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A216540
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).
9
0, 0, -1, -8, -45, -221, -1014, -4472, -19227, -81224, -338767, -1399320, -5736705, -23377770, -94804944, -382930847, -1541565610, -6188513994, -24784429501, -99058333803, -395227906723, -1574536914951, -6264614281978, -24896955293210, -98848880984490
OFFSET
1,4
COMMENTS
a(n) is equal to the rational part (with respect of the field Q(sqrt(13))) of the product sqrt(2(13-3*sqrt(13)))*X(2*n-1)/13, 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 5 for the argument 2Pi/13 defined by the relation A161905(n) + a(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.
It follows that s(2) + s(5) + s(6) = s(1)*s(3)*s(4) = sqrt((13 + 3*sqrt(13))/2) and s(2)*s(5)*s(6) = s(1) + s(3) - s(4) = sqrt((13 - 3*sqrt(13))/2).
a(n) is equal to the negated rational part (with respect of the field Q(sqrt(13))) of the product sqrt(2(13+3*sqrt(13)))*Y(2*n-1)/13, 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. Moreover we have A161905(n) - a(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
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
R. Witula and D. Slota, Quasi-Fibonacci numbers of order 13, (abstract) see p. 15.
FORMULA
G.f.: -x^3*(2*x-1)*(3*x-1)/(13*x^6-91*x^5+182*x^4-156*x^3+65*x^2-13*x+1). [Colin Barker, Sep 23 2012]
EXAMPLE
We note that: s(2)^3 + s(5)^3 + s(6)^3 = 2*(s(2) + s(5) + s(6)), s(2)^5 + s(5)^5 + s(6)^5 = 5* sqrt((13 + 3*sqrt(13))/2) - sqrt((13 - 3*sqrt(13))/2).
MATHEMATICA
LinearRecurrence[{13, -65, 156, -182, 91, -13}, {0, 0, -1, -8, -45, -221}, 30]
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Roman Witula, Sep 12 2012
EXTENSIONS
Better name from Joerg Arndt, Sep 17 2012
STATUS
approved