A216801 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).

2, -22, -117, -468, -1755, -6513, -24336, -91988, -351689, -1357408, -5277363, -20625774, -80909257, -318173258, -1253243498, -4941450657, -19495914360, -76945654032, -303737001009, -1199041027587, -4733273752831, -18683644465447, -73743457866962

Offset: 1

Author

Roman Witula, Sep 17 2012

Keywords

• sign
• easy

Comments

a(n) is equal to the rational part of the number 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.

Let us observe that all numbers of the form a(n)*13^(-floor((n+3)/6)) are integers.

We note that the sequence X(n) is defined "similarly" to sequence Y(n) in the comments to A216540. The only difference between them is in initial condition: X(2) = -Y(2).

Examples

			We have 4*a(3)=a(4), 4*a(4)=a(5)+a(3). The 3-valuation of a(n) for n=1,...,10 is contained in A167366. Moreover it can be obtained X(7) - 22*X(3) = 4*sqrt(2*(13-3*sqrt(13))), 4*X(5) - X(7) = 2*sqrt(26(13-3*sqrt(13))), and 15*X(5) - X(9) = 20*sqrt(26(13-3*sqrt(13))), which implies (15*X(5) - X(9))/(4*X(5) - X(7)) = 10.
		

References

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

Links

• Paolo Xausa, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (13,-65,156,-182,91,-13).

Crossrefs

Cf. A216540, A161905, A216861.

Programs

Mathematica

LinearRecurrence[{13, -65, 156, -182, 91, -13}, {2, -22, -117, -468, -1755, -6513}, 25] (* Paolo Xausa, Feb 23 2024 *)

Formula

G.f.: -x*(52*x^5-520*x^4+689*x^3-299*x^2+48*x-2) / (13*x^6-91*x^5+182*x^4-156*x^3+65*x^2-13*x+1). - Colin Barker, Jun 01 2013