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.
We note that a(n) = -A050185(n) for n=0,1,...,5 and a(6) + A050185(6) = -2. - Roman Witula, Sep 22 2012
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
Andrew Howroyd, Table of n, a(n) for n = 0..200
G. Dresden and Y. Li, Periodic Weighted Sums of Binomial Coefficients, arXiv:2210.04322 [math.NT], 2022.
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).
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
Roman Witula, Sep 11 2012
EXTENSIONS
Better name from Joerg Arndt, Sep 17 2012
STATUS
approved