A161905 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.

2, 4, 13, 52, 221, 949, 4056, 17186, 72163, 300482, 1241981, 5100758, 20833813, 84695026, 342920942, 1383646433, 5566235714, 22334785486, 89420529809, 357319721889, 1425447435997, 5678246483273, 22590565547134, 89775857333032, 356428030609222, 1413891596961194, 5604509198580578

Offset: 1

Author

Roman Witula, Sep 12 2012

Keywords

• sign
• easy

Comments

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.

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.

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.

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

Examples

			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
(4*X(7) - X(9))/(4*X(5) - X(7)) = 13*(sqrt(13) - 1) and
(4*X(11) - X(13))/(4*X(7) - X(9)) = 10*(sqrt(13) - 2)/(sqrt(13) - 1) = 5*(11 - sqrt(13))/6.
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.
		

References

• 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.
• R. Witula, On some applications of formulas for sums of the unimodular complex numbers, Wyd. Pracowni Komputerowej Jacka Skalmierskiego, Gliwice 2011 (in Polish).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• R. Witula and D. Slota, Quasi-Fibonacci numbers of order 13, (abstract) see p. 15.
• Index entries for linear recurrences with constant coefficients, signature (13,-65,156,-182,91,-13).

Crossrefs

Cf. A216605, A216486, A216597, A216508, A216540.

Programs

Mathematica

LinearRecurrence[{13,-65,156,-182,91,-13}, {2,4,13,52,221,949}, 30]
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 *)

Formula

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

Extensions

Better name from Joerg Arndt, Sep 17 2012