A272931 a(n) = 2^(n+1)*cos(n*arctan(sqrt(15))).

2, 1, -7, -11, 17, 61, -7, -251, -223, 781, 1673, -1451, -8143, -2339, 30233, 39589, -81343, -239699, 85673, 1044469, 701777, -3476099, -6283207, 7621189, 32754017, 2269261, -128746807, -137823851, 377163377, 928458781, -580194727, -4294029851, -1973250943

Offset

0,1

Comments

For n >= 1, |a(n)| is the unique odd positive solution y to 4^(n+1) = 15*x^2 + y^2. The value of x is |A106853(n-1)|. - Jianing Song, Jan 22 2019

Links

Colin Barker, Table of n, a(n) for n = 0..1000

Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.

Index entries for linear recurrences with constant coefficients, signature (1,-4).

Formula

Let a(x) = x/2 - i*sqrt(15)*x/2 and b(x) = x/2 + i*sqrt(15)*x/2, then:

a(n) = a(1)^n + b(1)^n.

a(n) = n! [x^n] exp(a(x)) + exp(b(x)).

a(n) = [x^n] (2 - x)/(4*x^2 - x + 1).

a(n) = Sum_{k=0..floor(n/2)} (-4)^k*n*(n - k - 1)!/(k!*(n - 2*k)!) for n >= 1.

For n >= 1, 15*a(n)^2 + A106853(n-1)^2 = 4^(n+1). - Jianing Song, Jan 22 2019

a(n) = a(n-1) - 4*a(n-2) for n>1. - Colin Barker, Jan 22 2019

a(n) = 2*A106853(n) - A106853(n-1). - R. J. Mathar, Aug 19 2022

Maple

seq(simplify(((1-I*sqrt(15))^n + (1+I*sqrt(15))^n)/2^n), n=0..32);

Mathematica

LinearRecurrence[{1, -4}, {2, 1}, 33]

Prog

(Sage)
[lucas_number2(i, 1, 4) for i in range(33)]
(PARI) Vec((2 - x) / (1 - x + 4*x^2) + O(x^40)) \\ Colin Barker, Jan 22 2019

Crossrefs

Cf. A087204, A002249, A106853.

Sequence in context: A075118 A100245 A275320 * A095137 A239104 A260259

Adjacent sequences: A272928 A272929 A272930 * A272932 A272933 A272934

Keyword

sign,easy

Author

Peter Luschny, May 11 2016

Status

approved