|
| |
|
|
A216597
|
|
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).
|
|
8
|
|
|
|
0, -1, -5, -22, -91, -364, -1430, -5564, -21541, -83200, -321100, -1239446, -4787770, -18514119, -71683040, -277913233, -1078918139, -4194134516, -16324764560, -63616690111, -248187382924, -969250588865, -3788814577730, -14823325196459, -58040165033110, -227415509487686
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,3
|
|
|
COMMENTS
|
a(n) is equal to the rational part of 2*X(2*n)/sqrt(13) (with respect of the field Q(sqrt(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 4 for the argument 2Pi/13 defined by the relation A216508(n) + a(n)*sqrt(13) = 2*X(2*n), where X(n) := s(2)^n + s(5)^n + s(6)^n, where s(j) := 2*sin(2*Pi*j/13).
I observe that all numbers of the form (a(6*n + k + 4) - 4*a(6*n + k + 3))*13^(-n), where k = 1,...,6, n = 0,1,... are integers. For example we have a(10)-4*a(9)=900*13 and a(11)-4*a(10)=266*13^2.
a(n) is equal to the negative rational part of 2*Y(2*n)/sqrt(13) (with respect of the field Q(sqrt(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. It can be proved that Y(n) = s(1)^n + s(3)^n + s(9)^n (we have s(9) = -s(4)), and 2*Y(2*n) = A216508(n) - a(n)*sqrt(13). - Roman Witula, Sep 24 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
|
|
|
|
FORMULA
|
G.f.: -x*(13*x^4 - 26*x^3 + 22*x^2 - 8*x + 1) / (13*x^6 - 91*x^5 + 182*x^4 - 156*x^3 + 65*x^2 - 13*x + 1). - Colin Barker, Jun 01 2013
a(n) = Sum_{k=0..n} (-1)^k*binomial(2*n,n+k)*(k|13), where (k|13) represents the Legendre symbol. - Greg Dresden, Oct 09 2022
|
|
|
EXAMPLE
|
We have s(2)^4 + s(5)^4 + s(6)^4 + sqrt(13) = s(2)^2 + s(5)^2 + s(6)^2 = (13 - sqrt(13))/2.
We note that 2*a(1) - a(2) = 1, 4*a(2) - a(3) = 2, 4*a(3) - a(4) = 3, 4*a(4) = a(5) and 4*a(n) - a(n+1) < 0 for every n = 5,6,...
|
|
|
MATHEMATICA
|
LinearRecurrence[{13, -65, 156, -182, 91, -13}, {0, -1, -5, -22, -91, -364}, 30]
|
|
|
PROG
|
(PARI) concat([0], Vec(-x*(13*x^4 -26*x^3 +22*x^2 -8*x +1) / (13*x^6 -91*x^5 +182*x^4 -156*x^3 +65*x^2 -13*x +1) + O(x^30))) \\ Andrew Howroyd, Feb 25 2018
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
sign,easy
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
