by the relation a(n) + A216540(n)*sqrt(13) = sqrt(2*(13-3*sqrt(13))/13)*X(2*n-1), where X(n) := s(2)^n + s(5)^n + s(6)^n, and s(j) := 2*sin(2*Pi*j/13), j=1,2,...,6.
%C We note that all numbers a(n+1)-4*a(n) for n=3,4,..., are divisible by 13. For example we have a(4)=4*a(3), a(5)-4*a(4)=13, a(6)-4*a(5)=5*13, a(7)-4*a(6)=20*13, and a(10)-4*a(9)=70*13^2.
%C a(n) is also equal to the rational part (with respect to the field Q(sqrt(13))) of the product sqrt(2*(13+3*sqrt(13))/13)*Y(2*n-1), where Y(n) = sqrt((13 - 3*sqrt(13))/2)*Y(n-1) + sqrt(13)*Y(n-2) - sqrt((13 + 3*sqrt(13))/2)*Y(n-3), with Y(0)=3, Y(1)=sqrt((13 - 3*sqrt(13))/2), and Y(2)=(13 + sqrt(13))/2. Let us observe that a(n) - A216540(n)*sqrt(13) = sqrt(2*(13+3*sqrt(13))/13)*Y(2*n-1) and Y(n) = s(1)^n + s(3)^n + s(9)^n (we have s(9) = -s(4)). - _Roman Witula_, Sep 22 2012
%D R. Witula and D. Slota, Quasi-Fibonacci numbers of order 13, Thirteenth International Conference on Fibonacci Numbers and their Applications, Congressus Numerantium, 201 (2010), 89-107.
%D R. Witula, On some applications of formulas for sums of the unimodular complex numbers, Wyd. Pracowni Komputerowej Jacka Skalmierskiego, Gliwice 2011 (in Polish).
%H Vincenzo Librandi, <a href="/A161905/b161905.txt">Table of n, a(n) for n = 1..1000</a>
%H R. Witula and D. Slota, <a href="https://www.mathstat.dal.ca/fibonacci/abstracts.pdf">Quasi-Fibonacci numbers of order 13</a>, (abstract) see p. 15.
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (13,-65,156,-182,91,-13).
%F G.f.: -x*(-2 + 22*x - 91*x^2 + 169*x^3 - 130*x^4 + 26*x^5) / (1 - 13*x + 65*x^2 - 156*x^3 + 182*x^4 - 91*x^5 + 13*x^6). - _R. J. Mathar_, Sep 18 2012
%e It can be shown that 4*X(5) - X(7) = sqrt(26*(13+3*sqrt(13))), 4*X(7) - X(9) = 13*(sqrt(13) - 1)*sqrt(2*(13 + 3*sqrt(13)))/4, and 4*X(11) - X(13) = 130*(sqrt(13) - 2)*sqrt(2*(13 + 3*sqrt(13)))/4, which implies
%e (4*X(7) - X(9))/(4*X(5) - X(7)) = 13*(sqrt(13) - 1) and
%e (4*X(11) - X(13))/(4*X(7) - X(9)) = 10*(sqrt(13) - 2)/(sqrt(13) - 1) = 5*(11 - sqrt(13))/6.
%e We also have a(6) - a(3) - a(1) = 4000, a(9) - 2*a(4) - a(3) + 3*a(1) = 300000, and a(11) - a(5) + a(4) - 2*a(2) - a(1) = 5100000.
%t LinearRecurrence[{13,-65,156,-182,91,-13}, {2,4,13,52,221,949}, 30]
%t CoefficientList[Series[(2-22 x+91 x^2-169 x^3+130 x^4-26 x^5)/(1-13 x+ 65 x^2- 156 x^3+182 x^4-91 x^5+13 x^6),{x,0,40}],x] (* _Harvey P. Dale_, Jun 05 2021 *)
%Y Cf. A216605, A216486, A216597, A216508, A216540.
%K sign,easy
%O 1,1
%A _Roman Witula_, Sep 12 2012
%E Better name from _Joerg Arndt_, Sep 17 2012
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